Transcript for:
Understanding Functions and Applications

This video is about the exponent rules, rules that govern expressions like two to the fifth, or x to the n. two to the fifth is just shorthand for two times two times two times two times two, written five times. And similarly x to the n is just x multiplied by itself. And times when we write these expressions, the number on the bottom that's being multiplied by itself is called the base. And the number at the top telling us how many times we're multiplying the base by itself is called the exponent. Sometimes the exponent is also called the power. The product rule says that a 5x to the power of n times x to the power of m, that's the same thing as x to the n plus m power. In other words, when I multiply two expressions with the same base, then I can add their exponents. For example, if I have two cubed times two to the fourth, that's equal to two to the seventh. And that makes sense, because two cubed times two to the fourth, means I multiply two by itself three times. And then I multiply that by two multiplied by itself four times. And in the end, I have to multiplied by itself seven times, which is two to the seventh, I'm just adding up the number of times as multiplied in each piece, to get the number of times as multiplied total. The quotient rule says that if I have x to the n power divided by x to the m power, that's equal to x to the n minus m power. In other words, if I divide two expressions with the same base, then I can subtract their exponents. For example, three to the six divided by three squared is going to be three to the six minus two, or three to the fourth. And this makes sense, because three to the six means I multiply three by itself six times, and then I divide that by three multiplied by itself twice. So when I cancel out threes, I have four threes left, notice that I have to subtract the number of threes on the bottom, from the number of threes at the top to get my number of threes remaining. That's why subtract my exponents. The power rule tells us if I have x to the n power raised to the m power, that's the same thing as x to the n times M power. In other words, when I raise a power to a power, I get to multiply the exponents. For example, five to the fourth cubed is equal to five to the four times three, or five to the 12th. And this makes sense, because five to the fourth cubed can be thought of as five to the fourth times five to the fourth, times five to the fourth, expanding this out some more, that's five times five times five times five times the same thing times the same thing again. So I have three groups of four, or five, which is a total of three times four, or 12. fives. The next rule involves what happens when I raise a number, or a variable to the zeroeth power, it turns out that anything to the zeroeth power is equal to one. Usually, this is just taken as a definition. But here's why it makes sense to me. If you have something like two cubed divided by two cubed, Well, certainly that has to equal one, anything divided by itself is just one. But using the quotient rule, we know that this is the same thing as two to the three minus three, because when we divide two things with the same base, we get to subtract their exponents. Therefore, this is the same thing as two to the zero. So two to the zero has to equal one in order to make it work with the quotient rule. And the same argument shows that anything to the zero power has to be equal to one. What happens when we take something to a negative power, x to the n is equal to one over x to the n. Again, most people just take this as a definition of a negative exponent. But here's why it makes sense. If I take something like five to seven times five to the negative seven, then buy the product roll the SAS to equal five to the seven plus negative seven, which is five to the zero, and we just said that that is equal to one. Now I have the equation of five to the seventh times five to the negative seventh equals one, if I divide both sides by five, the seventh, I get that five to the negative seventh has to equal one over five to the seventh. So that's where this rule about negative exponents comes from. That has to be true in order to be consistent with the product rule. Finally, let's look at a fractional exponents. What does an expression like x to the one over N really mean? Well, it means the nth root of x, for example, 64 to the 1/3 power means the cube root of 64, which happens to be four, and nine to the one half means the square root of nine, which is usually written without that little superscript up there. Now, the square root of nine is just three. fractional exponents also makes sense. For example, if I have five to the 1/3, and I cube that, then by the power role, that's equal to five to the 1/3, times three, which is just five to the one or five. So in other words, five to the 1/3, is the number that when you cube it, you get five. And that's exactly what's meant by the cube root of five, the cube root of five is also a number that when you cube it, you get five. The next rule tells us we can distribute an exponent over a product. In other words, if we have a product, x times y, all raised to the nth power, that's equal to x to the n times y to the N. For example, five times seven, all raised to the third power is equal to five cubed times seven cubed. And this makes sense, because five times seven, all raised to the cube power can be expanded as five times seven, times five times seven, times five times seven. But if I rearrange the order of multiplication, this is the same thing as five times five times five, times seven times seven times seven, or five cubed times seven cubed. Similarly, we could distribute an exponent over a quotient, if we have the quotient X over Y, all raised to the n power, that's the same as x to the n over y to the N. For example, two sevenths raised to the fifth power is the same thing as two to the fifth over seven to the fifth. This makes sense, because two sevens to the fifth can be expanded as two sevens multiply by itself five times, which can be written rewritten as two multiplied by itself five times divided by seven multiplied by itself five times, and that's two to the fifth, over seven to the fifth, as wanted. We've seen that we can distribute an exponent over multiplication, and division. But be careful, because we cannot distribute next bowknot. over addition, or subtraction, for example, a plus b to the n is not generally equal to a to the n plus b to the n, a minus b to the n is not generally equal to a to the n minus b to the n. And if you're not sure, just try an example with numbers. For example, two plus three squared is not the same thing as two squared plus three squared, and two minus three squared is definitely not equal to two squared minus three squared. In this video, I gave eight exponent rules, which I'll list again here. There's the product rule, the quotient rule, the power rule, the zero exponent, the negative exponent, the fractional exponent, and the two rules involving distributing exponents. Across multiplication, and division. In another video, I'll use these exponent rules to rewrite and simplify expressions involving exponents. In this video, I'll work out some examples of simplifying expressions using exponent rules. I'll start by reviewing the exponent rules. The product rule says that when you multiply two expressions with the same base, you add the exponents. The quotient rule says that when you divide two expressions with the same base, you subtract the exponents. The power rule says that when you take a power to a power, you multiply the exponents. The power of zero rule says that anything to the zero power is one, as long as the base is not zero. Since zero to the zero is undefined, it doesn't make sense. negative exponents to evaluate x to the minus n, we take the reciprocal one over x to the n. To evaluate a fractional exponent, like x to the one over n, we take the nth root of x, we can distribute an exponent over a product, a times b to the n is equal to A to the N times b to the n. And we can distribute an exponent over a quotient a over b to the n is a to the n over b to the n. In the rest of this video, we'll use these exponent rules to simplify expressions. For our first example, we want to simplify three times x to the minus two divided by x to the fourth, there's several possible ways to proceed. For example, we could use the negative exponent rule dr x to the minus two as one over x squared, all that gets divided by x to the fourth, still, notice that we only take the reciprocal of the x squared, the three stays where it is. And that's because the exponent of negative two only applies to the x not to the three. Now if we think of three as three over one, we have a product of two fractions and our numerator. And so we evaluate that by taking the product of the numerators times the product of the denominators, which is three over x squared, all divided by x to the fourth, I can think of x to the fourth as x to the fourth over one. So now I have a fraction of a fraction, which I can evaluate by multiplying by the reciprocal, that simplifies to three times one divided by x squared times x to the fourth, which is three over x to the six, using the product rule. Since x squared times x to the fourth is equal to x to the two plus four, or x to the six. An alternate way of solving this problem is to start by using the quotient rule, I can rewrite this as three times x to the minus two over x to the fourth, and by the quotient rule, that's three times x to the minus two minus four, or three times x to the minus six. Now using the negative exponent roll, x to the minus six is one over x to the six. And this product of fractions simplifies to three over x to the six, the same answer I got before. The second problem can be solved in similar ways. Please pause the video and try it before going on. One way to simplify would be to use the negative exponent rule first, and rewrite y to the minus five as one over wide the fifth. thinking of this as a fraction, divided by a fraction, I can multiply by the reciprocal and get four y cubed y to the fifth over one by the product rule. The numerator here is four y to the eight. And so my final answer is just for one of the eight. Alternatively, I could decide to use the quotient rule first. As in the previous problem, I can write this as for y cubed minus negative five by the quotient rule. And so that's for y to the eighth as before. I'd like to show you one more method to solve these two problems, kind of a shortcut method before we To go on, that shortcut relies on the principle that a negative exponent in the numerator corresponds to a positive exponent in the denominator. For example, the x to the negative two in the numerator here, after some manipulations became an X to the positive two in the denominator. Furthermore, a negative exponent in the denominator is equivalent to a positive exponent in the numerator. That's what happened when we had the y to the negative five in the denominator, and translated into a y to the positive five in the numerator. Sometimes people like to talk about this principle, by saying that you can pass a factor across the fraction bar by switching the sine of the exponent that is making a positive exponent negative, or a negative exponent positive. Let's see how this principle gives us a shortcut for solving these two problems. In the first problem, 3x to the minus two over x to the four, we can move the negative exponent in the numerator and make it a positive exponent the denominator, so we get three over x to the four plus two or x to the six. In the second example, for y cubed over y to the minus five, we can change the Y to the minus five in the denominator into a y to the five in the numerator and get our final answer of four y to the three plus five or eight. We'll use this principle again in the next problems. In this example, notice that I have I have y's in the numerator and the denominator, and also Z's in the numerator and the denominator. In order to simplify, I'm going to try to get all my y's either in the numerator or the denominator. And similarly for the Z's. Since I have more y's in the denominator, let me move this y to the three downstairs and make it a y to the negative three. I'm using the principle here that a positive exponent in the numerator corresponds to a negative exponent in the denominator. Now since I have a positive exponent, z in the numerator and a negative exponent denominator, and I want to get rid of negative exponents, I'm going to pass the Z's to the numerator as e to the minus two in the denominator, it comes as e to the plus two in the numerator. Notice that my number seven doesn't move. And when I do any of these manipulations, because it doesn't have an exponent, and the exponent of negative two, for example, only applies to the Z not to the seven. Now that I've got all my Z's in the numerator and all my y's in the denominator, it's easy to clean this up using the product rule. And I have my simplified expression. In this last example, we have a complicated expression raised to a fractional power. I'm going to start by simplifying the expression inside the parentheses. I can bring all my y's downstairs and all my x's upstairs and get rid of negative exponents at the same time. In other words, I can rewrite this as 25x to the fourth, I'll bring the Y to the minus five downstairs and make it y to the fifth on the denominator, bring the x to the minus six upstairs and make it x to the sixth the numerator, and then I still have the Y cubed on the denominator, all that's raised to the three halves power. Using the product rule, I can rewrite the expression on the inside of the parentheses as 25x to the 10th over y to the eighth. Recall that we're allowed to distribute an exponent across a product or across a quotient. When I distribute my three halves power, I get 25 to the three halves times x to the 10th to the three halves divided by y to the eighth to the three halves. Now the power rule tells me when I have a power to a power, I get to multiply the exponents. So I can rewrite this as 25 to the three halves times x to the 10 times three halves of our y to the eight times three halves. In other words, 25 to the three halves times x to the 15th over y to the 12th. Finally, I need to rewrite 25 to the three halves. Since three halves can be thought of as three times one half, or as one half times three, I can write 25 to the three halves as 25 to the three times one half, or as 25 to the one half times three. Well, using the power rule in reverse, I can think of this as 25 cubed to the one half, or as 25 to the one half cubed. Since when I take a power to a power, I multiply the exponents 25 cubed to the one half might be hard to evaluate, since 25 cubed is a huge number, but 25 to the one half is just the square root of 25. So I have the square root of 25 cubed, or five cubed, which is 125. Therefore, my original expression is going to be 120 5x to the 15 over y to the 12th. In this video, we use the exponent rules to simplify complicated expressions. This video goes through a few tricks for simplifying expressions with radicals in them. Recall that this notation means the nth root of x, so this notation here means the cube root of eight, the number that when you cube it, you get eight, that number would be two, when we write the root sign without a little number, that just means the square root so the two is implied. In this case, the square root of 25 is five since five squared is 25. Let's start by reviewing some rules for radical expressions. First, if we have the radical of a product, we can rewrite that as the product of two radicals. For example, the square root of nine times 16 is the same thing as the square root of nine times the square root of 16, you can check that both of these evaluate to 12. Similarly, it's possible to distribute a radical sine across division, the radical of a divided by b is the same thing as the radical of A divided by the radical of B. For example, the cube root of 64 over eight is the same thing as the cube root of 64 over the cube root of eight, and you can check that both of these evaluated to you have to be a little bit careful though, because it's not okay to distribute a radical sign across addition. In general, the nth root of a plus b is not equal to the nth root of A plus the nth root of b. And similarly, it's not okay to distribute a radical across subtraction. If you're ever in doubt, you can always check with simple examples. For example, the square root of one plus one is not the same thing as the square root of one plus the square root of one, the right side evaluates to one plus one or two, and the left side is square root of two and the irrational number. The second expression to show that that fails, I don't think it'll work to use the square root of one minus one it'll actually hold in that case, but I can show it's false by using say the square root of two minus one, which does not equal the square root of two minus the square root of one. You might notice that these Rules for Radicals, the ones that hold and the ones that don't hold, remind you of rules for exponents. And that's no coincidence. Because radicals can be written in terms of exponents. For example, if we look at the first rule, we can rewrite this, the nth root of A times B is the same thing as the one of our nth power. And by exponent rules, I can distribute an exponent across multiplication. And so this radical rule can be restated completely in terms of an exponent rule. Similarly, the second rule can be restated in terms of exponents as a or b to the one over n is equal to A to the one over n divided by b to the one over n. We can use the relationship between radicals and the exponents to rewrite a to the m over n. a to the m over N is the same thing as a to the m with the N through taken. That's also the same thing As the nth root of A, all taken to the nth power to see whether that's true, think about exponent rules. So a to the m over N is the same thing as a to the m, taken to the one over nth power. That's because when we take a power to a power, we multiply exponents, and M times one over n is equal to M over n. But a one over nth power is the same thing as an nth root. And therefore, this expression is the same thing as this expression. And that proves the first equivalence. The second equivalence can we prove similarly, by writing a to the m over n as a to the one over n times M. Again, this is works because when I take the power to the power, I'm multiply exponents, one over n times M is the same thing as M over n. But now, these two expressions are the same, because the one over nth power is the same as the nth root. One mnemonic for remembering these relationships is flower over root. So flour is like power, and root is like root, so that tells us we can write a fractional exponent, the M becomes the power, and the n becomes the root in either of these two orders. Now let's use these rules in some examples. If we want to compute 25 to the negative three halves power, well, first I'll use my exponent rule to rewrite that negative exponent as one over 25 to the three halves power. Next, I'll use the power of a root mnemonic to rewrite this as 25 to the third power square rooted, or, as 25 square rooted to the third power. I wrote the two's there for the square root for emphasis, but most of the time, people will omit this and just write the square root without a little number there. Now, I could use either of these two equivalent expressions to continue, but I'd rather use this one because it's easier to compute without a calculator. The square root of 25 is just five, five cubed is 125. So my answer is one over 125. If I tried to compute the cube of 25, first, I'd get a huge number. In general, it's usually easier to compute the route before the power when you're working without a calculator. Now let's do an example simplifying a more complicated expression with with exponent cynet. I want to take the square root of all this stuff. And since I don't really like negative exponents, I'm first going to rewrite this as the square root of 60x squared y to the sixth over z to the 11th. So I'll change that negative exponent to a positive exponent by moving this, this factor to the denominator. Now, when you're asked to simplify radical expression, that generally means to pull as much as possible, out of the radical side. To pull things out of the square root side, I'm going to factor my numbers and try to rewrite everything in terms of squares as much as possible. Since the square root of a square, those two operations undo each other. So I'll show you what I mean first, our factor 60. So 60 is going to be two squared times three times five. And I'll just copy everything over for now. Now I'll break things up into squares as much as possible. So I've got a two squared, and three times five, I've already got an x squared, I write the wider the six as y squared times y squared times y squared. And I'll write the Z to the 11th as z squared times z squared, I guess five times 12345 times when extra z, that should add up to z to the 11th. I add all those exponents together. Now I know that I can distribute my radical sign across multiplication and division. So I'll write this with a zillion different radicals here. And every time I see the square root of something squared, I can just cancel those square roots and the squares out and get what's what's left here. So So after doing that cancellation, I get two times the square root of three times five times x times y times y times y over z times itself, I guess five times times the square root of z. And now I can clean that up with exponents. I'll write that as the square root of 15 I guess two times a squared of 15 times x times y cubed over z to the fifth the square root of z. I'm gonna leave this example as is. But sometimes people prefer to rewrite radical expressions without radical signs in the denominator. That's called rationalizing the denominator. I won't do it here, but I'll show you how to do it in the next example. This example asks us to rationalize the denominator, that means to rewrite as an equivalent expression without radical signs in the denominator. To get rid of the radical sine and the denominator, I want to multiply my denominator by square root of x. But I can't just multiply the denominator willy nilly by something unless I multiply the numerator by the same thing. So then just multiply my expression by one in a fancy forum and I don't change the value of my expression. Now, if I just multiply together numerators 3x squared of x and multiply denominators squared of x 10 squared of x is the square root of x squared squared of x squared is just x. Now I can cancel my access from the numerator denominator, and my final answer is three times the square root of x. I rationalize my denominator, and the process got a nicer looking expression. In this video, we went over the Rules for Radicals. And we simplified some radical expressions by working with fractional exponents, pulling things out of the radical sign and rationalizing the denominator. This video goes over some common methods of factoring. Recall that factoring an expression means to write it as a product. So we could factor the number 30, by writing it as six times five, we could factor it more completely by writing it as two times three times five. As another example, we could factor the expression x squared plus 5x plus six by writing it as x plus two times x plus three. In this video, I'll go over how I get from here to here, how I know how to do the factoring. But for right now, I just want to review how I can go backwards how I can check that the factoring is correct. And that's just by multiplying out or distributing. If I distribute x plus two times x plus three, then I multiply x by x, that gives me x squared, x times three gives me 3x. Two times x gives me 2x. And two times three gives me six. So that simplifies to x squared plus 5x plus six, which checks out with what I started with. So you can think of factoring as the opposite of distributing out. And you can always check your factoring by distributing or multiplying out a bit of terminology, when I think of an expression as a sum of a bunch of things, then the things I sum up are called the terms. But if I think of the same expression as a product of things, then the things that I multiply together are called factors. Now let's get started on techniques of factoring. When I have to factor something, I always like to start by pulling out the greatest common factor, the greatest common factor means the largest thing that divides each of the terms. In this first example, the largest thing that divides both 15 and 25x is five. So the GCF is five. So I pull the five out, and then I divide each of the terms by that number. And so I get three plus 5x. Pause the video for a moment and see if you can find the greatest common factor of x squared y and y squared x cubed. The biggest thing that divides both x squared y and y squared x cubed is going to be x squared times y. One way to find this is to look for the power of x that smallest in each of these terms. So that's x squared. And the power of y that smallest in each of these terms is just y to the one or y. Now if I factor out the x squared y from each of the terms, that's like dividing each term by x squared y, if I divide the first term by x squared y, I just get one. If I divide the second term by x squared y, I'm going to be left with an X and a Y. I'll write this out on the side just to make it more clear, y squared x cubed over x squared y. That's like two y's on the end. Three x's on the top and two x's and a y on the bottom. So I'm left with just an X and a Y. So I'll write the x, y here, and I factored my expression. As always, I can check my answer by multiplying out. So if I multiply out my factored expression, I get x squared y is the first term and the second term, I get x there. Now let's see three X's multiplied together and two y's multiplied together. And that checks out with what I started with. The next technique of factoring, I'd like to go over his factoring by grouping. In this example, notice that we have four terms, factoring by grouping is a handy method to look at. If you have four terms in your expression, you need to factor in order to factor by grouping, I'm first going to factor out the greatest common factor of the first two terms, and then separately, factor out the greatest common factor of the last two terms. The greatest common factor of x cubed, and 3x squared is x squared. So I factor out the x squared, and I get x plus three. And now the greatest common factor of forex and 12 is just four. So I factor out the four from those two terms. Notice that the factor of x plus three now appears in both pieces. So I can factor out the greatest common factor of x plus three, and I'll factor it out on the left side instead of the right. And now I have an x squared from this first piece, and I have a four from this second piece. And that completes my factoring by grouping. You might wonder if we could factor further by factoring the expression x squared plus four. But in fact, as we'll see later, this expression, which is a sum of two squares, x squared plus two squared does not factor any further over the integers. Next, we'll do some factoring of quadratics. a quadratic is an expression with a squared term, just a term with x in it, and a constant term with no x's in it. I'd like to factor this expression as a product of x plus or minus some number times x plus or minus some other number. The key idea is that if I can find those two numbers, then if I were to distribute out this expression, those two numbers would have to multiply to give me my constant term of eight. And these two numbers would end up having to add to give me my negative six, because when I multiply out, this number will be a coefficient of x, and this number will be also another coefficient of x, they'll add together to the negative six. So if I look at all the pairs of numbers that multiply together to give me eight, so that could be one and eight, two, and four, four and two, but that's really the same thing as I had before. And that's sort of the same thing I had before. I shouldn't forget the negatives, I could have negative one, negative eight, or I could have negative two, negative four, those alternate multiply together to give me eight. Now I just have to find, see if there's a pair of these numbers that add to negative six, and it's not hard to see that these ones will work. So now I can write out my factoring as that would be x minus two times x minus four. And it's always a good idea to check by multiplying out, I'm going to get x squared minus 4x minus 2x, plus eight. And that works out to just what I want. Now this second examples a bit more complicated, because now my leading coefficient, my coefficient of x squared is not just one, it's the number 10. Now, there are lots of different methods for approaching a problem like this. And I'm just going to show you one method, my favorite method that uses factoring by grouping, but but to start out, I'm going to multiply my coefficient of x squared by my constant term, so I'm multiplying 10 by negative six, that gives me negative 60. And I'll also take my coefficient of x the number 11. And write that down here. Now I'm going to look for two numbers that multiply to give me negative 60. And add to give me 11. You might notice that this is exactly what we were doing in the previous problem. It's just here, we didn't have to multiply the coefficient of x squared by eight, because the coefficient of x squared was just one. So to find the two numbers that multiply to negative 60 and add to 11, you might just be able to come up with them in your head thinking about it, but if not, you can figure it out. Pretty simple. thematically by writing out all the factors, pairs of factors that multiply to negative 60. So I can start with negative one and 60, negative two and 30, negative three and 20. And keep going like this until I have found factors that actually add together to give me the number 11. And, and now that I look at it, I've already found them. 15 minus four, gives me 11. So I don't have to continue with my chart of factors. Now, once I found those factors, I write out my expression 10x squared, but instead of writing 11x, I write negative 4x plus 15x. Now I copy down the negative six, notice that negative 4x plus 15x equals 11x. That's how I chose those numbers. And so this expression is evaluates is the same as, as this expression, I haven't changed my expression. But I have turned it into something that I can apply factoring by grouping on Look, I've got four terms here. And so if I factor out my greatest common factor of my first two terms, that's let's see, I think it's 2x. So I factor out the 2x, I get 5x minus two, and then I factor out the greatest common factor of 15x and negative six, that would be three, and I get a 5x minus two, again, this is working beautifully. So I have a 5x minus two in each part. And so I put the 5x minus two on the right, and I put what's left from these terms in here. So that's 2x plus three. And I have factored my expression. There are a couple special kinds of expressions that appear frequently, that it's handy to just memorize the formula for. So the first one is the difference of squares. If you see something of the form a squared minus b squared, then you can factor that as a plus b times a minus b. And let's just check that that works. If I do a plus b times a minus b and multiply that out, I get a squared minus a b plus b A minus B squared, and those middle two terms cancel out. So it gives me back the difference of squares just like I want it. So for this first example, I if I think of x squared minus 16 as x squared minus four squared, then I can see that's a difference of squares. And I can immediately write it as x plus four times x minus four. And the second example, nine p squared minus one, that's the same thing as three p squared minus one squared. So that's three p plus one times three p minus one. Notice that if I have a sum of squares, for example, x squared plus four, which is x squared plus two squared, then that does not factor. The difference of squares formula doesn't apply. And there is no formula that applies for a sum of squares. There is, however, a formula for both a difference of cubes and a sum of cubes. The difference of cubes formula, a cubed minus b cubed is a minus b times a squared plus a b plus b squared. The formula for the sum of cubes is pretty much the same, you just switch the negative and positive sign here in here. So that gives us a plus b times a squared minus a b plus b squared. As usual, you can check these formulas by multiplying out. Let's look at one example of using these formulas. Y cubed plus 27 is actually a sum of two cubes because it's y cubed plus three cubed. So I can factor it using the sum of cubes formula by plugging in y for a and three for B. That gives me y plus three times y squared minus y times three plus three squared. And I can clean that up a little bit to read y plus three times y squared minus three y plus nine. So in this video, we went over several methods of factoring. We did factoring out the greatest common factor. We did factoring by grouping. We did factoring quadratics. And we did a difference of squares. And we did a difference and a psalm of cubes and more complicated problems, you may need to apply several these techniques in order to get through a single problem. For example, you might need to start by pulling out a greatest common factor and then Add to a factoring of quadratics, or something similar. Now go forth and factor. This video gives some additional examples of factoring. Please pause the video and decide which of these first five expressions factor and which one does not. The first expression can be factored by pulling out a common factor of x from each term. So that becomes x times x plus one. The second example can be factors as a difference of two squares, since x squared minus 25 is something squared, minus something else squared. And we know that anytime we have something like a squared minus b squared, that's a plus b times a minus b. So we can factor this as X plus five times x minus five. The third one is a sum of two squares, there's no way to factor a sum of two squares over real numbers. So this is the one that does not factor. just for completeness, let's look at the next to this next one does factor by grouping. When we factor by grouping, we pull up the biggest common factor out of the first two terms, that would be an x squared, that becomes x squared times x plus two. And then we factor as much as we can add the next two terms, that would be a three times x plus two. Notice that the x plus two factor now occurs in both of the resulting terms. So we can pull that x plus two out and get x plus two times x squared plus three, we can't factor any further because x squared plus three doesn't factor. Finally, we have a quadratic, this also factors. And I like to factor these also using a factoring by grouping trick. So first, what I do is I multiply the coefficient of x squared and the constant term, five times eight is 40. I'll write that on the top of my x. Now I take the coefficient of the x term, that's negative 14, and I write that on the bottom part of the x. Now I'm looking for two numbers that multiply to 40 and add to negative 14. Sometimes I can just guess numbers like this, but if not, I start writing out factors of 40. So factors of 40, I could do one times 40. Well, now I just noticed, I'm trying to add to a negative number. So if I use two positive factors, there's no way there's going to add to a negative number. It's better for me to use negative numbers, that factor 40, a negative times a negative still multiplies to 40. But they have a chance of adding to a negative number. So but negative one and negative 40, of course, don't work, they don't add to negative 14, they add to negative 41. So let me try some other factors. The next biggest number that divides 40, besides one is two, so I'll try negative two and negative 20. Those add to negative 22. That doesn't work. Next one that divides 40 would be four, so I'll try negative four and negative 10. Aha, we have a winner. So negative four plus negative 10 is negative 14, negative four times negative 10 is positive 40. We've got it. Alright, so the next step is to use factoring by grouping, we're going to first split up this negative 14x as negative 4x minus 10x. And carry down the eight and the 5x squared. Notice that this works, because I picked negative four and negative 10 to add up to negative 14, so so negative 4x minus 10x, will add up to negative 14x. So I've got the same expression, just just expand it out a little bit. Now I have four terms, I can do factoring by grouping, so I can group the first two terms and factor out the biggest thing I can that will be n x times 5x minus four. And now I'll factor the biggest thing I can out of these two numbers, including the the negative. So that becomes, let's see, I can factor out a negative two and that becomes 5x minus four since negative two times minus four is eight. All right, I've got the same 5x plus four in both my terms, so factoring by grouping is going swimmingly, I can factor out the 5x minus four from both those terms and I get the x minus two, and I factored this quadratic. If I want to, of course, I can always check my work by distributing out by multiplying out. So a check here would be multiplying 5x times x is 5x. squared 5x minus 10 to minus two is minus 10x minus four times x is minus 4x. And minus four times minus two is plus eight. So let's see, this does check out to exactly what it should be. So that was the method of factoring a quadratic. And all of these factors except for the sum of squares. So we saw that factoring by grouping is handy for factoring this expression here. It was also handy for factoring the quadratic indirectly, after splitting up the middle term into two terms. So how can you tell when a an expression is is appropriate to factor by grouping, there's, there's an easy way to tell that it might be a candidate, and that's that it has four terms. So if you see four terms, or in the case of quadratic, you can split it up into four terms, then that's a good candidate for factoring by grouping because you can group the first two terms group the second two terms, factoring by grouping all always work on, on expressions with four terms. But, but that's like the first thing to look for. So let's just review what are the same main techniques of factoring. We saw these on the previous page, we saw there was pull out common factors. There's difference of squares. There's factoring by grouping. There's factoring quadratics. And one more that I didn't mention is factoring sums and differences of cubes. that uses the formulas, aq minus b cubed is a minus b times a squared plus a b plus b squared. And a cubed plus b cubed is a plus b times a squared minus a b plus b squared. One important tip when factoring, I always recommend doing this first, pull out the common factors first. That'll simplify things and making the rest of factoring easier. One more tip is that you might need to do several these factoring techniques in one problem, for example, you might have to first pull out a common factor, then factor a difference of squares. And then you might notice that one of your factors is itself a difference of squares, and you have to apply a difference of squares again, so don't stop when you factor a little bit, keep factoring as far as you can go. Here are some extra examples of factoring quadratics. For you to practice, please pause the video and give these a try. For the first one, let's multiply two times negative 14. That gives us negative 28. And then we'll bring the three down in the bottom of the x. Now we're looking for two numbers that multiply to negative 28 and add to three. Well, to multiply two numbers to get a negative 28, we'll need one of them to be negative and one of them to be positive. So to be one, negative 128, or one, negative 28, those don't work. Let's see negative 214 or two, negative 14, those don't work. Hey, I just noticed the positive number had better be bigger than the negative number, so they add to a positive number. Let's see what comes next. How about negative four times seven, four times negative seven, I think four, negative four times seven will work. So I'll write those here at negative four, seven, copy down the two z squared. And I'll split up the three z into negative four z plus seven z and then minus 14. Now factoring by grouping, pull out a to z, that becomes z minus two, pull out a seven and that becomes z minus two again, looking good. I've got two z plus seven times z minus two as my factored expression. My second expression, I could work at the same way, drawing my axe and factoring by grouping that kind of thing. But it's actually going to be easier if I notice first that I can pull out a common factor from all of my terms, though, that'll make things a lot simpler to deal with. So notice that a five divides each of these terms, and in fact, I'm going to go ahead and pull out the negative five because I don't like having negatives in front of my squared term. So I'm going to pull out a common factor of negative five. Again, it would work if I forgot to do this, but it would be a lot more complicated. So negative five v squared, this becomes minus, this becomes plus nine V, since nine times negative five is negative 45. And this becomes minus 10 cents native 10 times negative five is positive 50. Now I can start my x and my factoring by grouping, or I can use kind of a shortcut method, which you may have seen before. So I can just put these here, and then I know that whatever numbers go here, they're gonna have to multiply to the negative 10. And they're gonna have to add to the nine. So that would be plus 10, and a minus one will do the trick. Those are all my factoring examples for today. I hope you enjoy your snow morning and have a chance to spend some time working in ALEKS. Bye. This video is about working with rational expressions. A rational expression is a fraction usually with variables in it, something like x plus two over x squared minus three is a rational expression. In this video, we'll practice adding, subtracting, multiplying and dividing rational expressions and simplifying them to lowest terms. We'll start with simplifying to lowest terms. Recall that if you have a fraction with just numbers in it, something like 21 over 45, we can reduce it to lowest terms by factoring the numerator and factoring the denominator and then canceling common factors. So in this example, the three is cancel, and our fraction reduces to seven over 15. If we want to reduce a rational expression with the variables and add to lowest terms, we proceed the same way. First, we'll factor the numerator, that's three times x plus two, and then factor the denominator. In this case of factors 2x plus two times x plus two, we could also write that as x plus two squared. Now we cancel the common factors. And we're left with three over x plus two. Definitely a simpler way of writing that rational expression. Next, let's practice multiplying and dividing. Recall that if we multiply two fractions with just numbers in them, we simply multiply the numerators and multiply the denominators. So in this case, we would get four times two over three times five or 8/15. If we want to divide two fractions, like in the second example, then we can rewrite it as multiplying by the reciprocal of the fraction on the denominator. So here, we get four fifths times three halves, and that gives us 12 tenths. But actually, we could reduce that fraction to six fifths, we use the same rules when we compute the product or quotient of two rational expressions with the variables. And then here, we're trying to divide two rational expressions. So instead, we can multiply by the reciprocal. I call this flipping and multiplying. And now we just multiply the numerators. And multiply the denominators. It might be tempting at this point to multiply out to distribute out the numerator and the denominator. But actually, it's better to leave it in this factored form and factored even more completely. That way, we'll be able to reduce the rational expression to cancel the common factors. So let's factor even more the x squared plus x factors as x times x plus one, and x squared minus 16. And that's a difference of two squares, that's x plus four times x minus four, the denominator is already fully factored, so we'll just copy it over. And now we can cancel common factors here and here, and we're left with x times x minus four. This is our final answer. Adding and subtracting fractions is a little more complicated because we first have to find a common denominator. A common denominator is an expression that both denominators divided into, it's usually best of the long run to use the least common denominator, which is the smallest expression that both denominators divided into. In this example, if we just want a common denominator, we could use six times 15, which is 90 because both six and 15 divided evenly into 90. But if we want the least common denominator, the best way to do that is to factor the two denominators. So six is two times 315 is three times five, and then put together only the factors we need for Both six and 50 into divider numbers. So if we just use two times three times five, which is 30, we know that two times three will divide it, and three times five will also divide it. And we won't be able to get a denominator any smaller, because we need the factors two, three, and five, in order to ensure both these numbers divided. Once we have our least common denominator, we can rewrite each of our fractions in terms of that denominator. So seven, six, I need to get a 30 in the denominator, so I'm going to multiply that by five over five, and multiply by the factors that are missing from the current denominator in order to get my least common denominator of 30. For the second fraction, for 15th 15 times two is 30. So I'm going to multiply by two over two, I can rewrite this as 3530 s minus 8/30. And now that I have a common denominator, I can just subtract my two numerators. And I get 27/30. If I factor, I can reduce this to three squared over two times five, which is nine tenths. The process for finding the sum of two rational expressions with variables in them follows the exact same process. First, we have to find the least common denominator, I'll do that by factoring the two denominators. So 2x plus two factors as two times x plus 1x squared minus one, that's a difference of two squares. So that's x plus one times x minus one. Now for the least common denominator, I'm going to take all the factors, I need to get an expression that each of these divides into, so I need the factor two, I need the factor x plus one, and I need the factor x minus one, I don't have to repeat the factor x plus one I just need to have at one time. And so I will get my least common denominator two times x plus one times x minus one, I'm not going to bother multiplying this out, it's actually better to leave it in factored form to help me simplify later. Now I can rewrite each of my two rational expressions by multiplying by whatever's missing from the denominator in terms of the least common denominator. So what I mean is, I can rewrite three over 2x plus two, I'll write the 2x plus two is two times x plus one, I'll write it in factored form. And then I noticed that compared to the least common denominator, I'm missing the factor of x minus one. So I multiply the numerator and the denominator by x minus one, I just need it in the denominator. But I can't get away with just multiplying by the denominator without changing my expression, I have to multiply by it on the numerator and the denominator. So I'm just multiplying by one and a fancy form and not changing the value here. So now I do the same thing for the second rational expression, I'll I'll write the denominator in factored form to make it easier to see what's missing from the denominator. What's missing in this denominator, compared to my least common denominator is just the factor two, so multiply the numerator and the denominator by two. Now I can rewrite everything. So the first rational expression becomes three times x minus one over two times x plus 1x minus one, and the second one becomes five times two over two times x plus 1x minus one, notice that I now have a common denominator. So I can just add together my numerators. So I get three times x minus one plus 10 over two times x plus 1x minus one. I'd like to simplify this. And the best way to do that is to leave the denominator in factored form. But I do have to multiply out the numerator so that I can add things together. So I get 3x minus three plus 10 over two times x plus 1x minus one, or 3x plus seven over two times x plus 1x minus one. Now 3x plus seven doesn't factor. And there's therefore no factors that I can cancel out. So this is already reduced. As much as it can be. This is my final answer. In this video, we saw how to simplify rational expressions to lowest terms by factoring and canceling common factors. We also saw how to multiply rational expressions by multiplying the numerator and multiplying the denominator, how to divide rational expressions by flipping and multiplying and how to add and subtract rational expressions by writing them in terms of the least common denominator This video is about solving quadratic equations. a quadratic equation is an equation that contains the square of the variable, say x squared, but no higher powers of x. The standard form for a quadratic equation is the form a x squared plus b x plus c equals zero, where A, B and C represent real numbers. And a is not zero so that we actually have an x squared term. Let me give you an example. 3x squared plus 7x minus two equals zero is a quadratic equation in standard form, here a is three, B is seven, and C is minus two. The equation 3x squared equals minus 7x plus two is also a quadratic equation, it's just not in standard form. The key steps to solving quadratic equations are usually to write the equation in standard form, and then either factor it or use the quadratic formula, which I'll show you later in this video. Let's start with the example y squared equals 18 minus seven y. Here our variable is y, and we need to rewrite this quadratic equation in standard form, we can do this by subtracting 18 from both sides and adding seven y to both sides. That gives us the equation y squared minus 18 plus seven y equals zero. And I can rearrange a little bit to get y squared plus seven y minus 18 equals zero. Now I've got my equation in standard form. Next, I'm going to try to factor it. So I need to look for two numbers that multiply to negative 18 and add to seven. two numbers that work are nine and negative two, so I can factor my expression on the left as y plus nine times y minus two equals zero. Now, anytime you have two quantities that multiply together to give you zero, either the first quantity has to be zero, or the second quantity has to be zero, or I suppose they both could be zero. In some situations, this is really handy, because that means that I know that either y plus nine equals zero, or y minus two equals zero. So I can as my next step, set my factors equal to zero. So y plus nine equals zero, or y minus two equals zero, which means that y equals negative nine, or y equals two. It's not a bad idea to check that those answers actually work by plugging them into the original equation, negative nine squared, does that equal 18 minus seven times negative nine, and you can work out that it does. And similarly, two squared equals 18 minus seven times two. In the next example, let's find solutions of the equation w squared equals 121. This is a quadratic equation, because it's got a square of my variable W, I can rewrite it in standard form by subtracting 121 from both sides. Notice that A is equal to one b is equal to zero because there's no w term, and c is equal to negative 121 in the standard form, a W squared plus BW plus c equals zero. Next step, I'm going to try to factor this expression. So since 121, is 11 squared, this is a difference of two squares, and it factors as w plus 11, times w minus 11 is equal to zero. If I set the factors equal to zero, I get w plus 11 equals zero or w minus 11 equals zero. So w equals minus 11, or w equals 11. In this example, I could have solved the equation more simply, I could have instead said that if W squared is 121, and then w has to equal plus or minus the square root of 121. In other words, W is plus or minus 11. If you saw the equation this way, it's important to remember the plus or minus since minus 11 squared equals 121, just like 11 squared does. Now let's find the solutions for the equation. x times x plus two equals seven. Some people might be tempted to say that, oh, if two numbers multiply to equal seven, then one of them better equal one and the other equals seven or maybe negative one and negative seven. But that's faulty reasoning in this case, because x and x plus two don't have to be whole numbers. They could be crazy. fractions or even irrational numbers. So instead, let's rewrite this equation in standard form. To do that, I'm first going to multiply out. So x times x is x squared x times two is 2x. That equals seven, and I'll subtract the seven from both sides to get x squared plus 2x minus seven is zero. Now I'm looking to factor it. So I need two numbers that multiply to negative seven and add to two, since the only way to factor negative seven is as negative one times seven or seven times negative one, it's easy to see that there are no whole numbers that will do will work. So there's no way to factor this expression over the integers. Instead, let's use the quadratic equation. So we have our leading coefficient of x squared is one. So A is one, B is two, and C is minus seven. And we're going to plug that into the equation quadratic equation, which goes x equals negative b plus or minus the square root of b squared minus four, I see all over two different people have different ways of remembering this formula, I'd like to remember it by seeing it x equals negative b plus or minus the square root of b squared minus four, I see oh over to a, but you can use any pneumonic you like. Anyway, plugging in here, we have x equals negative two plus or minus the square root of two squared minus four times one times negative seven support and remember the negative seven there to all over two times one. Now two squared is four, and four times one times negative seven is negative 28. So this whole quantity under the square root sign becomes four minus negative 28, or 32. So I can rewrite this as x equals negative two plus or minus the square root of 32, all over two. Since 32, is 16 times two and 16 is a perfect square, I can rewrite this as negative two plus or minus the square root of 16 times the square root of two over two, which is negative two plus or minus four times the square root of two over two. Next, I'm going to split out my fraction as negative two over two plus or minus four square root of two over two, and then simplify those fractions. This becomes negative one plus or minus two square root of two. So my answers are negative one plus two square root of two and negative one minus two square root of two. And if I need a decimal answer for any reason, I could work this out on my calculator. As our final example, let's find all real solutions for the equation, one half y squared equals 1/3 y minus two. I'll start as usual by putting it in standard form. So that gives me one half y squared minus 1/3 y plus two equals zero, I could go ahead and start trying to factor or use the quadratic formula right now. But I find fractional coefficients kind of annoying. So I'd like to get rid of them. By doing what I call clearing the denominator, that means I'm going to multiply the whole entire equation by the least common denominator. In this case, the least common denominator is two times three or six. So I'll multiply the whole equation by six have to make sure I multiplied both sides of the equation, but in this case, six times zero is just zero. And when I distribute the six, I get three y squared minus two y plus 12 equals zero. Now, I could try to factor this, but I think it's easier probably just to plunge in and use the quadratic formula. So I get x equals negative B, that's negative negative two or two, plus or minus the square root of b squared, minus four times a times c, all over to a. Working out the stuff in a square root sign, negative two squared is four. And here we have, let's see 144. So this simplifies to x equals to plus or minus the square root of four minus 144. That's negative 140. All of our sex. Well, if you're concerned about that negative number under the square root sign, you should be we can't take the square root of a negative number and get an N get a real number is our answer. There's no real number whose square is a negative number. And therefore, our conclusion is we have no real solutions to this quadratic equation. In this video, we solve some quadratic equations by first writing them in standard form and then Either factoring or using the quadratic formula. In some examples, factoring doesn't work, it's not possible to factor the equation. But in fact, using the quadratic formula will always work even if it's also possible to solve it by factoring. So you can't really lose by using the quadratic formula. It's just sometimes it'll be faster to factor instead. This video is about solving rational equations. A rational equation, like this one is an equation that has rational expressions in that, in other words, an equation that has some variables in the denominator. There are several different approaches for solving a rational equation, but they all start by finding the least common denominator. In this example, the denominators are x plus three and x, we can think of one as just having a denominator of one. Since the denominators don't have any factors in common, I can find the least common denominator just by multiplying them together. My next step is going to be clearing the denominator. By this, I mean that I multiply both sides of my equation by this least common denominator, x plus three times x, I multiply on the left side of the equation, and I multiply by the same thing on the right side of the equation. Since I'm doing the same thing to both sides of the equation, I don't change the the value of the equation. Multiplying the least common denominator on both sides of the equation is equivalent to multiplying it by all three terms in the equation, I can see this when I multiply out, I'll rewrite the left side the same as before, pretty much. And then I'll distribute the right side to get x plus three times x times one plus x plus three times x times one over x. So I've actually multiplied the least common denominator by all three terms of my equation. Now I can have a blast canceling things. The x plus three cancels with the x plus three on the denominator. The here are nothing cancels out because there's no denominator, and here are the x in the numerator cancels with the x in the denominator. So I can rewrite my expression as x squared equals x plus three times x times one plus x plus three. Now I'm going to simplify. So I'll leave the x squared alone on this side, I'll distribute out x squared plus 3x plus x plus three, hey, look, the x squared is cancel on both sides. And so I get zero equals 4x plus three, so 4x is negative three, and x is negative three fourths. Finally, I'm going to plug in my answer to check. This is a good idea for any kind of equation. But it's especially important for a rational equation because occasionally for rational equations, you'll get what's called extraneous solution solutions that don't actually work in your original equation because they make the denominator zero. Now, in this example, I don't think we're going to get any extraneous equations because negative three fourths is not going to make any of these denominators zero, so it should work out fine when I plug in. If I plug in, I get this, I can simplify the denominator here, negative three fourths plus three, three is 12 fourths, as becomes nine fourths. And this is one I'll flip and multiply to get minus four thirds. So here, I can simplify my complex fraction, it ends up being negative three nights, and one minus four thirds is negative 1/3. So that all seems to check out. And so my final answer is x equals negative three fourths. This next example looks a little trickier. And it is, but the same approach will work. First off, find the least common denominator. So here, my denominators are c minus five, c plus one, and C squared minus four c minus five, I'm going to factor that as C minus five times c plus one. Now, my least common denominator needs to have just enough factors to that each of these denominators divided into it. So I need the factor c minus five, I need the factor c plus one. And now I've already got all the factors I need for this denominator. So here is my least common denominator. Next step is to clear the denominators. So I do this by multiplying both sides of the equation by my least common denominator. In fact, I can just multiply each of the three terms by this least common denominator I went ahead and wrote my third denominator in factored form to make it easier to see what cancels. Now canceling time dies, this dies. And both of those factors die. cancelling out the denominator is the whole point of multiplying by the least common denominator, you're multiplying by something that's big enough to kill every single denominator, so you don't have to deal with denominators anymore. Now I'm going to simplify by multiplying out. So I get, let's see, c plus one times four c, that's four c squared plus four c, now I get minus just c minus five, and then over here, I get three c squared plus three, I can rewrite the minus quantity c minus five is minus c plus five. And now I can subtract the three c squared from both sides to get just a C squared over here, and the four c minus c that becomes a three C. And finally, I can subtract the three from both sides to get c squared plus three c plus two equals zero. got myself a quadratic equation that looks like a nice one that factors. So this factors to C plus one times c plus two equals zero. So either c plus one is zero, or C plus two is zero. So C equals negative one, or C equals negative two. Now let's see, we need to still check our answers. Without even going to the trouble of calculating anything, I can see that C equals negative one is not going to work, because if I plug it in to this denominator here, I get a denominator zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't actually satisfy my original equation. And so I can just cross it right out, C equals negative two. I can go if I go ahead, and that doesn't make any of my denominators zero. So if I haven't made any mistakes, it should satisfy my original equation, but, but I'll just plug it in to be sure. And after some simplifying, I get a true statement. So my final answer is C equals negative two. In this video, we saw the couple of rational equations using the method of finding the least common denominator and then clearing the denominator, we cleared the denominator by multiplying both sides of the equation by the least common denominator or equivalently. multiplying each of the terms by that denominator. There's another equivalent method that some people prefer, it still starts out the same, we find the least common denominator, but then we write all the fractions over that least common denominator. So in this example, we'd still use the least common denominator of x plus three times x. But our next step would be to write each of these rational expressions over that common denominator by multiplying the top and the bottom by the appropriate things. So one, in order to get the common denominator of x plus 3x, I need to multiply the top and the bottom by x plus three times x, one over x, I need to multiply the top and the bottom just by x plus three since that's what's missing from the denominator x. Now, if I simplify a little bit, let's say this is x squared over that common denominator, and here I have just x plus three times x over that denominator, and here I have x plus three over that common denominator. Now add together my fractions on the right side, so they have a common denominator. So this is x plus three times x plus x plus three. And now I have two fractions that have that are equal that have the same denominator, therefore their numerators have to be equal also. So the next step is to set the numerators equal. So I get x squared is x plus three times x plus x plus three. And if you look back at the previous way, we solve this equation, you'll recognize this equation. And so from here on we just continue as before. When choosing between these two methods, I personally tend prefer the clear the denominators method, because it's a little bit less writing, you don't have to get rid of those denominators earlier, you don't have to write them as many times. But some people find this one a little bit easier to remember, a little easier to understand either of these methods is fine. One last caution, don't forget at the end, to check your solutions and eliminate any extraneous solutions. These will be solutions that make the denominators of your original equations go to zero. This video is about solving radical equations, that is equations like this one that have square root signs in them, or cube roots or any other kind of radical. When I see an equation with a square root in it, I really want to get rid of the square root. But it'll be easiest to get rid of the square root. If I first isolate the square root. In other words, I want to get the term with the square root and that on one side of the equation by itself, and everything else on the other side of the equation. If I start with my original equation, x plus the square root of x equals 12. And I subtract x from both sides, then that does isolate the square root term on the left side with everything else on the right. Once I've isolated the term with the square root, I want to get rid of the square root. And I'll do that by squaring both sides of my equation. So I'll take the square root of x equals 12 minus x and square both sides. Now the square root of x squared is just x, taking the square root and then squaring those operations undo each other to work out 12 minus x squared, write it out and distribute 12 times 12 is 144 12 times minus x is minus 12x. I get another minus 12x from here. And finally minus x times minus x is positive x squared. So I can combine my minus 12 x's, that's minus 24x. And now I can subtract x from both sides to get zero equals 144 minus 25x plus x squared. That's a quadratic equation, I'll rewrite it in a little bit more standard form here. So now I've got a familiar quadratic equation with no radical signs left in my equation, I'll just proceed to solve it like I usually do for a quadratic, I'll try to factor it. So I'm going to look for two numbers that multiply to 144 and add to minus 25. I know I'm going to need negative numbers to get to negative 25. And in fact, I'll need two negative numbers. So they still multiply to a positive number. So I'll start listing some factors of 144, I could have negative one and a negative 144, negative two, and negative 72, negative four, negative 36, and so on. Once I've listed out the possible factors, it's not hard to find the two that add to negative 25. So that's negative nine and negative 16. So now I can factor in my quadratic equation as x minus nine times x minus 16 equals zero, that means that x minus nine is zero or x minus 16 is zero. So x equals nine or x equals 16. I'm almost done. But there's one last very important step. And that's to check the Solutions so that we can eliminate any extraneous solutions and extraneous solution as a solution that we get that does not actually satisfy our original equation and extraneous solutions can happen when you're solving equations with radicals in them. So let's first check x equals nine. If we plug in to our original equation, we get nine plus a squared of nine and we want that to equal 12. Well, the square root of nine is three, and nine plus three does indeed equal 12. So that solution checks out. Now, let's try x equals 16. Plugging in, we get 16 plus a squared of 16. And that's supposed to equal 12. Well, that says 16 plus four is supposed to equal 12. But that most definitely is not true. And so x equals 16. Turns out to be extraneous solution, and our only solution is x equals nine This next equation might not look like an equation involving radicals. But in fact, we can think of a fractional exponent as being a radical in disguise. Let's start by doing the same thing we did on the previous problem by isolating This time, we'll isolate the part of the equation that involves the fractional exponent. So I'll start with the original equation, two times P to the four fifths equals 1/8. And I'll divide both sides by two or equivalently, I can multiply both sides by one half, that gives me p to the four fifths equals 1/16. And I've effectively isolated the part of the equation with the fractional exponent as much as possible. Now, in the previous example, the next step was to get rid of the radical. In this example, we're going to get rid of the fractional exponent. And I'm going to actually do this in two stages. First, I'm going to raise both sides to the fifth power. That's because when I take an exponent to an exponent, I'm multiply my exponents. And so that becomes just p to the fourth equals 1/16 to the fifth power. Now, I'm going to get rid of the fourth power by raising both sides to the 1/4 power, or by taking the fourth root, there's something that you need to be careful about though, when taking an even root, or the one over an even number power, you always have to consider plus or minus your answer. It's kind of like when you write x squared equals four, and you take the square root of both sides, x could equal plus or minus the square root of four, right, x could equal plus or minus two, since minus two squared is four, just as well as two squared. So that's why when you take an even root, or a one over an even power of both sides, you always need to include the plus or minus sign, when it's an odd root or one over an odd power, you don't need to do that. If you had something like x cubed equals negative eight, then x equals the cube root of negative eight, which is negative two would be your only solution, you don't need to do the plus or minus because positive two wouldn't work. So that aside, explains why we need this plus or minus power. And now p to the four to the 1/4. When I raise a power to a power, I multiply by exponents, so that's just p to the one, which is equal to plus or minus 1/16 to the fifth power to the 1/4 power. Now I just need to simplify this expression, I don't really want to raise 1/16 to the fifth power, because 16 to the fifth power is like a really huge number. So I think I'm actually going to rewrite this first as P equals plus or minus 1/16. I'll write it back as the 5/4 power again. And as I continue to solve using my exponent rules, I'm going to prefer to write this as 1/16 to the fourth root to the fifth power, because it's going to be easier to take the fourth root, let's see the fourth root of 1/16 is the same thing as the fourth root of one over the fourth root of 16. Raise all that to the fifth power. fourth root of one is just one and the fourth root of 16 is to raise that to the fifth power, that's just going to be one to the fifth over two to the fifth, which is plus or minus 130 seconds. The last step is to check answers. So I have the two answers p equals 130 seconds, and P equals minus 130 seconds. And if I plug those both in 1/32 to the fourth fifth power. That gives me two times one to the fourth power over 32 to the fourth power, which is two times one over 32/5 routed to the fourth power. fifth root of 32 is two Raisa to the fourth power, I get 16. So this is two times 1/16, which is 1/8, just as we wanted in the original equation up here. Similarly, we can check that the P equals negative 1/32 actually does satisfy the equation, I'll leave that step to the viewer. So our two solutions are p equals one over 32, and P equals minus one over 32. I do want to point out an alternate approach to getting rid of the fractional exponent, we could have gotten rid of it all in one fell swoop by raising both sides of our equation to the five fourths power. five fourths is the reciprocal of four fifths. So when I use my exponent rules, and say that when I raise the power to the power, I multiply my exponents, that gives me p to the four fifths times five fourths is plus or minus 1/16 to the five fourths, in other words, P to the One Power, which is just P is plus or minus 1/16, to the five fourths, so that's an alternate and possibly faster way to get the solution. Once again, the plus or minus comes from the fact that when we take the 5/4 power, we're really taking an even root a fourth root, and so we need to consider both positive and negative answers. This video is about solving radical equations, that is equations like this one that have square root signs in them, or cube roots or any other kind of radical. In this video, we solved radical equations by first isolating the radical sign, or the fractional exponent, and then removing the radical sine or the fractional exponent by either squaring both sides or taking the reciprocal power of both sides. This video is about solving equations with absolute values in them. Recall that the absolute value of a positive number is just the number, but the absolute value of a negative number is its opposite. In general, I think of the absolute value of a number as representing its distance from zero on the number line, the number four, and the number of negative four are both at distance for from zero, and so the absolute value of both of them is four. Similarly, if I write the equation, the absolute value of x is three, that means that x has to be three units away from zero on the number line. And so X would have to be either negative three, or three. Let's start with the equation three times the absolute value of x plus two equals four. I'd like to isolate the absolute value part of the equation, I can do this by starting with my original equation, subtracting two from both sides and dividing both sides by three. Now I'll think in terms of distance on a number line, the absolute value of x is two thirds means that x is two thirds away from zero. So x could be here for here at negative two thirds or two thirds. And the answer to my equation is x is negative two thirds, or two thirds, I can check my answers by plugging in three times the added value of negative two thirds plus two, I need to check it that equals four. Well, the absolute value of negative two thirds is just two thirds. So this is three times two thirds plus two, which works out to four. Similarly, if I plug in positive two thirds, it also works out to give me the correct answer. The second example is a little different, because the opposite value sign is around a more complicated expression, not just around the X. I would start by isolating the absolute value part. But it's already isolated. So I'll just go ahead and jump to thinking about distance on the number line. So on my number line, the whole expression 3x plus two is supposed to be at a distance of four from zero. So that means that 3x plus two is here at four or 3x plus two though is it negative four, all right, those as equations 3x plus two equals four, or 3x plus two is minus four. And then I can solve. So this becomes 3x equals two, or x equals two thirds. And over here, I get 3x equals minus six, or x equals minus two. Finally, I'll check my answers. I'll leave it to you to verify that they both work. A common mistake on absolute value equations is to get rid of the absolute value signs like we did here, and then just solve for one answer, instead of solving for both answers. Another mistake sometimes people make is, once they get the first answer, they just assume that the negative of that works also. But that doesn't always work. In the first example, our two answers were both the negatives of each other. But in our second examples, or two answers, were not just the opposites of each other one was two thirds and the other was negative two. In this third example, let's again, isolate the absolute value part of the equation. So starting with our original equation, we can subtract 16 from both sides, and divide both sides by five or equivalently, multiply by 1/5. Now let's think about distance on the number line, we have an absolute value needs to equal negative three. So that means whatever is inside the absolute value sign needs to be at distance negative three from zero, well, you can't be at distance negative three from zero. Another way of thinking about this is you can't have the absolute value of something and end up with a negative number if the value is always positive, or zero. So this equation doesn't actually make sense, and there are no solutions to this equation. In this video, we solved absolute value equations. In many cases, an absolute value equation will have two solutions. But in some cases, it'll have no solutions. And occasionally, it'll have just one solution. This video is about interval notation, and easy and well known way to record inequalities. Before dealing with interval notation, it is important to know how to deal with inequalities. Our first example of an inequality is written here, all numbers between one and three, not including one and three. First, we used to write down our variable x, that it is important to locate the key values for this problem that is one and three. Now here we see that x is between these two numbers, meaning we will have one inequality statement on each side of the variable. Here it says not including one and three, which means that we do not have an or equal to sign beneath each inequality. Here we will put one the lowest key value, and here we will put three the highest key value. Next, we're going to graph this inequality on a number line. Here we write our key values one, and three. Because it is not including one, and three, we have an empty circle around each number. Because it isn't the numbers between we have a line connecting them. The last step of this problem is writing this inequality in interval notation. writing things in interval notation is kind of like writing an ordered pair. Here we will put one and then here we will put three. Next we need to put brackets around these numbers. For this problem, it is not including one in three, which means we will use soft brackets. However, if it were including one and three, we will use hard brackets. It is important to note for interval notation, that the smallest value always goes on the left and the biggest value always goes on the right. You also include a comma between your two key values. Now let's work on problem B. Once again we write our variable x. Again it is between negative four and two but this time it is including a negative four into this requires us to have the or equal to assign below the inequality. Then we put our lowest key value here and their highest here. The number line graph for this problem is slightly different. We still have our key values negative four and two. But instead of an open circle, like we have right here, we instead use a closed circle, representing that it is including negative four and two, you complete this with a line in between. Last we write this in interval notation. In the last problem, I said that for numbers, including the outside values, we would use these hard brackets. Now we are actually using this. So we have or hard bracket on each side because isn't including foreign zoo, then we put our smallest value on the left, and our highest value on the right, with the comma in between. You now know how to correctly write these two types of intervals. Now we are going to practice transforming these from inequality notation to interval notation, and vice versa. This is slightly more difficult than our previous examples of having two soft brackets or two hard brackets, we'll have two different types. Now on the left side will have a hard bracket because it is including the three. However, on the right side, it is a soft bracket because it is it it is not including the one that we put a comma in the middle, a lower key value and our higher key value. For the next problem, we're taking an equation already written an interval notation and putting it back into inequality notation. For the second problem, we can see are key values as being five and negative infinity. Now, this sounds a little bit weird, but let's just set up or an equality, we have our variable x, which is less than because of the soft bracket five, and greater than without the or equal to because it has a soft bracket there to infinity. But because x is always greater than negative infinity, we can take out this part, leaving us with x is less than five. It is important to note that a soft bracket always accompanies infinity. This is because infinity is not a real number, so we cannot include it just go as far up to it as we can. The next problem is a little tricky, because we don't see another key value over here. But let's just start off the equation as a soft bracket on this side due to the absence of an order equal to sign. And negative 15 is the lower key value. On the other side, we put the highest possible number infinity, and then close it off with a soft bracket. We can see the relation of this to the previous problem. And how it goes into this problem when x is greater than a key value instead of less than a key value. Once again, we have our sauce bracket for infinity, which is always true. Part D brings up an important point about which number goes on the left. For all of our other problems, we have had the inequalities pointing left instead of rights. When seeing Part D, you might think oh, four is on the left, so it would go here, and zero is on the right who would go here. But that is not true. When writing inequalities, you must always have the lower value on the left, which for this problem is zero. To fix this, we can simply flip this inequality. So it reads like this, it is still identical just written in an easier form to transform it into interval notation, then we can see that zero has a soft bracket, because it is not including zero, but r comma four or other key value and a hard backup because it is or equal to four. For interval notation, you must always have the smaller number on the left side. This was our video on interval notation, an alternative way to write inequalities. This video is about solving inequalities that have absolute value signs in them. Let's look at the inequality absolute value of x is less than five on the number line. Thinking of absolute value as distance. This means that the distance between x and zero is less than five units. So x has to live somewhere in between negative five and five We can express this as an inequality without absolute value signs by saying negative five is less than x, which is less than five. Or we can use interval notation, soft bracket negative five, five, soft bracket. Both of these formulations are equivalent to the original one, but don't involve the absolute value signs. In the second example, we're looking for the values of x for which the absolute value of x is greater than or equal to five. on the number line, this means that the distance of x from the zero, it's got to be bigger than or equal to five units. A distance bigger than five units means that x has to live somewhere over here, or somewhere over here, where it's farther than five units away from zero. Course x could also have a distance equal to five units. So I'll fill in that dot and shade in the other parts of the number line that satisfy my inequality. Now I can rewrite the inequality without the absolute value symbols by saying that x is less than or equal to negative five, or x is greater than or equal to five. I could also write this in interval notation, soft bracket, negative infinity, negative five, hard bracket. And the second part is hard bracket five infinity soft bracket, I combine these with a u for union. Because I'm trying to describe all these points on the number line together with all these other points. Let's take this analysis a step further with a slightly more complicated problem. Now I want the absolute value of three minus two t to be less than four. And an absolute value less than four means a distance less than four on the number line. But it's not the variable t that lives in here at a distance of less than four from zero, it's the whole expression, three minus two t. So three minus two t, live somewhere in here. And I can rewrite this as an inequality without absolute value signs by saying negative four is less than three minus two t is less than four. Now I have a compound inequality that I can solve the usual way. First, I subtract three from all three sides to get negative seven is less than negative two t is less than one. And now I'll divide all three sides by negative two. Since negative two is a negative number, this reverses the directions of the inequalities. Simplifying, I get seven halves is greater than t is greater than negative one half. So my final answer on the number line looks like all the stuff between negative a half and seven halves. But not including the endpoints, and an interval notation, I can write this soft bracket negative a half, seven, half soft bracket, please pause the video and try the next problem on your own. thinking in terms of distance, this inequality says that the distance between the expression three minus two t and zero is always bigger than four. Let me draw this on the number line. If three minus two t has a distance bigger than four from zero, then it can't be in this region that's near zero, it has to be on the outside in one of these two regions. That is three minus two t is either less than negative four, or three minus two t is bigger than four. I solve these two inequalities separately, the first one, subtracting three from both sides, I get negative two t is less than negative seven divided by negative two, I get T is bigger than seven halves. And then on the other side, I get negative two t is greater than one. So t is less than negative one half, I had to flip inequalities in the last step due to dividing by a negative number. So let's check this out on the number line again, the first piece says that t is greater than seven halves. I'll draw that over here. And the second piece says that t is less than negative one half. I'll draw that down here. Because these two statements are joined with an or I'm looking for the t values that are in this one, or in this one. That is I want both of these regions put together. So an interval notation, this reads negative infinity to negative one half soft bracket union soft bracket Seven halves to infinity. This last example looks more complicated. But if I simplify first and isolate the absolute value part, it looks pretty much like the previous ones. So I'll start by subtracting seven from both sides. And then I'll divide both sides by two. Now I'll draw my number line. And I'm looking for this expression for x plus five, to always have distance greater than or equal to negative three from zero, wait a second, distance greater than equal to negative three, well, distance is always greater than or equal to negative three is always greater than equal to zero. So this, in fact, is always true. And so the answer to my inequality is all numbers between negative infinity and infinity. In other words, all real numbers. Once solving absolute value inequalities, it's good to think about distance. And absolute value of something that's less than a number means that whatever's inside the absolute value signs is close to zero. On the other hand, an absolute value is something and being greater than a number means that whatever's inside the absolute value sign is far away from zero, because its distance from zero is bigger than that certain number. drawing these pictures on the number line is a helpful way to rewrite the absolute value and equality as an inequality that doesn't contain an absolute value sign. In this case, it would be negative three is less than x plus two is less than three. And in the other case, it would be either x plus two is less than negative three, or x plus two is greater than three. This video is about solving linear inequalities. Those are inequalities like this one that involve a variable here x, but don't involve any x squared or other higher power terms. The good news is, we can solve linear inequalities, just like we solve linear equations by distributing, adding and subtracting terms to both sides and multiplying and dividing by numbers on both sides. The only thing that's different is that if you multiply or divide by a negative number, then you need to reverse the direction of the inequality. For example, if we had the inequality, negative x is less than negative five, and we wanted to multiply both sides by negative one to get rid of the negative signs, we'd have to also switch or reverse the inequality. With this caution in mind, let's look at our first example. Since our variable x is trapped in parentheses, I'll distribute the negative five to free it from the parentheses. That gives me negative 5x minus 10 plus three is greater than eight. Negative 10 plus three is negative seven, so I'll rewrite this as negative 5x minus seven is greater than eight. Now I'll add seven to both sides to get negative 5x is greater than 15. Now I'd like to divide both sides by negative five. Since negative five is a negative number that reverses the inequality. So I get x is less than 15 divided by negative five. In other words, x is less than negative three. If I wanted to graph this on a number line, I just need to put down a negative three, an open circle around it, and shade in to the left, I use an open circle, because x is strictly less than negative three and can't equal negative three. If I wanted to write this in interval notation, I've added a soft bracket negative infinity, negative three soft bracket. Again, the soft bracket is because the negative three is not included. This next example is an example of a compound inequality. It has two parts. Either this statement is true, or this statement is true. I want to find the values of x that satisfy either one. I'll solve this by working out each part separately, and then combining them at the end. For the inequality on the left, I'll copy it over here. I'm going to add four to both sides. then subtract x from both sides and then divide both sides by two. I didn't have to reverse the inequality sign because I divided by a positive number on the right side, I'll copy the equation over, subtract one from both sides, and divide both sides by 696 is the same as three halves. Now I'm looking for the x values that make this statement. row four, make this statement true. Let me graph this on a number line. x is less than or equal to negative two, means I put a filled in circle there and graph everything to the left. X is greater than three halves means I put a empty circle here and shaded and everything to the right. My final answer includes both of these pieces, which I'll reshade in green, because x is allowed to be an either one or the other. Finally, I can write this in interval notation. The first piece on the number line can be described as soft bracket negative infinity, negative two hard bracket. And the second piece can be described as soft bracket three halves, infinity soft bracket to indicate that x can be in either one of these pieces, I use the union side, which is a U. That means that my answer includes all x values in here together with all x values in here. This next example is also a compound inequality. This time I'm joined by an ad, the and means I'm looking for all y values that satisfy both this and this at the same time. Again, I can solve each piece separately on the left, to isolate the Y, I need to multiply by negative three halves on both sides. So that gives me Why is less than negative 12 times negative three halves, the greater than flip to a less than because I was multiplying by negative three halves, which is a negative number. By clean up the right side, I get why is less than 18. On the right side, I'll start by subtracting two from both sides. And now I'll divide by negative four, again, a negative number, so that flips the inequality. So that's why is less than three over negative four. In other words, y is less than negative three fourths. Again, I'm looking for the y values that make both of these statements true at the same time. Let me graph this on the number line, the Y is less than 18. I can graph that by drawing the number 18. I don't want to include it. So I use an empty circle and I shaded everything to the left the statement y is less than negative three fourths, I need to draw a negative three fourths. And again, I don't include it, but I do include everything on the left. Since I want the y values for which both of these statements are true, I need the y values that are in both colored blue and colored red. And so that would be this part right here. I'll just draw it above so you can see it easily. So that would be all the numbers from negative three fourths and lower, not including negative three, four, since those are the parts of the number line that have both red and blue colors on them. In interval notation, my final answer will be soft bracket negative infinity to negative three fourths soft bracket. As my final example, I have an inequality that has two inequality signs in it negative three is less than or equal to 6x minus two is less than 10. I can think of this as being a compound inequality with two parts, negative 3x is less than or equal to 6x minus two. And at the same time 6x minus two is less than 10. I could solve this into pieces as before. But instead, it's a little more efficient to just solve it all at once, by doing the same thing to all three sides. So as a first step, I'll add two to all three sides. That gives me negative one is less than or equal to 6x is less than 12. And now I'm going to divide all three sides by six to isolate the x. So that gives me negative one, six is less than or equal to x is less than two. If we solved it, instead, in two pieces above, we'd end up with the same thing because we get negative one six is less than or equal to x from this piece, and we'd get x is less than two on this piece. And because of the and statement, that's the same thing as saying negative one, six is less than or equal to x, which is less than two. Either way we do it. Let's see if what it looks like on the number line. So on the number line, we're looking for things that are between two and negative one six, including the negative one, six But not including the to interval notation, we can write this as hard bracket, negative 162 soft bracket. In this video, we solve linear inequalities, including some compound inequalities, joined by the conjunctions and, and or, remember when we're working with and we're looking for places on the number line where both statements are true. That is, we're looking for the overlap on the number line. In this case, the points on the number line that are colored both red and blue at the same time. When we're working with oral statements, we're looking for places where either one or the other statement is true or both on the number line, this corresponds to points that are colored either red or blue, or both. And in this picture, that will actually correspond to the entire number line. In this video, we'll solve inequalities involving polynomials like this one, and inequalities involving rational expressions like this one. Let's start with a simple example. Maybe a deceptively simple example. If you see the inequality, x squared is less than four, you might be very tempted to take the square root of both sides and get something like x is less than two as your answer. But in fact, that doesn't work. To see why it's not correct, consider the x value of negative 10. Negative 10 satisfies the inequality, x is less than two since negative 10 is less than two. But it doesn't satisfy the inequality x squared is less than four, since negative 10 squared is 100, which is not less than four. So these two inequalities are not the same. And it doesn't work to solve a quadratic inequality just to take the square root of both sides, you might be thinking part of why this reasoning is wrong, as we've ignored the negative two option, right? If we had the equation, x squared equals four, then x equals two would just be one option, x equals negative two would be another solution. So somehow, our solution to this inequality should take this into account. In fact, a good way to solve an inequality involving x squares or higher power terms, is to solve the associated equation first. But before we even do that, I like to pull everything over to one side, so that my inequality has zero on the other side. So for our equation, I'll subtract four from both sides to get x squared minus four is less than zero. Now, I'm going to actually solve the associated equation, x squared minus four is equal to zero, I can do this by factoring to x minus two times x plus two is equal to zero. And I'll set my factors equal to zero, and I get x equals two and x equals minus two. Now, I'm going to plot the solutions to my equation on the number line. So I write down negative two and two, those are the places where my expression x squared minus four is equal to zero. Since I want to find where x squared minus four is less than zero, I want to know where this expression x squared minus four is positive or negative, a good way to find that out is to plug in test values. So first, a plug in a test value in this area of the number line, something less than negative to say x equals negative three. If I plug in negative three into x squared minus four, I get negative three squared minus four, which is nine minus four, which is five, that's a positive number. So at negative three, the expression x squared minus four is positive. And in fact, everywhere on this region of the number line, my expression is going to be positive, because it can jump from positive to negative without going through a place where it's zero, I can figure out whether x squared minus four is positive or negative on this region, and on this region of the number line by plugging in test value similar way, evaluate the plug in between negative two and two, a nice value is x equals 00 squared minus four, that's negative four and negative number. So I know that my expression x squared minus four is negative on this whole interval. Finally, I can plug in something like x equals 10, something bigger than two, and I get 10 squared minus four. Without even computing that I can tell that that's going to be a positive number. And that's all that's important. Again, since I want x squared minus four to be less than zero, I'm looking for the places on this number line where I'm getting negatives. So I will share that in on my number line. It's in here, not including the endpoints. Because the endpoints are where my expression x squared minus four is equal to zero, and I want it strictly less than zero, I can write my answer as an inequality, negative two is less than x is less than two, or an interval notation as soft bracket negative two to soft bracket. Our next example, we can solve similarly, first, we'll move everything to one side, so that our inequality is x cubed minus 5x squared minus 6x is greater than or equal to zero. Next, we'll solve the associated equation by factoring. So first, I'll write down the equation. Now I'll factor out an x. And now I'll factor the quadratic. So the solutions to my equation are x equals 0x equals six and x equals negative one, I'll write the solutions to the equation on the number line. So that's negative one, zero, and six. That's where my expression x times x minus six times x plus one is equal to zero. But I want to find where it's greater than or equal to zero. So again, I can use test values, I can plug in, for example, x equals negative two, either to this version of expression, or to this factored version. Since I only care whether my answer is positive or negative, it's sometimes easier to use the factored version. For example, when x is negative two, this factor is negative. But this factor, x minus six is also negative when I plug in negative two for x. Finally, x plus one, when I plug in negative two for x, that's negative one, that's also negative. And a negative times a negative times a negative gives me a negative number. If I plug in something, between negative one and zero, say x equals negative one half, then I'm going to get a negative for this factor, a negative for this factor, but a positive for this third factor. Negative times negative times positive gives me a positive for a test value between zero and six, let's try x equals one. Now I'll get a positive for this factor a negative for this factor, and a positive for this factor. positive times a negative times a positive gives me a negative. Finally, for a test value bigger than six, we could use say x equals 100. That's going to give me positive, positive positive. So my product will be positive. Since I want values where my expression is greater than or equal to zero, I want the places where it equals zero. And the places where it's positive. So my final answer will be close bracket negative one to zero, close bracket union, close bracket six to infinity. As our final example, let's consider the rational inequality, x squared plus 6x plus nine divided by x minus one is less than or equal to zero. Although it might be tempting to clear the denominator and multiply both sides by x minus one, it's dangerous to do that, because x minus one could be a positive number. But it could also be a negative number. And when you multiply both sides by a negative number, you have to reverse the inequality. Although it's possible to solve the inequality this way, by thinking of cases where x minus one is less than zero or bigger than zero, I think it's much easier just to solve the same way as we did before. So we'll start by rewriting so that we move all terms to the left and have zero on the right. Well, that's already true. So the next step would be to solve the associated equation. That is x squared plus 6x plus nine over x minus one is equal to zero. That would be where the numerator is 0x squared plus 6x plus nine is equal to zero. So we're x plus three squared is zero, or x equals negative three. There's one extra step we have to do for rational expressions. And that's we need to find where the expression does not exist. That is, let's find where the denominator is zero. And that said, x equals one. I'll put all those numbers on the number line. The places where am I rational expression is equal to zero, and the place where my rational expression doesn't exist, then I can start in with test values. For example, x equals minus four, zero and to work. If I plug those values into this expression here, I get a negative answer a negative answer and a positive answer. The reason I need to conclude the values on my number line where my denominators zero is because I can my expression can switch from negative to positive by passing through a place where my rational expression doesn't exist, as well as passing by passing flew through a place where my rational expression is equal to zero. Now I'm looking for where my original expression was less than or equal to zero. So that means I want the places on the number line where my expression is equal to zero, and also the places where it's negative. So my final answer is x is less than one, or an interval notation, negative infinity to one. In this video, we solved polynomial and rational inequalities by making a number line and you think test values to make a sine chart. The distance formula can be used to find the distance between two points, if we know their coordinates, if this first point has coordinates, x one, y one, and the second point has coordinates x two, y two, then the distance between them is given by the formula, the square root of x two minus x one squared plus y two minus y one squared. This formula actually comes from the Pythagorean Theorem, let me draw a right triangle with these two points as two of its vertices, then the length of this side is the difference between the 2x coordinates. Similarly, the length of this vertical side is the difference between the two y coordinates. Now that Pythagoras theorem says that for any right triangle, with sides labeled A and B, and hypotony is labeled C, we have that a squared plus b squared equals c squared. Well, if we apply that to this triangle here, this hypotony News is the distance between the two points that we're looking for. So but Tyrion theorem says, the square of this side length, that is x two minus x one squared, plus the square of this side length, which is y two minus y one squared has to equal the square of the hypothesis, that is d squared. taking the square root of both sides of this equation, we get the square root of x two minus x one squared plus y two minus y one squared is equal to d, we don't have to worry about using plus or minus signs when we take the square root because distance is always positive. So we've derived our distance formula. Now let's use it as an example. Let's find the distance between the two points, negative one five, for negative one, five, maybe over here, and for two, this notation just means that P is the point with coordinates negative one five, and q is the point with coordinates for two. We have the distance formula, let's think of P being the point with coordinates x one y one, and Q being the point with coordinates x two y two. But as we'll see, it really doesn't matter which one is which, plugging into our formula, we get d is the square root of n x two minus x one. So that's four minus negative one squared, and then we add y two minus y one, so that's two minus five squared. Working out the arithmetic a little bit for minus negative one, that's four plus one or five squared, plus negative three squared. So that's the square root of 25 plus nine are the square root of 34. Let's see what would have happened if we'd call this first point, x two y two instead, and a second. Why'd x one y one, then we would have gotten the same distance formula, but we would have taken negative one minus four and added the difference of y's squared. So that's why two five minus two squared. That gives us the square root of negative five squared plus three squared, which works out to the same thing. In this video, we use the distance formula to find the distance between two points. When writing down the distance formula, sometimes students forget whether this is a plus or a minus and whether these are pluses or minuses. One way to remember is to think that the distance formula comes from the Pythagorean Theorem. That's why there's a plus in here. And then a minus in here, because we're finding the difference between the coordinates to find the lengths of the sides. The midpoint formula helps us find the coordinates of the midpoint of a line segment, as long as we know the coordinates of the endpoints. So let's call the coordinates of this endpoint x one, y one, and the coordinates of this other endpoint x two, y two, the x coordinate of the midpoint is going to be exactly halfway in between the x coordinates of the endpoints. To get a number halfway in between two other numbers, we just take the average. Similarly, the y coordinate of this midpoint is going to be exactly halfway in between the y coordinates of the endpoints. So the y coordinate of the midpoint is going to be the average of those y coordinates. So we see that the coordinates of the midpoint are x one plus x two over two, y one plus y two over two. Let's use this midpoint formula in an example, we want to find the midpoint of the segment between the points, negative one, five, and four to me draw the line segment between them. So visually, the midpoint is going to be somewhere around here. But to find its exact coordinates, we're going to use x one plus x two over two, and y one plus y two over two, where this first point is coordinates, x one, y one, and the second point has coordinates x two y two, it doesn't actually matter which point you decide is x one y one, which one is x two, y two, the formula will still give you the same answer for the midpoint. So let's see, I take my average of my x coordinates. So that's negative one, plus four over two, and the average of my Y coordinates, so that's five plus two over two, and I get three halves, seven halves, as the coordinates of my midpoint. In this video, we use the midpoint formula to find the midpoint of a line segment, just by taking the average of the x coordinates and the average of the y coordinates. This video is about graphs and equations of circles. Suppose we want to find the equation of a circle of radius five, centered at the point three, two will look something like this. For any point, x, y on the circle, we know that the distance of that point xy from the center is equal to the radius that is five from the distance formula, that distance of five is equal to the square root of the difference of the x coordinates. That's x minus three squared plus the difference of the y coordinates, that's y minus two squared. And if I square both sides of that equation, I get five squared is this square root squared. In other words, 25 is equal to x minus three squared plus y minus two squared. Since the square root and the squared undo each other. A lot of times people will write the X minus three squared plus y minus two squared on the left side of the equation and the 25 on the right side of the equation. This is the standard form for the equation of the circle. The same reasoning can be generalized Find the general equation for a circle with radius R centered at a point HK. For any point with coordinates, x, y on the circle, the distance between the point with coordinates x, y and the point with coordinates HK is equal to the radius r. So we have that the distance is equal to r, but by the distance formula, that's the square root of the difference between the x coordinates x minus h squared plus the difference in the y coordinates y minus k squared. squaring both sides as before, we get r squared is the square root squared, canceling the square root and the squared, and rearranging the equation, that gives us x minus h squared plus y minus k squared equals r squared. That's the general formula for a circle with radius r, and center HK. Notice that the coordinates h and k are subtracted here, the two squares are added because they are in the distance formula, and the radius is squared on the other side, if you remember this general formula, that makes it easy to write down the equation for a circle, for example, if we want the equation for a circle, of radius six, and center at zero, negative three, then we have our equal six, and h k is our zero negative three. So plugging into the formula, we get x minus zero squared, plus y minus negative three squared equals six squared, or simplified this is x squared plus y plus three squared equals 36. Suppose we're given an equation like this one, and we want to decide if it's the equation of a circle, and if so what's the center was the radius. Well, this equation matches the form for a circle, x minus h squared plus y minus k squared equals r squared. If we let h, b five, Why be negative sex since that way, subtracting a negative six is like adding a six, five must be our r squared. So that means that R is the square root of five. So this is our radius. And our center is the point five, negative six. And this is indeed the equation for a circle, which we could then graph by putting down the center and estimating the radius, which is a little bit more than two. This equation might not look like the equation of a circle, but it actually can be transformed. To look like the equation of a circle, we're trying to make it look like something of the form x minus h squared plus y minus k squared equals r squared. First, I'd like to get rid of the coefficients in front of the x squared and the y squared, so I'm going to divide everything by nine. This gives me x squared plus y squared plus 8x minus two y plus four equals zero. Next, I'm going to group the x terms together the x squared and the 8x. So write x squared plus 8x. I'll group the y terms together the Y squared minus two y. And I'll subtract the four over to the other side. This still doesn't look much like the equation of a circle. But as the next step, I'm going to do something called completing the square, I'm going to take this coefficient of x eight divided by two and then square it. In other words, eight over two squared, that's four squared, which is 16. I'm going to add 16 to both sides of my equation. So the 16 here, and on the other side, now I'm going to do the same thing to the coefficient of y, negative two divided by two is negative one, square that and I get one. So now I'm going to add a one to both sides, but I'll put it near the y terms. So add a one there. And then to keep things balanced, I have to add it to the other side. On the right side, I have the number 13. On the left side, I can wrap up this expression x squared plus 8x plus 16 into x plus four squared. To convince you, that's correct. If we multiply out x plus four times x plus four, we get x squared plus 4x plus 4x plus 16, or x squared plus 8x plus 16, the same thing we have right here. Similarly, we can wrap up y squared minus two y plus one as y minus one squared. Again, I'll work out the distribution down here, that's y squared minus y minus y plus one, or y squared minus two y plus one, just like we have up here. If you're wondering how I knew to use four and minus one, the four came from taking half of eight, and the minus one came from taking half the negative two. Now we have an equation for a circle and standard form. And we can read off the center, which is negative four, negative one, and the radius, which is the square root of 13. It might seem like magic that this trick of taking half of this coefficient and squaring it and adding it to both sides lets us wrap up this expression into this expression a perfect square. But to see why that works, let's take a look at what happens if we expand out x minus h squared, the thing that we're trying to get to. So if we expand that out, x minus h squared, which is x minus h times x minus h is, which multiplies out to x squared minus HX minus h x plus h squared, or x squared minus two h x plus h squared. Now if we start out with this part, with some x squared term and some coefficient of x, we're trying to decide what to add on in order to wrap it up into x minus h squared. The thing to add on is h squared here, which comes from half of this coefficient squared, right, because half of negative two H is a negative H, square that you get the H squared. And then when we do wrap it up, it's half of this coefficient of x, that appears right here. This trick of completing the square is really handy for turning an equation for a circle in disguise, into the standard equation for the circle. In this video, we found the standard equation for a circle x minus h squared plus y minus k squared equals r squared, where r is the radius, and h k is the center. We also showed a method of completing the square. When you have an equation for a circle in disguise, completing the square will help you rewrite it into the standard form. This video is about graphs and equations of lines. Here we're given the graph of a line, we want to find the equation, one standard format for the equation of a line is y equals mx plus b. here am represents the slope, and B represents the y intercept, the y value where the line crosses the y axis, the slope is equal to the rise over the run. Or sometimes this is written as the change in y values over the change in x values. Or in other words, y two minus y one over x two minus x one, where x one y one and x two y two are points on the line. While we could use any two points on the line, to find the slope, it's convenient to use points where the x and y coordinates are integers, that is points where the line passes through grid points. So here would be one convenient point to use. And here's another convenient point to use. The coordinates of the first point are one, two, and the next point, this is let's say, five, negative one. Now I can find the slope by looking at the rise over the run. So as I go through a run of this distance, I go through a rise of that distance especially gonna be a negative rise or a fall because my line is pointing down. So let's see counting off squares. This is a run of 1234 squares and a rise of 123. So negative three, so my slope is going to be negative three over four. I got that answer by counting squares, but I could have also gotten it by Looking at the difference in my y values over the difference in my x values, that is, I could have done negative one minus two, that's from my difference in Y values, and divide that by my difference in x values, which is five minus one, that gives me negative three over four, as before. So my M is negative three fourths. Now I need to figure out the value of b, my y intercept, well, I could just read it off the graph, it looks like approximately 2.75. But if I want to be more accurate, I can again use a point that has integer coordinates that I know it's exact coordinates. So either this point or that point, let's try this point. And I can start off with my equation y equals mx plus b, that is y equals negative three fourths x plus b. And I can plug in the point one, two, for my x and y. So that gives me two equals negative three fourths times one plus b, solving for B. Let's see that's two equals negative three fourths plus b. So add three fourths to both sides, that's two plus three fourths equals b. So b is eight fourths plus three fourths, which is 11. Four switches actually, just what I eyeballed it today. So now I can write out my final equation for my line y equals negative three fourths x plus 11 fourths by plugging in for m and b. Next, let's find the equation for this horizontal line. a horizontal line has slope zero. So if we think of it as y equals mx plus b, m is going to be zero. In other words, it's just y equals b, y is some constant. So if we can figure out what that that constant y value is, it looks like it's two, let's see this three, three and a half, we can just write down the equation directly, y equals 3.5. For a vertical line, like this one, it doesn't really have a slope. I mean, if you tried to do the rise over the run, there's no run. So you'd I guess you'd be dividing by zero and get an infinite slope. But But instead, we just think of it as an equation of the form x equals something. And in this case, x equals negative two, notice that all of the points on our line have the same x coordinate of negative two and the y coordinate can be anything. So this is how we write the equation for a vertical line. In this example, we're not shown a graph of the line, we're just get told that it goes through two points. But knowing that I go through two points is enough to find the equation for the line. First, we can find the slope by taking the difference in Y values over the difference in x values. So that's negative three minus two over four minus one, which is negative five thirds. So we can use the standard equation for the line, this is called the slope intercept form. And we can plug in negative five thirds. And we can use one point, either one will do will still get the same final answer. So let's use one two and plug that in to get two equals negative five thirds times one plus b. And so B is two plus five thirds, which is six thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five thirds x plus 11 thirds. This is method one. method two uses a slightly different form of the equation, it's called the point slope form and it goes y minus y naught is equal to m times x minus x naught, where x naught y naught is a point on the line and again is the slope. So we calculate the slope the same way by taking a difference in Y values over a difference in x values. But then, we can simply plug in any point. For example, the point one two will work we can plug one in for x naught and two in for Y not in this point slope form. That gives us y minus Two is equal to minus five thirds x minus one. Notice that these two equations, while they may look different, are actually equivalent, because if I distribute the negative five thirds, and then add the two to both sides, I get the same equation as above. So we've seen two ways of finding the equation for the line using the slope intercept form. And using the point slope form. In this video, we saw that you can find the equation for a line, if you know the slope. And you know one point, you can also find the equation for the line if you know two points, because you can use the two points to get the slope and then plug in one of those points. To figure out the rest of the equation. We saw two standard forms for the equation of a line the slope intercept form y equals mx plus b, where m is the slope, and B is the y intercept. And the point slope form y minus y naught equals m times x minus x naught, where m again is the slope and x naught y naught is a point on the line. This video is about finding parallel and perpendicular lines. Suppose we have a line of slope three fourths, in other words, the rise over the run is three over four, than any other line that's parallel to this line will have the same slope, the same fries over run. So that's our first fact to keep in mind, parallel lines have the same slopes. Suppose on the other hand, we want to find a line perpendicular to our original line with its original slope of three fourths. A perpendicular line, in other words, align at a 90 degree angle to our original line will have a slope, that's the negative reciprocal of the opposite reciprocal of our original slope. So we take the reciprocal of three for us, that's four thirds, and we make it its opposite by changing it from positive to negative. So I'll write this as a principle that perpendicular lines have opposite reciprocal slopes, to get the hang of what it means to be an opposite reciprocal, let's look at a few examples. So here's the original slope, and this will be the opposite reciprocal, which would represent the slope of our perpendicular line. So for example, if one was to the opposite reciprocal, reciprocal of two is one half opposite means I change the sign, if the first slope turned out to be, say, negative 1/3, the reciprocal of 1/3 is three over one or just three. And the opposite means I change it from a negative to a positive, so my perpendicular slope would be would be three. One more example, if I started off with a slope of, say, seven halves, then the reciprocal of that would be two sevenths. And I change it to an opposite reciprocal, by changing the positive to a negative. Let's use these two principles in some examples. In our first example, we need to find the equation of a line that's parallel to this slump, and go through the point negative three two. Well, in order to get started, I need to figure out the slope of this line. So let me put this line into a more standard form the the slope intercept form. So I'll start with three y minus 4x plus six equals zero, I'm going to solve for y to put it in this this more standard form. So I'll say three y equals 4x minus six and then divide by three to get y equals four thirds x minus six thirds is divided all my turns by three, I can simplify that a little bit y equals four thirds x minus two. Now I can read off the slope of my original line, and one is four thirds. That means my slope of my new line, my parallel line will also be four thirds. So my new line will have the equation y equals four thirds ax plus b for some B, of course, B will probably not be negative two like it was for the first line, it'll be something else determined by the fact that it goes through this point negative three to, to figure out what b is, I plug in the point negative three to four, negative three for x, and two for y. So that gives me two equals four thirds times negative three plus b, and I'll solve for b. So let's see. First, let me just simplify. So two equals, this is negative 12 thirds plus b, or in other words, two is negative four plus b. So that means that B is going to be six, and by my parallel line will have the equation y equals four thirds x plus six. Next, let's find the equation of a line that's perpendicular to a given line through and go through a given point. Again, in order to get started, I need to find what the slope of this given line is. So I'll rewrite it, well, first, I'll just copy it over 6x plus three, y equals four. And then I can put it in the standard slope intercept form by solving for y. So three, y is negative 6x plus four, divide everything by three, I get y equals negative six thirds x plus four thirds. So y is negative 2x, plus four thirds. So my original slope of my original line is negative two, which means my perpendicular slope is the opposite reciprocal, so I take the reciprocal of of negative two, that's negative one half and I change the sign so that gives me one half as the slope of my perpendicular line. Now, my new line, I know is going to be y equals one half x plus b for some B, and I can plug in my point on my new line, so one for y and four for x, and solve for b, so I get one equals two plus b, So b is negative one. And so my new line has equation one half x minus one. These next two examples are a little bit different, because now we're looking at a line parallel to a completely horizontal line, let me draw this line y equals three, so the y coordinate is always three, which means that my line will be a horizontal line through that height at height y equals three, if I want something parallel to this line, it will also be a horizontal line. Since it goes through the point negative two, one, C, negative two, one has to go through that point there, it's gonna have always have a y coordinate of one the same as the y coordinate of the point it goes through, so my answer will just be y equals one. In the next example, we want a line that's perpendicular to the line y equals four another horizontal line y equals four, but perpendicular means I'm going to need a vertical line. So I need a vertical line that goes through the point three, four. Okay, and so I'm going to draw a vertical line there. Now vertical lines have the form x equals something for the equation. And to find what x equals, I just need to look at the x coordinate of the point I'm going through this point three four, so that x coordinate is three and all the points on this, this perpendicular and vertical line have X coordinate three so my answer is x equals three. In this video will use the fact that parallel lines have the same slope and perpendicular lines have opposite reciprocal slopes, to find the equations of parallel and perpendicular lines. This video introduces functions and their domains and ranges. A function is a correspondence between input numbers usually the x values and output numbers, usually the y values, that sends each input number to exactly one output number. Sometimes a function is thought of as a rule or machine in which you can feed in x values as input and get out y values as output. So, non mathematical example of a function might be the biological mother function, which takes as input any person and it gives us output their biological mother. This function satisfies the condition that each input number object person in this case, get sent to exactly one output per Because if you take any person, they just have one biological mother. So that rule does give you a function. But if I change things around and just use the the mother function, which sends to each person, their mother, that's no longer going to be a function because there are some people who have more than one mother, right, they could have a biological mother and adopted mother, or a mother and a stepmother, or any number of situations. So since there's, there's at least some people who you would put it as input, and then you'd get like more than one possible output that violates this, this rule of functions, that would not be a function. Now, most of the time, we'll work with functions that are described with equations, not in terms of mothers. So for example, we can have the function y equals x squared plus one. This can also be written as f of x equals x squared plus one. Here, f of x is function notation that stands for the output value of y. Notice that this notation is not representing multiplication, we're not multiplying f by x, instead, we're going to be putting in a value for x as input and getting out a value of f of x or y. For example, if we want to evaluate f of two, we're plugging in two as input for x either in this equation, or in that equation. Since F of two means two squared plus one, f of two is going to equal five. Similarly, f of five means I plug in five for x, so that's gonna be five squared plus one, or 26. Sometimes it's useful to evaluate a function on a more complicated expression involving other variables. In this case, remember, the functions value on any expression, it's just what you what you get when you plug in that whole expression for x. So f of a plus three is going to be the quantity a plus three squared plus one, we could rewrite that as a squared plus six a plus nine plus one, or a squared plus six a plus 10. When evaluating a function on a complex expression, it's important to keep the parentheses when you plug in for x. That way, you evaluate the function on the whole expression. For example, it would be wrong to write f of a plus three equals a plus three squared plus one without the parentheses, because that would imply that we were just squaring the three and not the whole expression, a plus three. Sometimes a function is described with a graph instead of an equation. In this example, this graph is supposed to represent the function g of x. Not all graphs actually represent functions. For example, the graph of a circle doesn't represent a function. That's because the graph of a circle violates the vertical line test, you can draw a vertical line and intersect the graph in more than one point. But our graph, it left satisfies the vertical line test, any vertical line intersects the graph, and at most one point, that means is a function, because every x value will have at most one y value that corresponds to it. Let's evaluate g have to note that two is an x value. And we'll use the graph to find the corresponding y value. So I look for two on the x axis, and find the point on the graph that has that x value that's right here. Now I can look at the y value of that point looks like three, and therefore gf two is equal to three. If I try to do the same thing to evaluate g of five, I run into trouble. Five is an x value, I look for it on the x axis, but there's no point on the graph that has that x value. Therefore, g of five is undefined, or we can say it does not exist. The question of what x values and y values make sense for a function leads us to a discussion of domain and range. The domain of a function is all possible x values that makes sense for that function. The range is the y values that make sense for the function. In this example, we saw that the x value of five didn't have a corresponding y value for this function. So the x value of five is not in the domain of our function J. To find the x values in the domain, we have to look at the x values that correspond to points on the graph. One way to do that is to take the shadow or projection of the graph onto the x axis and see what x values are hit. It looks like we're hitting all x values starting at negative eight, and continuing up to four. So our domain is the x's between negative eight, and four, including those endpoints. Or we can write this in interval notation as negative eight comma four with square brackets. To find the range of the function, we look for the y values corresponding to points on this graph, we can do that. By taking the shadow or projection of the graph onto the y axis, we seem to be hitting our Y values from negative five, up through three. So our range is wise between negative five and three are an interval notation negative five, three with square brackets. If we meet a function that's described as an equation instead of a graph, one way to find the domain and range are to graph the function. But it's often possible to find the domain at least more quickly, by using algebraic considerations. We think about what x values that makes sense to plug into this expression, and what x values need to be excluded, because they make the algebraic expression impossible to evaluate. Specifically, to find the domain of a function, we need to exclude x values that make the denominator zero. Since we can't divide by zero, we also need to exclude x values that make an expression inside a square root sign negative. Since we can't take the square root of a negative number. In fact, we need to exclude values that make the expression inside any even root negative because we can't take an even root of a negative number, even though we can take an odd root like a cube root of a negative number. Later, when we look at logarithmic functions, we'll have some additional exclusions that we have to make. But for now, these two principles should handle all functions We'll see. So let's apply them to a couple examples. For the function in part A, we don't have any square root signs, but we do have a denominator. So we need to exclude x values that make the denominator zero. In other words, we need x squared minus 4x plus three to not be equal to zero. If we solve x squared minus 4x plus three equal to zero, we can do that by factoring. And that gives us x equals three or x equals one, so we need to exclude these values. All other x values should be fine. So if I draw the number line, I can put on one and three and just dig out a hole at both of those. And my domain includes everything else on the number line. In interval notation, this means my domain is everything from negative infinity to one, together with everything from one to three, together with everything from three to infinity. In the second example, we don't have any denominator to worry about, but we do have a square root sign. So we need to exclude any x values that make three minus 2x less than zero. In other words, we can include all x values for which three minus 2x is greater than or equal to zero. Solving that inequality gives us three is greater than equal to 2x. In other words, x is less than or equal to three halves, I can draw this on the number line, or write it in interval notation. Notice that three halves is included, and that's because three minus 2x is allowed to be zero, I can take the square root of zero, that's just zero. And that's not a problem. Finally, let's look at a more complicated function that involves both the square root and denominator. Now there are two things I need to worry about. I need the denominator to not be equal to zero, and I need the stuff inside the square root side to be greater than or equal to zero. from our earlier work, we know the first condition means that x is not equal to three, and x is not equal to one. And the second condition means that x is less than or equal to three halves. Let's try both of those conditions on the number line. x is not equal to three and x is not equal to one means we've got everything except those two dug out points. And the other condition x is less than or equal to three halves means we can have three halves and everything To the left of it. Now to be in our domain and to be legit for our function, we need both of these conditions to be true. So I'm going to connect these conditions with N and N, that means we're looking for a numbers on the number line that are colored both red and blue. So I'll draw that above in purple. So that's everything from three halves to one, I have to dig out one because one was a problem for the denominator. And then I can continue for all the things that are colored both both colors red and blue. So my final domain is going to be, let's see negative infinity, up to but not including one together with one, but not including it to three halves, and I include three half since that was colored, both red and blue. Also, in this video, we talked about functions, how to evaluate them, and how to find domains and ranges. This video gives the graphs of some commonly used functions that I call the toolkit functions. The first function is the function y equals x, let's plot a few points on the graph of this function. If x is zero, y zero, if x is one, y is one, and so on y is always equal to x doesn't have to just be an integer, it can be any real number, and we'll connecting the dots, we get a straight line through the origin. Let's look at the graph of y equals x squared. If x is zero, y is zero. So we go through the origin again, if x is one, y is one, and x is negative one, y is also one, the x value of two gives a y value of four and the x value of negative two gives a y value of four also connecting the dots, we get a parabola. That is this, this function is an even function. That means it has mirror symmetry across the y axis, the left side it looks like exactly like the mirror image of the right side. That happens because when you square a positive number, like two, you get the exact same y value as when you square it's a mirror image x value of negative two. The next function y equals x cubed. I'll call that a cubic. Let's plot a few points when x is zero, y is zero. When x is one, y is one, when x is negative one, y is negative one, two goes with the point eight way up here and an x value of negative two is going to give us negative eight. If I connect the dots, I get something that looks kind of like this. This function is what's called an odd function, because it has 180 degree rotational symmetry occur around the origin. If I rotate this whole graph by 180 degrees, or in other words, turn the paper upside down, I'll get exactly the same shape. The reason this function has this odd symmetry is because when I cube a positive number, and get its y value, I get n cube the corresponding negative number to get its y value, the negative number cubed gives us exactly the negative of the the y value we get with cubing the positive number. Let's look at the next example. Y equals the square root of x. Notice that the domain of this function is just x values bigger than or equal to zero because we can't take the square root of a negative number. Let's plug in a few x values. X is zero gives y is 0x is one square root of one is one, square root of four is two, and connecting the dots, I get a function that looks like this. The absolute value function is next. Again, if we plug in a few points, x is zero goes with y equals 0x is one gives us one, the absolute value of negative one is 122 is on the graph, and the absolute value of negative two is to I'm ending up getting a V shaped graph. It also has even or a mirror symmetry. Y equals two to the x is what's known as an exponential function. That's because the variable x is in the exponent. If I plot a few points, two to the zero is one, two to the one is two, two squared is four, two to the minus one is one half. I'll plot these on my graph, you know, let me fill in a few more points. So let's see. two cubed is eight. That's way up here and negative two gives Maybe 1/4 1/8 connecting the dots, I get something that's shaped like this. You might have heard the expression exponential growth, when talking about, for example, population growth, this is function is represents exponential growth, it grows very, very steeply. Every time we increase the x coordinate by one, we double the y coordinate. We could also look at a function like y equals three dx, or sometimes y equals e to the x, where E is just a number about 2.7. These functions look very similar. It's just the bigger bass makes us rise a little more steeply. Now let's look at the function y equals one over x. It's not defined when x is zero, but I can plug in x equals one half, one over one half is to whenever one is one, and one or two is one half. By connect the dots, I get this shape in the first quadrant, but I haven't looked at negative values of x, when x is negative, whenever negative one is negative one, whenever negative half is negative two, and I get a similar looking piece in the third quadrant. This is an example of a hyperbola. And it's also an odd function, because it has that 180 degree rotational symmetry, if I turn the page upside down, it'll look exactly the same. Finally, let's look at y equals one over x squared. Again, it's not defined when x is zero, but I can plug in points like one half or X, let's see one over one half squared is one over 1/4, which is four. And one over one squared, one over two squared is a fourth, it looks pretty much like the previous function is just a little bit more extreme rises a little more steeply, falls a little more dramatically. But for negative values of x, something a little bit different goes on. For example, one over negative two squared is just one over four, which is a fourth. So I can plot that point there, and one over a negative one squared, that's just one. So my curve for negative values of x is gonna lie in the second quadrant, not the third quadrant. This is an example of an even function because it has perfect mirror symmetry across the y axis. These are the toolkit functions, and I recommend you memorize the shape of each of them. That way, you can draw at least a rough sketch very quickly without having to plug in points. That's all for the graphs of the toolkit functions. If we change the equation of a function, then the graph of the function changes or transforms in predictable ways. This video gives some rules and examples for transformations of functions. To get the most out of this video, it's helpful if you're already familiar with the graph of some typical functions, I call them toolkit functions like y equals square root of x, y equals x squared, y equals the absolute value of x and so on. If you're not familiar with those graphs, I encourage you to watch my video called toolkit functions first, before watching this one. I want to start by reviewing function notation. If g of x represents the function, the square root of x, then we can rewrite these expressions in terms of square roots. For example, g of x minus two is the same thing as the square root of x minus two. g of quantity x minus two means we plug in x minus two, everywhere we see an x, so that would be the same thing as the square root of quantity x minus two. In this second example, I say that we're subtracting two on the inside of the function, because we're subtracting two before we apply the square root function. Whereas on the first example, I say that the minus two is on the outside of the function, we're doing the square root function first and then subtracting two. In this third example, g of 3x, we're multiplying by three on the inside of the function. To evaluate this in terms of square root, we plug in the entire 3x for x and the square root function. That gives us the square root of 3x. In the next example, we're multiplying by three on the outside of the function. This is just three times the square root of x. Finally, g of minus x means the square root of minus x. Now, this might look a little odd because we're not used to taking the square root of a negative number. But remember that if x itself is negative, like negative to the negative x will be negative negative two or positive two. So we're really be taking the square root of a positive number in that case, let me record which of these are inside and which of these are outside of my function. In this next set of examples, we're using the same function g of x squared of x. But this time, we're starting with an expression with square roots in it, and trying to write it in terms of g of x. So the first example, the plus 17 is on the outside of my function, because I'm taking the square root of x first, and then adding 17. So I can write this as g of x plus 17. In the second example, I'm taking x and adding 12 first, then I take the square root of the whole thing. Since I'm adding the 12 to x first that's on the inside of my function. So I write that as g of the quantity x plus 12. Remember that this notation means I plug in the entire x plus 12, into the square root sign, which gives me exactly square root of x plus 12. And this next example, undoing the square root first and then multiplying by negative 36. So i minus 36, multiplications, outside my function, I can rewrite this as minus 36 times g of x. Finally, in this last example, I take x multiplied by a fourth and then apply the square root of x. So that's the same thing as g of 1/4 X, my 1/4 X is on the inside of my function. In other words, it's inside the parentheses when I use function notation. Let's graph the square root of x and two transformations have this function, y equals the square root of x goes to the point 0011. And for two, since the square root of four is two, it looks something like this. In order to graph y equals the square root of x minus two, notice that the minus two is on the outside of the function, that means we're going to take the square root of x first and then subtract two. So for example, if we start with the x value of zero, and compute the square root of zero, that's zero, then we subtract two to give us a y value of negative two, an x value of one, which under the square root function had a y value of one now has a y value that's decreased by two, one minus two is negative one. And finally, an x value of four, which under the square root function had a y value of two now has a y value of two minus two or zero, its y value is also decreased by two. If I plot these new points, zero goes with negative two, one goes with negative one, and four goes with zero, I have my transformed graph. Because I subtracted two on the outside of my function, my y values were decreased by two, which brought my graph down by two units. Next, let's look at y equals the square root of quantity x minus two. Now we're subtracting two on the inside of our function, which means we subtract two from x first and then take the square root. In order to get the same y value of zero as we had in our blue graph, we need our x minus two to be zero, so we need our original x to be two. In order to get the y value of one that we had in our blue graph, we need to be taking the square root of one, so we need x minus two to be one, which means that we need to start with an x value of three. And in order to reproduce our y value of two from our original graph, we need the square root of x minus two to be two, which means we need to start out by taking the square root of four, which means our x minus two is four, so our x should be up there at six. If I plot my x values, with my corresponding y values of square root of x minus two, I get the following graph. Notice that the graph has moved horizontally to the right by two units. To me, moving down by two units, makes sense because we're subtracting two, so we're decreasing y's by two units, but the minus two on the inside that kind of does the opposite of what I expect, I might expected to to move the graph left, I might expect the x values to be going down by two units, but instead, it moves the graph to the right, because the x units have to go up by two units in order to get the right square root when I then subtract two units again, the observations we made for these transformations of functions hold in general, according to the following rules. First of all, numbers on the outside of the function like In our example, y equals a squared of x minus two, those numbers affect the y values. And a result in vertical motions, like we saw, these motions are in the direction you expect. So subtracting two was just down by two. If we were adding two instead, that would move us up by two numbers on the inside of the function. That's like our example, y equals the square root of quantity x minus two, those affect the x values and results in a horizontal motion, these motions go in the opposite direction from what you expect. Remember, the minus two on the inside actually shifted our graph to the right, if it had had a plus two on the inside, that would actually shift our graph to the left. Adding results in a shift those are called translations, but multiplying like something like y equals three times the square root of x, that would result in a stretch or shrink. In other words, if I start with the square root of x, and then when I graph y equals three times the square root of x, that stretches my graph vertically by a factor of three. Like this, if I want to graph y equals 1/3, times the square root of x, that shrinks my graph vertically by a factor of 1/3. Finally, a negative sign results in reflection. For example, if I start with the graph of y equals the square root of x, and then when I graph y equals the square root of negative x, that's going to do a reflection in the horizontal direction because the negative is on the inside of the square root sign. A reflection in the horizontal direction means a reflection across the y axis. If instead, I want to graph y equals negative A squared of x, that negative sign on the outside means a vertical reflection, a reflection across the x axis. Pause the video for a moment and see if you could describe what happens in these four transformations. In the first example, we're subtracting four on the outside of the function. adding or subtracting means a translation or shift. And since we're on the outside of the function affects the y value, so that's moving us vertically. So this transformation should take the square root of graph and move it down by four units, that would look something like this. In the next example, we're adding 12 on the inside, that's still a translation. But now we're moving horizontally. And so since we go the opposite direction, we expect we are going to go to the left by 12 units, that's going to look something like this. And the next example, we're multiplying by three and introducing a negative sign both on the outside of our function outside our function means we're affecting the y values. So in multiplication means we're stretching by a factor of three, the negative sign means we reflect in the vertical direction, here's stretching by a factor of three vertically, before I apply the minus sign, and now the minus sign reflects in the vertical direction. Finally, in this last example, we're multiplying by one quarter on the inside of our function, we know that multiplication means stretch or shrink. And since we're on the inside, it's a horizontal motion, and it does the opposite of what we expect. So instead of shrinking by a factor of 1/4, horizontally, it's actually going to stretch by the reciprocal, a factor of four horizontally. that'll look something like this. Notice that stretching horizontally by a factor of four looks kind of like shrinking vertically by a factor of one half. And that's actually borne out by the algebra, because the square root of 1/4 x is the same thing as the square root of 1/4 times the square root of x, which is the same thing as one half times the square root of x. And so now we can see algebraically that vertical shrink by a factor of one half is the same as a horizontal stretch by a factor of four, at least for this function, the square root function. This video gives some rules for transformations of functions, which I'll repeat below. numbers on the outside correspond to changes in the y values, or vertical motions. numbers on the inside of the function, affect the x values, and result in horizontal motions. Adding and subtracting corresponds to translations or shifts. multiplying and dividing by numbers corresponds to stretches and shrinks and putting in a negative sign. corresponds to a reflection, horizontal reflection, if the negative sign is on the inside, and a vertical reflection, if the negative sign is on the outside, knowing these basic rules about transformations empowers you to be able to sketch graphs of much more complicated functions, like y equals three times the square root of x plus two, by simply considering the transformations, one at a time. a quadratic function is a function that can be written in the form f of x equals A x squared plus bx plus c, where a, b and c are real numbers. And a is not equal to zero. The reason we require that A is not equal to zero is because if a were equal to zero, we'd have f of x is equal to b x plus c, which is called a linear function. So by making sure that a is not zero, we make sure there's really an x squared term which is the hallmark of a quadratic function. Please pause the video for a moment and decide which of these equations represent quadratic functions. The first function can definitely be written in the form g of x equals A x squared plus bx plus c. fact it's already written in that form, where a is negative five, B is 10, and C is three. The second equation is also a quadratic function, because we can think of it as f of x equals one times x squared plus zero times x plus zero. So it is in the right form, where A is one, B is zero, and C is zero. It's perfectly fine for the coefficient of x and the constant term to be zero for a quadratic function, we just need the coefficient of x squared to be nonzero, so the x squared term is preserved. The third equation is not a quadratic function. It's a linear function, because there's no x squared term. The fourth function might not look like a quadratic function. But if we rewrite it by expanding out the X minus three squared, let's see what happens. We get y equals two times x minus three times x minus three plus four. So that's two times x squared minus 3x minus 3x, plus nine plus four, continuing, I get 2x squared, minus 12 access, plus 18 plus four, in other words, y equals 2x squared minus 12x plus 22. So in fact, our function can be written in the right form, and it is a quadratic function. A function that is already written in the form y equals A x squared plus bx plus c is said to be in standard form. So our first example g of x is in standard form a function that's written in the format of the last function that is in the form of y equals a times x minus h squared plus k for some numbers, a, h, and K. That said to be in vertex form, I'll talk more about standard form and vertex form in another video. In this video, we identified some quadratic functions, and talked about the difference between standard form and vertex form. This video is about graphing quadratic functions. a quadratic function, which is typically written in standard form, like this, or sometimes in vertex form, like this, always has the graph that looks like a parabola. This video will show how to tell whether the problem is pointing up or down. How to find its x intercepts, and how to find its vertex. The bare bones basic quadratic function is f of x equals x squared, it goes to the origin, since f of zero is zero squared, which is zero, and it is a parabola pointing upwards like this. The vertex of a parabola is its lowest point if it's pointing upwards, and its highest point if it's pointing downwards. So in this case, the vertex is at 00. The x intercepts are where the graph crosses the x axis. In other words, where y is zero In this function, y equals zero means that x squared is zero, which happens only when x is zero. So the x intercept, there's only one of them is also zero. The second function, y equals negative 3x squared also goes to the origin. Since the functions value when x is zero, is y equals zero. But in this case, the parabola is pointing downwards. That's because thinking about transformations of functions, a negative sign on the outside reflects the function vertically over the x axis, making the problem instead of pointing upwards, reflecting the point downwards. The number three on the outside stretches the graph vertically by a factor of three. So it makes it kind of long and skinny like this. In general, a negative coefficient to the x squared term means the problem will be pointing down. Whereas a positive coefficient, like here, the coefficients, one means the parabola is pointing up. Alright, that roll over here. So if a is bigger than zero, the parabola opens up. And if the value of the coefficient a is less than zero, then the parabola opens down. In this second example, we can see again that the vertex is at 00. And the x intercept is x equals zero. Let's look at this third example. If we multiplied our expression out, we'd see that the coefficient of x squared would be to a positive number. So that means our parabola is going to be opening up. But the vertex of this parabola will no longer be at the origin. In fact, we can find the parabolas vertex by thinking about transformations of functions. Our function is related to the function y equals to x squared, by moving it to the right by three and up by four. Since y equals 2x squared has a vertex at 00. If we move that whole parabola including the vertex, right by three, and up by four, the vertex will end up at the point three, four. So a parabola will look something like this. Notice how easy it was to just read off the vertex when our quadratic function is written in this form. In fact, any parabola any quadratic function written in the form a times x minus h squared plus k has a vertex at h k. By the same reasoning, we're moving the parabola with a vertex at the origin to the right by H, and by K. That's why this form of a quadratic function is called the vertex form. Notice that this parabola has no x intercepts because it does not cross the x axis. For our final function, we have g of x equals 5x squared plus 10x plus three, we know the graph of this function will be a parabola pointing upwards, because the coefficient of x squared is positive. To find the x intercepts, we can set y equals zero, since the x intercepts is where our graph crosses the x axis, and that's where the y value is zero. So zero equals 5x squared plus 10x plus three, I'm going to use the quadratic formula to solve that. So x is negative 10 plus or minus the square root of 10 squared minus four times five times three, all over two times five. That simplifies to x equals negative 10 plus or minus the square root of 40 over 10, which simplifies further to x equals negative 10 over 10, plus or minus the square root of 40 over 10, which is negative one plus or minus two square root of 10 over 10, or negative one plus or minus square root of 10 over five. Since the square root of 10 is just a little bit bigger than three, this works out to approximately about negative two fifths and negative eight foot or so somewhere around right here. So our parabola is going to look something like this. Notice that it goes through crosses the x axis at y equals three, that's because when we plug in x equals zero, and two here we get y equal three, so the y intercept is at three. Finally, we can find the vertex. Since this function is written in standard form, On the Y equals a x plus a squared plus bx plus c form, and not in vertex form, we can't just read off the vertex like we couldn't before. But there's a trick called the vertex formula, which says that whenever you have a function in a quadratic function in standard form, the vertex has an x coordinate of negative B over two A. So in this case, that's an x coordinate of negative 10 over two times five or negative one, which is kind of like where I put it on the graph. To find the y coordinate of that vertex, I can just plug in negative one into my equation for x, which gives me at y equals five times negative one squared, plus 10, times negative one plus three, which is negative two. So I think I better redraw my graph a little bit to put that vertex down at a y coordinate of negative two where it's supposed to be. Let's summarize the steps we use to graph these quadratic functions. First of all, the graph of a quadratic function has the shape of a parabola. The parabola opens up, if the coefficient of x squared, which we'll call a is greater than zero and down if a is less than zero. To find the x intercepts, we set y equal to zero, or in other words, f of x equal to zero and solve for x, the find the vertex, we can either read it off as h k, if our function is in vertex form, or we can use the vertex formula and get the x coordinate of the vertex to be negative B over to a if our function is in standard form. To find the y coordinate of the vertex in this case, we just plug in the x coordinate and figure out what y is. Finally, we can always find additional points on the graph by plugging in values of x. In this video, we learned some tricks for graphing quadratic functions. In particular, we saw that the vertex can be read off as h k, if our function is written in vertex form, and the x coordinate of the vertex can be calculated as negative B over two A, if our function is in standard form. For an explanation of why this vertex formula works, please see the my other video. a quadratic and standard form looks like y equals A x squared plus bx plus c, where a, b and c are real numbers, and a is not zero. a quadratic function in vertex form looks like y equals a times x minus h squared plus k, where a h and k are real numbers, and a is not zero. When a functions in vertex form, it's easy to read off the vertex is just the ordered pair, H, okay. This video explains how to get from vertex form two standard form and vice versa. Let's start by converting this quadratic function from vertex form to standard form. That's pretty straightforward, we just have to distribute out. So if I multiply out the X minus three squared, I get minus four times x squared minus 6x plus nine plus one, distributing the negative four, I get negative 4x squared plus 24x minus 36 plus one. So that works out to minus 4x squared plus 24x minus 35. And I have my quadratic function now in standard form. Now let's go the other direction and convert a quadratic function that's already in standard form into vertex form. That is, we want to put it in the form of g of x equals a times x minus h squared plus k, where the vertex is at h k. To do this, it's handy to use the vertex formula. The vertex formula says that the x coordinate of the vertex is given by negative over to a, where A is the coefficient of x squared, and B is the coefficient of x. So in this case, we get an x coordinate of negative eight over two times two or negative two. To find the y coordinate of the vertex, we just plug in the x coordinate into our formula for a g of x. So that's g of negative two, which is two times negative two squared plus eight times negative two plus six. And that works out to be negative two by coincidence. So the vertex for our quadratic function has coordinates negative two, negative two. And if I want to write g of x in vertex form, it's going to be a times x minus negative two squared plus minus two. That's because remember, we subtract H and we add K. So that simplifies to g of x equals a times x plus two squared minus two. And finally, we just need to figure out what this leading coefficient as. But notice, if we were to multiply, distribute this out, then the coefficient of x squared would end up being a. So therefore, the coefficient of x squared here, which is a has to be the same as the coefficient of x squared here, which we conveniently also called a, in other words, are a down here needs to be two. So I'm going to write that as g of x equals two times x plus two squared minus two, lots of twos in this problem. And that's our quadratic function in vertex form. If I want to check my answer, of course, I could just distribute out again, I'd get two times x squared plus 4x, plus two, minus two, in other words, 2x squared plus 8x plus six, which checks out to exactly what I started with. This video showed how to get from vertex form to standard form by distributing out and how to get from standard form to vertex form by finding the vertex using the vertex formula. Suppose you have a quadratic function in the form y equals x squared plus bx plus c, and you want to find where the vertex is, when you graph it. The vertex formula says that the x coordinate of this vertex is that negative B over two A, this video gives a justification for where that formula comes from. Let's start with a specific example. Suppose I wanted to find the x intercepts and the vertex for this quadratic function. To find the x intercepts, I would set y equal to zero and solve for x. So that's zero equals 3x squared plus 7x minus five. And to solve for x, I use the quadratic formula. So x is going to be negative seven plus or minus the square root of seven squared minus four times three times negative five, all over two times three. That simplifies to negative seven plus or minus a squared of 109 over six, we could, I could also write this as negative seven, six, plus the square root of 109 over six, or x equals negative seven, six minus the square root of 109 over six, since the square root of 109 is just a little bit bigger than 10, this can be approximated by negative seven six plus 10, six, and negative seven, six minus 10, six. So pretty close to, I guess about what half over here and pretty close to about negative three over here, I'm just going to estimate so that I can draw a picture of the function. Since the leading coefficient three is positive, I know my parabola is going to be opening up and the intercepts are somewhere around here and here. So roughly speaking, it's gonna look something like this. Now the vertex is going to be somewhere in between the 2x intercepts, in fact, it's going to be by symmetry, it'll be exactly halfway in between the 2x intercepts. Since the x intercepts are negative seven, six plus and minus 100 squared of 109 over six, the number halfway in between those is going to be exactly negative seven, six, right, because on the one hand, I have negative seven six plus something. And on the other hand, I have negative seven six minus that same thing. So negative seven, six will be exactly in the middle. So my x coordinate of my vertex will be at negative seven sex. Notice that I got that number from the quadratic formula. More generally, if I want to find the x intercepts for any quadratic function, I set y equal to zero and solve for x using the quadratic formula, negative b plus or minus the square root of b squared minus four AC Oliver to a, b, x intercepts will be at these two values, but the x coordinate of the vertex, which is exactly halfway in between the 2x intercepts will be at negative B over two A. That's where the vertex formula comes from. And it turns out that this formula works even when there are no x intercepts, even when the quadratic formula gives us no solutions. That vertex still has the x coordinate, negative B over two A. And that's the justification of the vertex formula. This video is about polynomials and their graphs. We call that a polynomial is a function like this one, for example. His terms are numbers times powers of x. I'll start with some definitions. The degree of a polynomial is the largest exponent. For example, for this polynomial, the degree is for the leading term is the term with the largest exponent. In the same example, the leading term is 5x to the fourth, it's conventional to write the polynomial in descending order of powers of x. So the leading term is first. But the leading term doesn't have to be the first term. If I wrote the same polynomial as y equals, say, two minus 17x minus 21x cubed plus 5x. Fourth, the leading term would still be the 5x to the fourth, even though it was last. The leading coefficient is the number in the leading term. In this example, the leading coefficient is five. Finally, the constant term is the term with no x's in it. In our same example, the constant term is to please pause the video for a moment and take a look at this next example of polynomial figure out what's the degree the leading term, the leading coefficient and the constant term. The degree is again, four, since that's the highest power we have, and the leading term is negative 7x. to the fourth, the leading coefficient is negative seven, and the constant term is 18. In the graph of the polynomial Shown here are the three marks points are called turning points, because the polynomial turns around and changes direction at those three points, those same points can also be called local extreme points, or they can be called local maximum and minimum points. For this polynomial, the degree is four, and the number of turning points is three. Let's compare the degree and the number of turning points for these next three examples. For the first one, the degree is to and there's one turning point. For the second example, that agree, is three, and there's two turning points. And for this last example, the degree is four, and there's one turning point. For this first example, and the next two, the number of turning points is exactly one less than the degree. So you might conjecture that this is always true. But in fact, this is not always true. In this last example, the degree is four, but the number of turning points is one, not three. In fact, it turns out that while the number of turning points is not always equal to a minus one, it is always less than or equal to the degree minus one. That's a useful factor. Remember, when you're sketching graphs are recognizing graphs of polynomials. The end behavior of a function is how the ends of the function look, as x gets bigger and bigger heads towards infinity, or x gets goes through larger and larger negative numbers towards negative infinity. In this first example, the graph of the function goes down as x gets towards infinity as x goes towards negative infinity. I can draw this with two little arrows pointing Down on either side, or I can say in words, that the function is falling as we had left and falling also as we had right. In the second example, the graph rises to the left and rises to the right. In the third example, the graph falls to the left, but rises to the right. And in the fourth example, it rises to the left and falls to the right. If you study these examples, and others, you might notice there's a relationship between the degree of the polynomial the leading coefficients of the polynomials and the end behavior. Specifically, these four types of end behavior are determined by whether the degree is even or odd. And by whether the leading coefficient is positive or negative. When the degree is even, and the leading coefficient is positive, like in this example, where the leading coefficient is one, we have this sorts of n behavior rising on both sides. When the degree is even at the leading coefficient is negative, like in this example, we have the end behavior that's falling on both sides when the degree is odd, and the leading coefficient is positive, that's like this example, with the degree three and the leading coefficient three, then we have this sort of end behavior. And finally, when the degree is odd, and the leading coefficient is negative, like in this example, we have this sort of NBA havior. I like to remember this chart just by thinking of the most simple examples, y equals x squared, y equals negative x squared, y equals x cubed, and y equals negative x cubed. Because I know by heart what those four examples look like, then I just have to remember that any polynomial with a even degree and positive leading coefficient has the same end behavior as x squared. And similarly, any polynomial with even degree and negative leading coefficient has the same end behavior as negative x squared. And similar statements for x cubed and negative x cubed. We can use facts about turning points and behavior, to say something about the equation of a polynomial just by looking at this graph. In this example, because of the end behavior, we know that the degree is odd, we know that the leading coefficient must be positive. And finally, since there are 1234, turning points, we know that the degree is greater than or equal to five. That's because the number of turning points is less than or equal to the degree minus one. And in this case, the number of turning points we said was four. And so solving that inequality, we get the degree is bigger than equal to five. Put in some of that information together, we see the degree could be five, or seven, or nine, or any odd number greater than or equal to five. But it couldn't be for example, three or six. Because even numbers and and also numbers less than five are right out. This video gave a lot of definitions, including the definition of degree leading coefficients, turning points, and and behavior. We saw that knowing the degree and the leading coefficient can help you make predictions about the number of turning points and the end behavior, as well as vice versa. This video is about exponential functions and their graphs. an exponential function is a function that can be written in the form f of x equals a times b to the x, where a and b are any real numbers. As long as a is not zero, and B is positive. It's important to notice that for an exponential function, the variable x is in the exponent. This is different from many other functions we've seen, for example, quite a quadratic function like f of x equals 3x squared has the variable in the base, so it's not an exponential function. For exponential functions, f of x equals a times b dx. We require that A is not equal to zero, because otherwise, we would have f of x equals zero times b dx. Which just means that f of x equals zero. And this is called a constant function, not an exponential function. Because f of x is always equal to zero in an exponential function, but we require that B is bigger than zero, because otherwise, for example, if b is equal to negative one, say, we'd have f of x equals a times negative one to the x. Now, this would make sense for a lot of values of x. But if we try to do something like f of one half, with our Bs, negative one, that would be the same thing as a times the square root of negative one, which is not a real number, we'd get the same problem for other values of b, if the values of b were negative. And if we tried b equals zero, we'd get a kind of ridiculous thing a times zero to the x, which again is always zero. So that wouldn't count as an exponential either. So we can't use any negative basis, and we can't use zero basis, if we want an exponential function. The number A in the expression f of x equals a times b to the x is called the initial value. And the number B is called the base. The phrase initial value comes from the fact that if we plug in x equals zero, we get a times b to the zero, well, anything to the zero is just one. So this is equal to A. In other words, f of zero equals a. So if we think of starting out, when x equals zero, we get the y value of a, that's why it's called the initial value. Let's start out with this example, where y equals a times b to the x next special function, and we've set a equal to one and B equals a two. Notice that the y intercept, the value when x is zero, is going to be one. If I change my a value, my initial value, the y intercept changes, the function is stretched out. If I make the value of a go to zero, and then negative, then my initial value becomes negative, and my graph flips across the x axis. Let's go back to an a value of say one, and see what happens when we change B. Right now, the B value the basis two, if I increase B, my y intercept sticks at one, but my graph becomes steeper and steeper. If I put B back down close to one, my graph becomes more flat at exactly one, my graph is just a constant. As B gets into fractional territory, point 8.7 point six, my graph starts to slope the other way, it's decreasing now instead of increasing, but notice that the y intercept still hasn't changed, I can get it more and more steep as my B gets farther and farther away from one of course, when B goes to negative territory, my graph doesn't make any sense. So a changes the y intercept, and B changes the steepness of the graph. So that it's added whether it's increasing for B values bigger than one, and decreasing for values of the less than one. we'll summarize all these observations on the next slide. We've seen that for an exponential function, y equals a times b to the x, the parameter or number a gives the y intercept, the parameter B tells us how the graph is increasing or decreasing. Specifically, if b is greater than one, the graph is increasing. And if b is less than one, the graph is decreasing. The closer B is to the number one, the flatter the graph. So for example, if I were to graph y equals point two five to the x and y equals point four to the x, they would both be decreasing graphs, since the base for both of them is less than one. But point two five is farther away from one and point four is closer to one. So point four is going to be flatter. And point two five is going to be more steep. So in this picture, This red graph would correspond to point two five to the x, and the blue graph would correspond to point four to the x. For all these exponential functions, whether the graphs are decreasing or increasing, they all have a horizontal asymptote along the x axis. In other words, at the line at y equals zero, the domain is always from negative infinity to infinity, and the range is always from zero to infinity, because the range is always positive y values. Actually, that's true. If a is greater than zero, if a is less than zero, then our graph flips over the x axis, our domain stays the same, but our range becomes negative infinity to zero. The most famous exponential function is f of x equals e to the x. This function is also sometimes written as f of x equals x of x. The number E is Oilers number as approximately 2.7. This function has important applications to calculus and to some compound interest problems. In this video, we looked at exponential functions, functions of the form a times b to the x, where the variable is in the exponent, we saw that they all have the same general shape, either increasing like this, or decreasing like this, unless a is negative, in which case they flip over the x axis. They all have a horizontal asymptote at y equals zero, the x axis. In this video, we'll use exponential functions to model real world examples. Let's suppose you're hired for a job. The starting salary is $40,000. With a guaranteed annual raise of 3% each year, how much will your salary be after one year, two years, five years. And in general after two years, let me chart out the information. The left column will be the number of years since you're hired. And the right column will be your salary. When you start work, at zero years after you're hired, your salary will be $40,000. After one year, you'll have gotten a 3% raise. So your salary will be the original 40,000 plus 3% of 40,000.03 times 40,000. I can think of this first number as one at times 40,000. And I can factor out the 40,000 from both terms, to get 40,000 times one plus 0.03. we rewrite this as 40,000 times 1.03. This is your original salary multiplied by a growth factor of 1.03. After two years, you'll get a 3% raise from your previous year salary, your previous year salary was 40,000 times 1.03. But you'll add 3% of that, again, I can think of the first number is one times 40,000 times 1.03. And I can factor out the common factor of 40,000 times 1.03 from both terms, to get 40,000 times 1.03 times one plus 0.03. Let me rewrite that as 40,000 times 1.03 times 1.03 or 40,000 times 1.3 squared. We can think of this as your last year salary multiplied by the same growth factor of 1.03. After three years, a similar computation will give you that your new salary is your previous year salary times that growth factor of 1.03 that can be written as 40,000 times 1.03 cubed. And in general, if you're noticing the pattern, after two years, your salary should be 40,000 times 1.03 to the t power. In other words, your salary after two years is your original salary multiplied By the growth factor of 1.03, taken to the t power, let me write this as a formula s of t, where S of t is your salary is equal to 40,000 times 1.03 to the T. This is an exponential function, that is a function of the form a times b to the T, where your initial value a is 40,000 and your base b is 1.03. Notice that your base is the amount that your salary gets multiplied by each year. Given this formula, we can easily figure out what your salary will be after, for example, five years by plugging in five for T. I worked that out on my calculator to be $46,370.96 to the nearest cent. exponential functions are also useful in modeling population growth. The United Nations estimated that the world population in 2010 was 6.7 9 billion growing at a rate of 1.1% per year. Assuming that the growth rate stayed the same and will continue to stay the same, we'll write an equation for the population at t years after the year 2010 1.1%, written as a decimal is 0.011. So if we work out a chart as before, we see that after zero years since 2010, we have our initial population 6.7 9 billion, after one year, we'll take that 6.7 9 billion and add 1.1% of it. That is point 011 times 6.79. This works out to 6.79 times one plus point 011 or 6.79 times 1.011. Here we have our initial population of 6.7 9 billion, and our growth factor of 1.011. That's how much the population got multiplied by in one year. As before, we can work out that after two years, our population becomes 6.79 times 1.011 squared, since they got multiplied by 1.011, twice, and after two years, it'll be 6.79 times 1.011 to the t power. So our function that models population is going to be 6.79 times 1.011 to the T. Here, t represents time in years, since 2010. Just for fun, I'll plug in t equals 40. That's 40 years since 2010. So that's the year 2050. And I get 6.79 times 1.011 to the 40th power, which works out to 10 point 5 billion. That's the prediction based on this exponential model. The previous two examples were examples of exponential growth. This last example is an example of exponential decay. The drugs Seroquel is metabolized and eliminated from the body at a rate of 11% per hour. If 400 milligrams are given how much remains in the body. 24 hours later, I'll chart out my information where the left column will be time in hours since the dose was given, and the right column will be the number of milligrams of Seroquel still on the body. zero hours after the dose is given, we have the full 400 milligrams in the body. One hour later, we have the formula 100 milligrams minus 11% of it, that's minus point one one times 400. If I factor out the 400 from both terms, I get 400 times one minus 0.11 or 400 times point eight nine. The 400 represents the initial amount. The point eight nine I'll call the growth factor, even though in this case the quantity is decreasing, not growing. So really it's kind of a shrink factor. Let's see what happens after two hours. Now I'll have 400 times 0.89 my previous amount, it'll again get multiplied by point eight, nine, so that's going to be 400 times 0.89 squared. And in general, after t hours, I'll have 400 multiplied by this growth or shrinkage factor of point eight nine, raised to the t power. Since each hour, the amount of Seroquel gets multiplied by point eight, nine, that number less than one. All right, my exponential decay function as f of t equals 400 times 0.89 to the T, where f of t represents the number of milligrams of Seroquel in the body. And t represents the number of hours since the dose was given. To find out how much is in the body after 24 hours, I just plug in 24 for t. This works out to about 24.4 milligrams, I hope you notice the common form for the functions used to model all three of these examples. The functions are always in the form f of t equals a times b to the T, where a represented the initial amount, and B represented the growth factor. To find the growth factor B, we started with the percent increase or decrease. We wrote it as a decimal. And then we either added or subtracted it from one, depending on whether the quantity was increasing or decreasing. Let me show you that as a couple examples. In the first example, we had a percent increase of 3% on the race as a decimal, I'll call this r, this was point O three. And to get the growth factor, we added that to one to get 1.03. In the world population example, we had a 1.1% increase, we wrote this as point 011 and added it to one to get 1.011. In the drug example, we had a decrease of 11%. We wrote that as a decimal. And we subtracted the point one one from one to get 0.89. In fact, you can always write the growth factor B as one plus the percent change written as a decimal. If you're careful to make that percent change, negative when the quantity is decreasing and positive when the quantity is increasing. Since here one plus negative point 11 gives us the correct growth factor of point eight nine which is less than one as a good check. Remember that if your quantity is increasing, the base b should be bigger than one. And if the quantity is decreasing, then B should be less than one. exponential functions can also be used to model bank accounts and loans with compound interest, as we'll see in another video. This video is about interpreting exponential functions. An antique car is worth $50,000 now, and its value increases by 7%. Each year, let's write an equation to model its value x years from now. After one year, it's value is the 50,000 plus 0.07 times the 50,000. That's because its value has grown by 7% or point oh seven times 50,000. this can be written as 50,000 times one plus point O seven. Notice that adding 7% to the original value is the same as multiplying the original value by one plus point oh seven or by 1.07. After two years, the value will be 50,000 times one plus 0.07 squared, or 50,000 times 1.07 squared. That's because the previous year's value is multiplied again by 1.7. In general, after x years We have x, the antique cars value will be 50,000 times 1.07 to the x. That's because the original value of 50,000 gets multiplied by 1.07x times one time for each year. If we dissect this equation, we see that the number of 50,000 comes from the original value of the car. The one point is seven, which I call the growth factor comes from one plus point 707. The point O seven being the, the percent increase written as a decimal. So the form of this equation is an exponential equation, V of x equals a times b to the x, where A is the initial value, and B is the growth factor. But we could also write this as a times one plus r to the x, where A is still the initial value, but R is the percent increase written as a decimal. This same equation will come up in the next example, here, my Toyota Prius is worth only 3000. Now, and its value is decreasing by 5% each year. So after one year, its value will be 3000 minus 0.05 times 3000. That is 3000 times one minus 0.05. I can also write that as 3000 times 0.95. Decreasing the value by 5% is like multiplying the value by one minus point oh five, or by point nine, five. After two years, the value will be multiplied by point nine five again. So the value will be 3000 times point nine, five squared. And after x years, the value will be 3000 times point nine five to the x. So my equation for the value is 3000 times point nine five to the x. This is again, an equation of the form V of x equals a times b to the x, where here, a is 3000, the initial value, and B is point nine, five, I'll still call that the growth factor, even though we're actually declining value not growing. Now remember where this point nine five came from, it came from taking one and subtracting point 05, because of the 5%, decrease in value, so I can again, write my equation in the form a times well, this time times one minus r to the x, where R is our point 05. That's our percent decrease written as a decimal. Please take a moment to study this equation and the previous one. They say that when you have an exponential function, the number here is the initial value. If it's written in this form, B is your growth factor. But you can think of B as being either one minus r, where r is the percent decrease, or one plus r, where r is the percent increase. In this example, we're given a function f of x to model the number of bacteria in a petri dish x hours after 12 o'clock noon, we want to know what was the number of bacteria at noon, and by what percent, the number of bacteria is increasing every hour, we can see from the equation that the number of bacteria is increasing and not decreasing, because the base of the exponential function 1.45 is bigger than one. Notice that our equation f of x equals 12 times 1.45 to the X has the form of a times b to the x, or we can think of it as a times one plus r to the x. Here a is 12, b is 1.45 and r is 0.45. Based on this familiar form, we can recognize that the initial amount of bacteria is going to be 12 12,000. Since those are our units and this value 1.45 is our growth factor. What the number of bacteria is multiplied by each hour? Well are the point four five is the the rate of increase, in other words, a 45% increase each hour. So our answers to the questions are 12,045%. In this example, the population of salamanders is modeled by this exponential function, where x is the number of years since 2015. Notice that the number of salamanders is decreasing, because the base of our exponential function point seven, eight is less than one. So if we recognize the form of our exponential function, a times b to the x, or we can think of this as a times one minus r to the x, where A is our initial value, and r is our percent decrease written as a decimal. Our initial value is 3000. So that's the number of salamanders zero years after 2015. Our growth factor B is 0.78. But if I write that as one minus r, I see that R has to be one minus 0.78, or 0.22. In other words, our population is decreasing by 22% each year. In this video, we saw that exponential functions can be written in the form f of x equals a times b to the x, where A is the initial value. And B is the growth factor. We also saw that they can be written in the form a times one plus r to the x when the amount is increasing. And as a times one minus r to the x when the amount is decreasing. In this format, R is the percent increase or the percent decrease written as a decimal. So 15% increase will be an R value of 0.15 and a growth factor B of 1.15. Whereas a 12% decrease will be an R value of point one, two, and a B value of one minus point one, two, or 0.7. Sorry, 0.88. These observations help us quickly interpret exponential functions. For example, here, we have an initial value of 100 and a 15% increase. And here, we have an initial value of 50 and a 40%. decrease. exponential functions can be used to model compound interest for loans and bank accounts. Suppose you invest $200 in a bank account that earns 3% interest every year. If you make no deposits or withdrawals, how much money will you have accumulated after 10 years, because 3% of the money that's in the bank is getting added each year, the money in the bank gets multiplied by 1.03 each year. So after one year, the amount will be 200 times 1.3, after two years 200 times one point O three squared, and and after t years 200 times one point O three to the t power. So the function modeling the amount of money, I'll call it P of t is given by 200 times 1.03 to the T. More generally, if you invest a dollars at an annual interest rate of our for t years. In the end, you'll have P of t equals a times one plus r to the T here r needs to be written as a decimal, so 0.03 In our example, for the 3% annual interest rate. Going back to our specific example, After 10 years, the amount of money is going to be P of 10 which is 200 times 1.03 to the 10 which works out to 206 $68.78 to the nearest cent. In this problem, we've assumed that the interest accumulates once per year. But in the next few examples, we'll see what happens when the interest rate accumulates more frequently, twice a year, or every month. For example, let's deposit $300 in an account that earns 4.5% annual interest compounded semi annually, this means two times a year or every six months. A 4.5% annual interest rate compounded two times a year means that we're actually getting 4.5 over 2% interest, every time the interest is compounded. That is every six months. That's 2.25% interest every half a year. Note that 2.25% is the same as 0.02 to five as a decimal. So every time we earn interest, our money gets multiplied by 1.02 to five. Let me make a chart of what happens. After zero years, which is also zero half years, we have our original $300. for half a year, that's one half year, our money gets earns interest one time, so we multiply the 300 by 1.02 to five, after one year, that's two half years, our money earns interest two times. So we multiply 300 by 1.0225 squared. continuing this way, after 1.5 years, that's three half years, we have 300 times 1.02 to five cubed. And after two years or four half years, we have 300 times 1.02 to five to the fourth. In general, after t years, which is to T half years, our money will grow to 300 times 1.02 to five to the two t power. Because we've compounded interest to tee times our formula for our amount of money is P of t equals 300 times one point O two to five to the two t where t is the number of years. To finish the problem, after seven years, we'll have P of $7, which is 300 times 1.02 to five to the two times seven or 14 power. And that works out to $409.65 to the nearest cent. In this next example, we're going to take out a loan for $1,200 in annual interest rate of 6% compounded monthly. Although loans and bank accounts might feel different, they're mathematically the same. It's like from the bank's point of view, they're investing money in you and getting interest on their money from you. So we can work them out with the same kind of math 6% annual interest rate compounded monthly means you're compounding 12 times a year. So each time you compound interest, you're just going to get six over 12% interest. That's point 5% interest. And as a decimal, that's 0.005. Let's try it out again, what happens. Time is zero, of course, you'll have the original loan amount of 1200. After one year, that's 12 months, your loan has had interest added to it 12 times so it gets multiplied by 1.05 to the 12th. After two years, that's 24 months, it's had interest added to it 24 times, so it gets multiplied by 1.05 to the 24th power. Similarly, after three years or 36 months, your loan amount will be 1200 times 1.05 to the 36th power. And in general, after two years, that's 12 t months. So the interest will be compounded 12 t times and so we have to raise the 1.05 To the 12 t power. This gives us the general formula for the money owed is P of t equals 1200 times 1.05 to the 12 T, where T is the number of years. In particular, after three years, we'll have to pay back a total of 1200 times 1.05 to the 12 times three or 36 power, which works out to $1,436.02 to the nearest cent. These last two problems follow a general pattern. If A is the initial amount of the loan or bank account, and r is the annual interest rate, compounded n times per year, then our formula for the amount of money is going to be a times one plus r over n to the n t. This formula exactly matches what we did in this problem. First, we took the interest rate our which was 6%, and divided it by the number of compounding periods each year 12. We wrote that as a decimal and added one to it. That's where we got the 1.05 from. We raised this to not to the number of years, but to 12 times the number of years. That's the number of compounding periods per year, times the number of years. And we multiplied all that by the initial amount of money, which was 1200. This formula for compound interest is a good one to memorize. But it's also important to be able to reason your way through it, like we did in this chart. There's one more type of compound interest. And that's interest compounded continuously. You can think of continuous compounding as the limit of compounding more and more frequently 10 times a year, 100 times a year 1000 times a year a million times a year. In the limit, you get continuous compounding. The formula for continuous compounding is P of t equals a times e to the RT, where p of t is the amount of money, t is the time in years. A is the initial amount of money. And R is the annual interest rate written as a decimal. So 0.025 in this problem from the 2.5% annual interest rate. He represents the famous constant Oilers constant, which is about 2.718. So in this problem, we have P of t is 4000 times e to the 0.025 T. And after five years, we'll have P of five, which is 4000 times e to the 0.0 to five times five, which works out to $4,532.59 to the nearest cent. To summarize, if R represents the annual interest rate written as a decimal, that is 2% would be 0.02. And t represents the number of years and a represents the initial amount of money, then for just simple annual interest compounded once a year. Our formula is P of t is a times one plus r to the T for compound interest compounded n times per year. Our formula is P of t is a times one plus r over n to the n T. And for compound interest compounded continuously, we get P of t is a times e to the RT. In this video, we looked at three kinds of compound interest problems, simple annual interest, interest compounded and times per year, and continuously compounded interest. This video introduces logarithms. logarithms are a way of writing exponents. The expression log base a of B equals c means that a to the C equals b. In other words, log base a of B is the exponent that you raise a to to get BE THE NUMBER A It is called the base of the logarithm. It's also called the base when we write it in this exponential form. Some students find it helpful to remember this relationship. log base a of B equals c means a to the C equals b, by drawing arrows, a to the C equals b. Other students like to think of it in terms of asking a question, log base a of B, asks, What power do you raise a to in order to get b? Let's look at some examples. log base two of eight is three, because two to the three equals eight. In general, log base two of y is asking you the question, What power do you have to raise to to to get y? So for example, log base two of 16 is four, because it's asking you the question to what power equals 16? And the answer is four. Please pause the video and try some of these other examples. log base two of two is asking, What power do you raise to two to get two? And the answer is one. Two to the one equals two. log base two of one half is asking two to what power gives you one half? Well, to get one half, you need to raise two to a negative power. So that would be two to the negative one. So the answer is negative one. log base two of 1/8 means what power do we raise to two in order to get 1/8. Since 1/8, is one over two cubed, we have to raise two to the negative three power to get one over two cubed. So our exponent is negative three. And that's our answer to our log expression. Finally, log base two of one is asking to what power equals one. Well, anything raised to the zero power gives us one, so this log expression evaluates to zero. Notice that we can get positive negative and zero answers for our logarithm expressions. Please pause the video and figure out what these logs evaluate to. to work out log base 10 of a million. Notice that a million is 10 to the sixth power. Now we're asking the question, What power do we raise tend to to get a million? So that is what power do we raise 10 to to get 10 to the six? Well, of course, the answer is going to be six. Similarly, since point O one is 10 to the minus three, this log expression is the same thing as asking, what's the log base 10 of 10 to the minus three? Well, what power do you have to raise 10? to to get 10 to the minus three? Of course, the answer is negative three. Log base 10 of zero is asking, What power do we raise 10 to to get zero. If you think about it, there's no way to raise 10 to an exponent get zero. Raising 10 to a positive exponent gets us really big positive numbers. Raising 10 to a negative exponent is like one over 10 to a power that's giving us tiny fractions, but they're still positive numbers, we're never going to get zero. Even if we raise 10 to the zero power, we'll just get one. So there's no way to get zero and the log base 10 of zero does not exist. If you try it on your calculator using the log base 10 button, you'll get an error message. Same thing happens when we do log base 10 of negative 100. We're asking 10 to what power equals negative 100. And there's no exponent that will work. And more generally, it's possible to take the log of numbers that are greater than zero, but not for numbers that are less than or equal to zero. In other words, the domain of the function log base a of x, no matter what base you're using for a, the domain is going to be all positive numbers. A few notes on notation. When you see ln of x, that's called natural log, and it means the log base e of x where he is that famous number that's about 2.718. When you see log of x with no base at all, by convention, that means log base 10 of x And it's called the common log. Most scientific calculators have buttons for natural log, and for common log. Let's practice rewriting expressions with logs in them. log base three of one nine is negative two, can be rewritten as the expression three to the negative two equals 1/9. Log of 13 is shorthand for log base 10 of 13. So that can be rewritten as 10 to the 1.11394 equals 13. Finally, in this last expression, ln means natural log, or log base e, so I can rewrite this equation as log base e of whenever E equals negative one. Well, that means the same thing as e to the negative one equals one over e, which is true. Now let's go the opposite direction. We'll start with exponential equations and rewrite them as logs. Remember that log base a of B equals c means the same thing as a to the C equals b, the base stays the same in both expressions. So for this example, the base of three in the exponential equation, that's going to be the same as the base in our log. Now I just have to figure out what's in the argument of the log. And what goes on the other side of the equal sign. Remember that the answer to a log is an exponent. So the thing that goes in this box should be my exponent for my exponential equation. In other words, you. And I'll put the 9.78 as the argument of my log. This works, because log base three of 9.78 equals view means the same thing as three to the U equals 9.78, which is just what we started with. In the second example, the base of my exponential equation is E. So the base of my log is going to be the answer to my log is an exponent. In this case, the exponent 3x plus seven. And the other expression, the four minus y becomes my argument of my log. Let me check, log base e of four minus y equals 3x plus seven means e to the 3x plus seven equals four minus y, which is just what I started with. I can also rewrite log base e as natural log. This video introduced the idea of logs, and the fact that log base a of B equal c means the same thing as a to the C equals b. So log base a of B is asking you the question, What power exponent Do you raise a to in order to get b. In this video, we'll work out the graph, so some log functions and also talk about their domains. For this first example, let's graph a log function by hand by plotting some points. The function we're working with is y equals log base two of x, I'll make a chart of x and y values. Since we're working this out by hand, I want to pick x values for which it's easy to compute log base two of x. So I'll start out with the x value of one because log base two of one is zero, log base anything of one is 02 is another x value that's easy to compute log base two of two, that's asking, What power do I raise to two to get to one? And the answer is one. Power other powers of two are easy to work with. So for example, log base two of four that saying what power do I raise to two to get four, so the answer is two. Similarly, log base two of eight is three and log base two of 16 is four. Let me also work with some fractional values for X. If x is one half, then log base two of one half that saying what power do I raise to two to get one half? Well, that needs a power of negative one. It's also easy to compute by hand, the log base two of 1/4 and 1/8. log base two of 1/4 is negative two since two to the negative two is 1/4. And similarly, log base two of one eight is negative f3 I'll put some tick marks on my x and y axes. Please pause the video and take a moment to plot these points. Let's see, I have the point, one zero, that's here to one that's here, for two, that is here, and then eight, three, which is here. And the fractional x values, one half goes with negative one, and 1/4 with negative two 1/8 with negative three. And if I connect the dots, I get a graph that looks something like this. If I had smaller and smaller fractions, I would keep getting more and more negative answers when I took log base two of them, so my graph is getting more and more negative, my y values are getting more and more negative, as x is getting close to zero. Now I didn't draw any parts of the graph over here with negative X values, I didn't put any negative X values in my chart, that omission is no accident. Because if you try to take the log base two or base anything of a negative number, like say negative four or something, there's no answer. This doesn't exist because there's no power that you can raise to to to get a negative number. So there are no points on the graph for negative X values. And similarly, there are no points on the graph where x is zero, because you can't take log base two of zero, there's no power you can raise to two to get zero. I want to observe some key features of this graph. First of all, the domain is x values greater than zero. In interval notation, I can write that as a round bracket because I don't want to include zero to infinity, the range is going to be the y values, while they go all the way down into the far reaches of the negative numbers. And the graph gradually increases y value is getting bigger and bigger. So the range is actually all real numbers are an interval notation negative infinity to infinity. Finally, I want to point out that this graph has a vertical asymptote at the y axis, that is at the line x equals zero. I'll draw that on my graph with a dotted line. A vertical asymptote is a line that our functions graph gets closer and closer to. So this is the graph of y equals log base two of x. But if I wanted to graph say, y equals log base 10 of x, it would look very similar, it would still have a domain of X values greater than zero, a range of all real numbers and a vertical asymptote at the y axis, it will still go through the point one zero, but it would go through the point 10 one instead, because log base 10 of 10 is one, it would look pretty much the same, just a lot flatter over here. But even though it doesn't look like it with the way I've drawn it, it's still gradually goes up to n towards infinity. In fact, the graph of y equals log base neaa of X for a bigger than one looks pretty much the same, and has the same three properties. Now that we know what the basic log graph looks like, we can plot at least rough graphs of other log functions without plotting points. Here we have the graph of natural log of X plus five. And again, I'm just going to draw a rough graph. If I did want to do a more accurate graph, I probably would want to plot some points. But I know that roughly a log graph, if it was just like y equals ln of x, that would look something like this, and it would go through the point one zero, with a vertical asymptote along the y axis. Now if I want a graph, ln of x plus five, that just shifts our graph by five units, it'll still have the same vertical asymptote. Since the vertical line shifted up by five units, there's still a vertical line, but instead of going through one zero, it'll go through the point one, five. So I'll draw a rough sketch here. Let's compare our starting function y equals ln x and the transformed version y equals ln x plus five in terms of the domain, the range and the vertical asymptote. Our original function y equals ln x had a domain of zero to infinity Since adding five on the outside affects the y values, and the domain is the x values, this transformation doesn't change the domain. So the domain is still from zero to infinity. Now the range of our original y equals ln x was from negative infinity to infinity. Shifting up by five does affect the y values, and the range is talking about the y values. But since the original range was all real numbers, if you add five to all set of all real numbers, you still get the set of all real numbers. So in this case, the range doesn't change either. And finally, we already saw that the original vertical asymptote of the y axis x equals zero, when we shift that up by five units, it's still the vertical line x equals zero. In this next example, we're starting with a log base 10 function. And since the plus two is on the inside, that means we shift that graph left by two. So I'll draw our basic log function. Here's our basic log function. So I'll think of that as y equals log of x going through the point one, zero, here's its vertical asymptote. Now I need to shift everything left by two. So my vertical asymptote shifts left, and now it's at the line x equals negative two, instead of at x equals zero, and my graph, let's see my point, one zero gets shifted to, let's see negative one zero, since I'm subtracting two from the axis, and here's a rough sketch of the resulting graph. Let's compare the features of the two graphs drawn here. We're talking about domains, the original had a domain of from zero to infinity. But now I've shifted that left. So I've subtracted two from all my x values. And here's my new domain, which I can also verify just by looking at the picture. My range was originally from negative infinity to infinity, well, shifting left only affects the x value. So it doesn't even affect the range. So my range is still negative infinity to infinity. My vertical asymptote was originally at x equals zero. And since I subtract two from all x values, that shifts it to x equals negative two. In this last problem, I'm not going to worry about drawing this graph. I'll just use algebra to compute its domain. So let's think about what's the issue, when you're taking the logs of things? Well, you can't take the log of a negative number is zero. So whatever's inside the argument of the log function, whatever is being fed into log had better be greater than zero. So I'll write that down, we need to minus 3x to be greater than zero. Now it's a matter of solving an inequality to it's got to be greater than 3x. So two thirds is greater than x. In other words, x has to be less than two thirds. So our domain is all the x values from negative infinity to two thirds, not including two thirds. It's a good idea to memorize the basic shape of the graph of a log function. It looks something like this, go through the point one zero, and has a vertical asymptote on the y axis. Also, if you remember that you can't take the log of a negative number, or zero, then that helps you quickly compute domains for log functions. Whatever's inside the log function, you set that greater than zero, and solve. This video is about combining logs and exponents. Please pause the video and take a moment to use your calculator to evaluate the following four expressions. Remember, that log base 10 on your calculator is the log button. While log base e on your calculator is the natural log button, you should find that the log base 10 of 10 cubed is three. The log base e of e to the 4.2 is 4.2 10 to the log base 10 of 1000 is 1000. And eat the log base e of 9.6 is 9.6. In each case, the log and the exponential function with the same base undo each other and we're left with the exponent. In fact, it's true that for any base a the log base a of a to the x is equal to x, the same sort of cancellation happens if we do the exponential function in the log function with the same base in the opposite order. For example, we take 10 to the power of log base 10 of 1000, the 10 to the power and the log base 10 undo each other, and we're left with the 1000s. This happens for any base, say, a data log base a of x is equal to x. We can describe this by saying that an exponential function and a log function with the same base undo each other. If you're familiar with the language of inverse functions, the exponential function and log function are inverses. Let's see why these roles hold for the first log role. log base a of a dx is asking the question, What power do we raise a two in order to get a to the x? In other words, a two what power is a dx? Well, the answer is clearly x. And that's why log base a of a to the x equals x. For the second log rule, notice that the log base a of x means the power we raise a two to get x. But this expression is saying that we're supposed to raise a to that power. If we raise a to the power, we need to raise a two to get x, then we'll certainly get x. Now let's use these two rules. In some examples. If we want to find three to the log base three of 1.43 to the power and log base three undo each other, so we're left with 1.4. If we want to find ln of e iliacs. Remember that ln means log base e. So we're taking log base e of e to the x, well, those functions undo each other, and we're left with x. If we want to take 10 to the log of three z, remember that log without a base written implies that the base is 10. So really, we want to take 10 to the log base 10 of three z will tend to a power and log base 10 undo each other. So we're left with a three z. Finally, this last statement hold is ln of 10 to the x equal to x, well, ln means log base e. So we're taking log base e of 10 to the x, notice that the base of the log and the base of the exponential function are not the same. So they don't undo each other. And in fact, log base e of 10 to the x is not usually equal to x, we can check with one example. Say if x equals one, then log base e of 10 to the one, that's log base e of 10. And we can check on the calculator that's equal to 2.3. And some more decimals, which is not the same thing as one. So this statement is false, it does not hold. We need the basis to be the same for logs and exponents to undo each other. In this video, we saw that logs and exponents with the same base undo each other. Specifically, log base a of a to the x is equal to x and a to the log base a of x is also equal to x for any values of x and any base a. This video is about rules or properties of logs. The log rules are closely related to the exponent rules. So let's start by reviewing some of the exponent rules. To keep things simple, we'll write everything down with a base of two. Even though the exponent rules hold for any base. We know that if we raise two to the zero power, we get one, we have a product rule for exponents, which says that two to the M times two to the n is equal to two to the m plus n. In other words, if we multiply two numbers, then we add the exponents. We also have a quotient rule that says that two to the M divided by n to the n is equal to two to the m minus n. In words, that says that if we divide two numbers, then we subtract the exponents. Finally, we have a power rule that says if we take a power to a power, then we multiply the exponents. Each of these exponent rules can be rewritten as a log rule. The first rule, two to the zero equals one can be rewritten in terms of logs as log base two of one equals zero. That's because log base two of one equals zero mean To the zero equals one. The second rule, the product rule can be rewritten in terms of logs by saying log of x times y equals log of x plus log of y. I'll make these base two to agree with my base that I'm using for my exponent rules. In words, that says the log of the product is the sum of the logs. Since logs really represent exponent, this is saying that when you multiply two numbers together, you add their exponents, which is just what we said for the exponent version. The quotient rule for exponents can be rewritten in terms of logs by saying the log of x divided by y is equal to the log of x minus the log of y. In words, we can say that the log of the quotient is equal to the difference of the logs. Since logs are really exponents, another way of saying the same thing is that when you divide two numbers, you subtract their exponents. That's how we described the exponent rule above. Finally, the power rule for exponents can be rewritten in terms of logs by saying the log of x to the n is equal to n times log of x. Sometimes people describe this rule by saying when you take the log of an expression with an exponent, you can bring down the exponent and multiply. If we think of x as being some power of two, this is really saying when we take a power to a power, we multiply their exponents. That's exactly how we described the power rule above. It doesn't really matter if you multiply this exponent on the left side, or on the right side. But it's more traditional to multiply it on the left side. I've given these rules with the base of two, but they actually work for any base. To help you remember them, please take a moment to write out the log roles using a base of a you should get the following chart. Let's use the log rules to rewrite the following expressions as a sum or difference of logs. In the first expression, we have a log base 10 of a quotient. So we can rewrite the log of the quotient as the difference of the logs. Now we still have the log of a product, I can rewrite the log of a product as the sum of the logs. So that is log of y plus log of z. When I put things together, I have to be careful because here I'm subtracting the entire log expression. So I need to subtract both terms of this song on the make sure I do that by putting them in parentheses. Now I can simplify a little bit by distributing the negative sign. And here's my final answer. In my next expression, I have the log of a product. So I can rewrite that as the sum of two logs. I can also use my power rule to bring down the exponent T and multiply it in the front. That gives me the final expression log of five plus t times log of to one common mistake on this problem is to rewrite this expression as t times log of five times two. In fact, those two expressions are not equal. Because the T only applies to the two, not to the whole five times two, we can't just bring it down in front using the power wall. After all, the power rule only applies to a single expression raised to an exponent, and not to a product like this. And these next examples, we're going to go the other direction. Here we're given sums and differences of logs. And we want to wrap them up into a single log expression. By look at the first two pieces, that's a difference of logs. So I know I can rewrite it as the log of a quotient. Now I have the sum of two logs. So I can rewrite that as the log of a product. I'll clean that up a little bit and rewrite it as log base five of a times c over B. In my second example, I can rewrite the sum of my logs as the log of a product now, I will Like to rewrite this difference of logs as the log of a quotient. But I can't do it yet, because of that factor of two multiplied in front. But I can use the power rule backwards to put that two back up in the exponent. So I'll do that first. So I will copy down the ln of x plus one times x minus one, and rewrite this second term as ln of x squared minus one squared. Now I have a straightforward difference of two logs, which I can rewrite as the log of a quotient. I can actually simplify this some more. Since x plus one times x minus one is the same thing as x squared minus one. I can cancel factors to get ln of one over x squared minus one. In this video, we saw four rules for logs that are related to exponent rules. First, we saw that the log with any base of one is equal to zero. Second, we saw the product rule, the log of a product is equal to the sum of the logs. We saw the quotient rule, the log of a quotient is the difference of the logs. And we saw the power rule. When you take a log of an expression with an exponent in it, you can bring down the exponent and multiply it. It's worth noticing that there's no log rule that helps you split up the log of a psalm. In particular, the log of a psalm is not equal to the sum of the logs. If you think about logs and exponent rules going together, this kind of makes sense, because there's also no rule for rewriting the sum of two exponential expressions. Log rules will be super handy, as we start to solve equations using logs. Anytime you have an equation like this one that has variables in the exponent logs are the tool of choice for getting those variables down where you can solve for them. In this video, I'll do a few examples of solving equations with variables in the exponent. For our first example, let's solve for x the equation five times two to the x plus one equals 17. As my first step, I'm going to isolate the difficult spot, the part that has the variables and the exponent. In this example, I can do that by dividing both sides by five. That gives me two to the x plus one equals 17 over five. Next, I'm going to take the log of both sides, it's possible to take the log with any base, but I prefer to take the log base 10, or the log base e for the simple reason that my calculator has buttons for those logs. So in this example, I'll just take the log base 10. So I can omit the base because it's a base 10 is implied here. And that gives me this expression. As my next step, I'm going to use log roles to bring down my exponent and multiply it on the front. It's important to use parentheses here because the entire x plus one needs to get multiplied by log two. So that was my third step using the log roles. Now all my variables are down from the exponent where I can work with them, but I still need to solve for x. Right now x is trapped in the parentheses. So I'm going to free it from the parentheses by distributing. So I get x log two plus log two equals log of 17 fifths. Now I will try to isolate x by moving all my terms with x's in them to one side, and all my terms without x's in them to the other side. Finally, I factor out my x. Well, it's already kind of factored out, and I divide to isolate it. So I read out what I did. So far I distributed. I moved all the x terms on one side, and the terms without x's on the other side. And then I isolated the x by factoring out and dividing. We have an exact solution for x. This is correct, but maybe not so useful if you want a decimal answer. At this point, you can type in everything into your calculator using parentheses liberally to get a decimal answer about 0.765. It's always a good idea to check your work by typing in this decimal answer and seeing if the equation checks out. This next example is trickier because there are variables in the exponent in two places with two different bases. First step is normally to clean things up and simplify by isolating the tricky stuff. But in this example, there's nothing really to clean up or simplifier, or no way to isolate anything more than it already is. So we'll go ahead to the next step and take the log of both sides. Again, I'll go ahead and use log base 10. But we couldn't use log base e instead. Next, we can use log roles to bring down the exponents. This gives me 2x minus three in parentheses, log two equals x minus two, log five. Now I'm going to distribute things out to free the excess from the parentheses. That gives me 2x log two minus three log two equals x log five minus two log five. Now I need to group the x terms on one side, and the other terms that don't involve x on the other side. So I'll keep the 2x log two on the left, put the minus log x log five on the left, and that gives me the minus two log five that stays on the right and a plus three log two on the right. Finally, I need to isolate x by factoring and dividing by factoring I just mean I factor out the x from all the terms on the left, so that gives me x times the quantity to log two minus log five. And that equals this mess on the right. Now I can divide the right side by the quantity on the left side. When you type this into your calculator, I encourage you to type the whole thing in rather than type bits and pieces in because if you round off, you'll get a less accurate answer than if you type the whole thing in at once. In this example, when I type it in, I get a final answer of about 5.106. In this equation, we have the variable t in the exponent in two places. The letter E is not a variable, it represents the number e whose decimal approximation is about 2.7. Because there's already an E and the expression, it's going to be handy to use natural log in this problem instead of log base 10. But before we take the log of both sides, let's clean things up. by isolating the tricky parts, we could at least divide both sides by five. And that will give us either the negative 0.05 t is equal to three fifths it is 0.2 T. One way to proceed would now be to clean things out further by dividing both sides by either the point two t, but I'm going to take a different approach and go ahead and take the natural log of both sides. That gives me ln of e to the minus 0.05 t is equal to ln of three fifths e to the 0.2 t. Now on the left side, I can immediately use my log rule to bring down my exponent and get minus 0.05 t, L and E. But on the right side, I can't bring down the exponent yet because this e to the point two t is multiplied by three fifths. So before I can bring down the exponent, I need to split up this product using the product rule. So I can rewrite this as ln of three fifths plus ln e to the 0.2 T. And now I can bring down the exponent. So that third step was using log roles to ultimately bring down my exponents. Now ln of E is a really nice expression, ln IV means log base e of E. So that's asking what power do I raise E to in order to get e? And the answer is one. So anytime I have ln of E, I can just replace that with one. That's why using natural log is a little bit handier here than using log base 10. You can make that simplification. Next, I'm ready to solve for t. So I don't need to distribute here. But I do need to bring my T terms to one side of my terms without t to the other side. So let's see. I'll put my T terms on the left and my team turns without T on the right. And finally I'm going to isolate t by factoring and dividing. So by factoring I mean I factor out my T and now I can divide. Using my calculator, I can get a decimal answer of 2.04 Three, three. This video gave some examples of solving equations with variables in the exponent. And the key idea was to take the log of both sides and use the log properties to bring the exponents down. This video, give some examples of equations with logs in them like this one. In order to solve the equations like this one, we have to free the variable from the log. And we'll do that using exponential functions. My first step in solving pretty much any kind of equation is to simplify it and isolate the tricky part. In this case, the tricky part is the part with the log in it. So I can isolate it by first adding three to both sides. That gives me two ln 2x plus five equals four, and then I can divide both sides by two. Now that I've isolated the tricky part, I still need to solve for x, but x is trapped inside the log function. to free it, I need to somehow undo the log function. Well log functions and exponential functions undo each other. And since this is a log base e, I need to use the exponential function base e also. So I'm going to take e to the power of both sides. In other words, I'll take e to the ln 2x plus five, and that's going to equal e to the power of two. Now e to the ln of anything, I'll just write a for anything. That means e to the log base e of a, that you the power and log base e undo each other. So we just get a, I'm going to use that principle over here, e to the ln of 2x plus five, the E and the log base e undo each other. And we're left with 2x plus five on the left side. So 2x plus five is equal to E squared. And from there, it's easy to finish the problem and solve for x by subtracting five from both sides and then dividing by two. So I'll just write the third step here is to just say finish solving for x. There's one last step we need to do when solving equations with logs in them. And that's to check our answers. Because we may get extraneous solutions. an extraneous solution is a solution that comes out of the solving process, but doesn't actually satisfy the original equation. And those can happen for equations with logs in them, because we might get a solution that makes the argument of the log negative or zero, and we can't take the log of a negative numbers zero. So let's check and plug in the solution of E squared minus five over two. We'll plug that in for x in our original equation and see if that works. So let's see the twos cancel here. So I get two ln e squared minus five plus five, minus three, I want that to equal one. And now my fives cancel. And so I have two ln e squared minus three that I want to equal one. Well, I think this is going to work out because let's see, ln is log base e. and log base e of E squared that's asking the question, What power do I raise e two to get e squared? Well, I have to raise it to the power of two to get e squared. So this becomes two times two minus three. And does that equal one, four minus three does equal one. So that all checks out. And we didn't have any problem with taking the log of negative numbers zero, we didn't get any extraneous solutions. So this is our solution. The second equation is a little bit trickier, because there's a log into places. Now notice that this is a log where there's no base written. So a base 10 is implied. So I'm already thinking in order to undo a log base 10, I'm going to want to take a 10 to the power of both sides. Of course, it's still a good idea to isolate the tricky part, but there's nothing really to isolate here. So I'm going to say, can't do it here. So we'll just jump right to straight step two, and take 10 to the power of both sides. Okay, so that's going to give me 10 to the whole thing, log x plus three plus log x, that whole thing is in the exponent equals 10 to the One Power. Well, I know what to do with the right side 10 to the one is just 10. But what do I do with the 10 to these two things added up Up, while remembering my exponent rules, I know that when you add up the exponent, that's what happens when you multiply two things. So this is the same thing as 10 to the log x plus three times 10 to the log x, right, because when you multiply two things, you add the exponent, so these are the same. Okay, now we're in business, because 10 to the log base 10, those undo each other. And so this whole expression here simplifies to x plus three. Similarly, 10 to the log base, 10 of x is just x. So I'm multiplying x plus three by x, that's equal to 10. Now I have an equation I can deal with, it's a quadratic, so I'm going to first multiply out to make it look more like a quadratic, get everything to one side. So is equal to zero. And, and now I can either factor or use the quadratic formula. I think this one factors, it looks like X plus five times x minus two, so I'm going to get x is negative five, or x is two. So that was all the third step to finish solving for x. Finally, we need to check our solutions to make sure we haven't gotten some extraneous ones. So let's see if x equals negative five. If I plug that into my original equation, that says, I'm checking that log of negative five plus three, plus log of negative five, checking that's equal to one. Well, this is giving me a queasy feeling. And I hope it's giving you a queasy feeling too, because log of negative two does not exist, right, you can't take the log of a negative number. Same thing with log of negative five. So x equals negative five is an extraneous solution, it doesn't actually solve our original equation. Let's check the other solution, x equals two. So now we're checking to see if log of two plus three plus log of two is equal to one. Since there's no problem with taking logs of negative numbers, or zero here, this should work out fine. And just to finish checking, we can see let's see this is log of five plus log of two, we want that equal one. But using my log rules, the sum of two logs is the log of the product. So log of five times two, we want that to equal one. And that's just log base 10 of 10. And that definitely equals one because log base 10 of 10 says, What power do I raise tend to to get 10. And that power is one. So the second solution x equals two does check out. And that's our final answer. Before I leave this problem, I do want to mention that some people have an alternative approach. Some people like to start with the original equation, and then use log roles to combine everything into one log expression. So since we have the sum of two logs, we know that's the same as the log of a product, right, so we can rewrite the left side as log of x plus three times x, that equals one, then we do the same trick of taking 10 to the power of both sides. And as before, the 10 to the power and the log base 10 undo each other, and we get x plus three times x equals 10, just like we did before. In the first solution, we ended up using exponent rules to rewrite things. In the second alternative method, we use log rules to rewrite things. So the methods are really pretty similar, pretty equivalent, and they certainly will get us to the same answer. So we've seen a couple examples of equations with logs in them and how to solve them. And the key step is to use exponential functions to undo the log. In other words, take e to the power of both sides to undo natural log, and take 10 to the power of both sides. To undo log base 10. In this video, we'll use exponential equations in some real life applications, like population growth and radioactive decay. I'll also introduce the ideas of half life and doubling time. In this first example, let's suppose we invest $1,600 in a bank account that earns 6.5% annual interest compounded once a year. How many years will it take until the account has $2,000 in it if you don't make any further deposits or withdrawals since our money is earning six point 5% interest each year, that means that every year the money gets multiplied by 1.065. So after t years, my 1600 gets multiplied by 1.065 to the t power. I'll write this in function notation as f of t equals 1600 times 1.065 to the T, where f of t is the amount of money after t years. Now we're trying to figure out how long it will take to get $2,000 $2,000 is a amount of money. So that's an amount for f of t. And we're trying to solve for T the amount of time. So let me write out my equation and make a note that I'm solving for t. Now to solve for t, I want to first isolate the tricky part. So I'm going to the tricky part is the part with the exponential in it. So I'm going to divide both sides by 1600. That gives me 2000 over 1600 equals 1.065. To the tee, I can simplify this a little bit further as five fourths. Now that I've isolated the tricky part, my next step is going to be to take the log of both sides. That's because I have a variable in the exponent. And I know that if I log take the log of both sides, I can use log roles to bring that exponent down where I can solve for it. I think I'll use log base e this time. So I have ln fi force equals ln of 1.065 to the T. Now by the power rule for logs, on the right side, I can bring that exponent t down and multiply it in the front. Now it's easy to isolate t just by dividing both sides by ln of 1.065. Typing that into my calculator, I get that t is approximately 3.54 years. And the next example, we have a population of bacteria that initially contains 1.5 million bacteria, and it's growing by 12% per day, we want to find the doubling time, the doubling time means the amount of time it takes for a quantity to double in size. For example, the amount of time it takes to get from the initial 1.5 million bacteria to 3 million bacteria would be the doubling time. Let's start by writing an equation for the amount of bacteria. So if say, P of t represents the number of bacteria in millions, then my equation and T represents time in days, then P of t is going to be given by the initial amount of bacteria times the growth factor 1.12 to the T. That's because my population of bacteria is growing by 12% per day. That means the number of bacteria gets multiplied by one point 12. Since we're looking for the doubling time, we're looking for the t value when P of D will be twice as big. So I can set P of t to be three and solve for t. As before, I'll start by isolating the tricky part, taking the log bringing the T down. And finally solving for T. Let's see three over 1.5 is two so I can write this as ln two over ln 1.12. Using my calculator, that's about 6.12 days. It's an interesting fact that doubling time only depends on the growth rate, that 12% growth, not the initial population. In fact, I could have figured out the doubling time without even knowing how many bacteria were in my initial population. Let me show you how that would work. If I didn't know how many I started with, I could still write P of t equals a times 1.12 to the t where a is our initial population that I don't know what it is. Then if I want to figure out how long it takes for my population, to double Well, if I start with a and double that I get to a. So I'll set my population equal to two a. And I'll solve for t. Notice that my A's cancel. And so when I take the log of both sides and bring the T down and solve for t, I get the exact same thing as before, it didn't matter what the initial population was, I didn't even have to know what it was. In this next example, we're told the initial population, and we're told the doubling time, we're not told by what percent the population increases each minute, or by what number, we're forced to multiply the population by each minute. So we're gonna have to solve for that, I do know that I want to use an equation of the form y equals a times b to the T, where T is going to be the number of minutes, and y is going to be the number of bacteria. And I know that my initial amount a is 350. So I can really write y equals 350 times b to the T. Now the doubling time tells me that when 15 minutes have elapsed, my population is going to be twice as big, or 700. plugging that into my equation, I have 700 equals 350 times b to the 15. Now I need to solve for b. Let me clean this up a little bit by dividing both sides by 350. That gives me 700 over 350 equals b to the 15th. In other words, two equals b to the 15th. To solve for B, I don't have to actually use logs here, because my variables in the base not in the exponent, so I don't need to bring that exponent down. Instead, the easiest way to solve this is just by taking the 15th root of both sides or equivalently. The 1/15 power. That's because if I take B to the 15th to the 1/15, I multiply by exponents, that gives me B to the one is equal to two to the 1/15. In other words, B is two to the 1/15, which as a decimal is approximately 1.047294. I like to use a lot of decimals if I'm doing a decimal approximation and these kind of problems to increase accuracy. But of course, the most accurate thing is just to leave B as it is. And I'll do that and rewrite my equation is y equals 350 times two to the 1/15 to the T. Now I'd like to work this problem one more time. And this time, I'm going to use the form of the equation y equals a times e to the RT. This is called a continuous growth model. It looks different, but it's actually an equivalent form to this growth model over here. And I'll say more about why these two forms are equivalent at the end, I can use the same general ideas to solve in this form. So I know that my initial amount is 350. And I know again that when t is 15, my Y is 700. So I plug in 700 here 350 e to the r times 15. And I solve for r. Again, I'm going to simplify things by dividing both sides by 350. That gives me two equals e to the r times 15. This time my variable does end up being in the exponent. So I do want to take the log of both sides, I'm going to use natural log, because I already have an E and my problem. So natural log and E are kind of more harmonious the other than then common log with base 10 and E. So I take the log of both sides. Now I can pull the exponent down. So that's our times 15 times ln of E. Well Elena V is just one, right? Because Elena V is asking what power do I raise E to to get e that's just one. And so I get 15 r equals ln two. So r is equal to ln two divided by 15. Let me plug that back in to my original equation as e to the ln two over 15 times t. Now I claimed that these two equations were actually the same thing just looking different. And the way to see that is if I start with this equation, and I rewrite as e to the ln two over 15 like that to the tee. Well, I claim that this quantity right here is the same thing as my two to the 1/15. And in fact One way to see that is e to the ln two already to the 115. The Right, that's the same, because every time I take a power to a power, I multiply exponents. But what's Ed Oh, and two, E and ln undo each other, so that's just 350 times two to the 1/15 to the T TA, the equations are really the same. And these, you'll always be able to find two different versions of an exponential equation, sort of the standard one, a times b to the T, or the continuous growth one, a times e to the RT. In this last example, we're going to work with half life, half life is pretty much like doubling time, it just means the amount of time that it takes for a quantity to decrease to half as much as we originally started with, we're told that the half life of radioactive carbon 14 is 5750 years. So that means it takes that long for a quantity of radioactive carbon 14 to decay, so that you just have half as much left and the rest is nonradioactive form. So we're told a sample of bone that originally contained 200 grams of radioactive carbon 14 now contains only 40 grams, we're supposed to find out how old the sample is, is called carbon dating. Let's use the continuous growth model this time. So our final amount, so this is our amount of radioactive c 14 is going to be the initial amount times e to the RT, we could have used the other model to we could have used f of t equals a times b to the T, but I just want to use the continuous model for practice. So we know that our half life is 5750. So what that means is when t is 5750, our amount is going to be one half of what we started with. Let me see if I can plug that into my equation and figure out use that to figure out what r is. That's ours called the continuous growth rate. So I plug in one half a for the final amount, a is still the initial amount, e to the r and I have 5750 I can cancel my A's. And now I want to solve for r, r is in my exponent. So I do need to take the log of both sides to solve for it. I'm going to use log base e since I already have an E and my problem, log base e is more compatible with E then log base 10 is okay. Now, on the left side, I still have log of one half and natural log of one half on the right side, ln n e to a power those undo each other. So I'm left with R times 5750. Now I can solve for r, it's ln one half over 5750 could work that as a decimal, but it's actually more accurate just to keep it in exact form. So now I can rewrite my equation, I have f of t equals a times e to the ln one half over five 750 t. Now I can use that to figure out my problem. And my problem the bone originally contained 200 grams, that's my a, I want to figure out when it's going to contain only 40 grams, that's my final amount. And so I need to solve for t. I'll clean things up by dividing both sides by 200. Let's say 40 over 200 is 1/5. Now I'm going to take the ln of both sides. And ln and e to the power undo each other. So I'm left with Elena 1/5 equals ln of one half divided by five 750 T. And finally I can solve for t but super messy but careful use of my calculator gives me an answer of 13,351 years approximately. That kind of makes sense in terms of the half life because to get from 200 to 40 you have to decrease by half a little more than two times right or increasing by half once would get you to 100 decreasing to half again would get you to 50 a little more than 40 and two half lives is getting pretty close to 13,000 years. This video introduced a lot of new things it introduced continue Less growth model, which is another equivalent way of writing an exponential function. The relationship is that the B in this example, is the same thing as EDR. In that version, it also introduced the ideas of doubling time. And HalfLife the amount of time it takes a quantity to double or decreased to half in an exponential growth model. Recall that a linear equation is equation like for example, 2x minus y equals one. It's an equation without any x squared or y squared in it, something that could be rewritten in the form y equals mx plus b, the equation for a line a system of linear equations is a collection of two or more linear equations. For example, I could have these two equations. A solution to a system of equations is that an x value and a y value that satisfy both of the equations. For example, the ordered pair two three, that means x equals two, y equals three is a solution to this system. Because if I plug in x equals two, and y equals three into the first equation, it checks out since two times two minus three is equal to one. And if I plug in x equals two and y equals three into the second equation, it also checks out two plus three equals five. However, the ordered pair one for that is x equals one, y equals four is not a solution to the system. Even though this X and Y value work in the second equation, since one plus four does equal five, it doesn't work in the first equation, because two times one minus four is not equal to one. In this video, we'll use systematic methods to find the solutions to systems of linear equations. In this first example, we want to solve this system of equations, there are two main methods we could use, we could use the method of substitution, or we could use the method of elimination. If we use the method of substitution, the main idea is to isolate one variable in one equation, and then substitute it in to the other equation. For example, we can start with the first equation 3x minus two y equals four, and isolate the x by adding two y to both sides, and then dividing both sides by three. Think I'll rewrite that a little bit by breaking up the fraction into two fractions four thirds plus two thirds y. Now, I'm going to copy down the second equation 5x plus six y equals two. And I'm going to substitute in my expression for x. That gives me five times four thirds plus two thirds y plus six y equals two. And now I've got an equation with only one variable in it y. So I can solve for y as a number. First, I'm going to distribute the five so that gives me 20 thirds plus 10 thirds y plus six y equals two. And now I'm going to keep all my y terms, my terms with y's in them on the left side, but I'll move all my terms without y's in them to the right side. At this point, I could just add up all my fractions and solve for y. But since I don't really like working with fractions, I think I'll do the trick of clearing the denominators here. So I'm going to actually multiply both sides by my common denominator of three just to get rid of the denominators and not have to work with fractions. So let me write that down. Distributing the three, I get 10 y plus eight t and y equals six minus 20. Now add things together. So that's 28 y equals negative 14. So that means that y is going to be negative 14 over 28, which is negative one half. So I've solved for y. And now I can go back and plug y into either of my equations to solve for x, I plug it into my first equation. So I'm plugging in negative one half for y. That gives me 3x plus one equals four. So 3x equals three, which means that x is equal to one. I've solved my system of equations and gotten x equals one On y equals minus one half, I can also write that as an ordered pair, one, negative one half for my solution. Now let's go back and solve the same system, but use a different method, the method of elimination, the key idea to the method of elimination is to multiply each equation by a constant to make the coefficients of one variable match. Let me start by copying down my two equations. Say I'm trying to make the coefficient of x match. One way to do that is to multiply the first equation by five, and the second equation by three. That way, the coefficient of x will be 15 for both equations, so let me do that. So for the first equation, I'm going to multiply both sides by five. And for the second equation, I'm going to multiply both sides by three. That gives me for the first equation 15x minus 10, y equals 20. And for the second equation, 15x plus 18, y is equal to six. Notice that the equate the coefficients of x match. So if I subtract the second equation from the first, the x term will completely go away, it'll be zero times x, and I'll be left with let's see negative 10 y minus 18, y is going to give me minus 28 y. And if I do 20 minus six, that's going to give me 14. solving for y, I get y is 14 over minus 28, which is minus one half just like before. Now we can continue, like we did in the previous solution, and substitute that value of y into either one of the equations. I'll put it again in here. And my solution proceeds as before. So once again, I get the solution that x equals one and y equals minus one half. Before we go on to the next problem, let me show you graphically what this means. Here I've graphed the equations 3x minus two, y equals four, and 5x plus six y equals two. And we can see that these two lines intersect in the point with coordinates one negative one half, just like we predicted by solving equations algebraically. Let's take a look at another system of equations. I'm going to rewrite the first equation, so the x term is on the left side with the y term, and the constant term stays on the right. And I'll rewrite or copy down the second equation. Since the coefficient of x in the first equation is minus four, and the second equation is three, I'm going to try using the method of elimination and multiply the first equation by three and the second equation by four. That'll give me a coefficient of x of negative 12. And the first equation, and 12. And the second equation, those are equal and opposite, right, so I'll be able to add together my equations to cancel out access. So let's do that. My first equation becomes negative 12x plus 24, y equals three, and I'll put everything by three. And my second equation, I'll multiply everything by four. So that's 12x minus 24, y equals eight. Now something kind of funny has happened here, not only do the x coefficients match has in with opposite signs, but the Y coefficients do also. So if I add together my two equations, in order to cancel out the x term, I'm also going to cancel out the y term, and I'll just get zero plus zero is equal to three plus eight is 11. Well, that's a contradiction, we can't have zero equal to 11. And that shows that these two equations actually have no solution. Let's look at this situation graphically. If we graph our two equations, we see that they're parallel lines with the same slope. This might be more clear, if I isolate y in each equation, the first equation, I get y equals, let's see, dividing by eight that's the same thing as four eighths or one half x plus one eight. And the second equation, if I isolate y, let's say minus six y equals minus 3x plus two divided by minus six, that's y equals one half x minus 1/3. So indeed, they have the same slope. And so they're parallel with different In intercepts, and so they can have no intersection. And so it makes sense that we have no solution to our system of linear equations. This kind of system that has no solution is called an inconsistent system. In this third example, yet a third behavior happens. This time, I think I'm going to solve by substitution because I already have X with a coefficient of one. So it's really easy to just isolate X in the first equation, and then plug in to the second equation to get three times six minus five y plus 15, y equals 18. If I distribute out, I get the strange phenomenon that the 15 y's cancel and I just get 18 equals 18, which is always true. This is what's called a dependent system of linear equations. If you look more closely, you can see that the second equation is really just a constant multiple, the first equation is just three times every term is three times as big as the corresponding term and the first equation. So there's no new information in the second equation, anything, any x and y values that satisfy the first one will satisfy the second one. So this system of equations has infinitely many solutions. Any ordered pair x y, where X plus five y equals six, or in other words, X's minus five y plus six will satisfy this system of equations. That would include a y value of zero corresponding x value of six or a y value of one corresponding to an x value of one, or a y value of 1/3. Corresponding to an x value of 13 thirds just by plugging into this equation will work. Graphically, if I graph both of these equations, the lines will just be on top of each other, so I'll just see one line. In this video, we've solved systems of linear equations, using the method of substitution and the method of elimination. We've seen that systems of linear equations can have one solution. When the lines that the equations represent intersect in one point, they can be inconsistent, and have no solutions that corresponds to parallel lines, or they can be dependent and have infinitely many solutions that corresponds to the lines lying on top of each other. In this video, I'll work through a problem involving distance rate and time. The key relationship to keep in mind is that the rate of travel is the distance traveled divided by the time it takes to travel again. For example, if you're driving 60 miles an hour, that's your rate. And that's because you're going a distance of 60 miles in one hour. Sometimes it's handy to rewrite that relationship by multiplying both sides by T time. And that gives us that R times T is equal to D. In other words, distance is equal to rate times time. There's one more important principle to keep in mind. And that's the idea that rates add. For example, if you normally walk at three miles per hour, but you're walking on a moving sidewalk, that's going at a rate of two miles per hour, then your total speed of travel with respect to you know something stationary is going to be three plus two, or five miles per hour. All right, that is a formula as our one or the first rate per second rate is equal to the total rate. Let's use those two key ideas. distance equals rate times time, and rates add in the following problem. else's boat has a top speed of six miles per hour and still water. While traveling on a river at top speed. She went 10 miles upstream in the same amount of time she went 30 miles downstream, we're supposed to find the rate of the river currents. I'm going to organize the information in this problem into a chart. During the course of Elsa stay, there were two situations we need to keep in mind. For one period of time she was going upstream. And for another period of time she was going downstream. For each of those, I'm going to chart out the distance you traveled. The rate she went at and the time it took when she was going upstream. She went a total distance of 10 miles. When she was going downstream she went a longer distance of 30 miles. But the times to travel those two distances were the same. Since I don't know what that time was, I'll just give it a variable I'll call it T now Think about her rate of travel, the rate she traveled upstream was slower because she was going against the current and faster when she was going downstream with the current. We don't know what the speed of the current is, that's what we're trying to figure out. So maybe I'll give it a variable R. But we do know that in still water also can go six miles per hour. And when she's going downstream, since she's going with the direction of the current rates should add, and her rate downstream should be six plus R, that's her rate, and still water plus the rate of the current. On the other hand, when she's going upstream, then she's going against the current, so her rate of six miles per hour, we need to subtract the rate of the current from that. Now that we've charted out our information, we can turn it into equations using the fact that distance equals rate times time, we actually have two equations 10 equals six minus R times T, and 30 is equal to six plus R times T. Now that we've converted our situation into a system of equations, our next job is to solve the system of equations. In this example, I think the easiest way to proceed is to isolate t in each of the two equations. So in the first equation, I'll divide both sides by six minus r, and a second equation R divided by six plus R. That gives me 10 over six minus r equals T, and 30 over six plus r equals t. Now if I set my T variables equal to each other, I get, I get 10 over six minus r is equal to 30 over six plus R. I'm making progress because now I have a single equation, the single variable that I need to solve. Since the variable R is trapped in the denominator, I'm going to proceed by clearing the denominator. So I'm going to multiply both sides by the least common denominator, that is six minus r times six plus R. Once I cancel things out, I get that the six plus r times 10 is equal to 30 times six minus r, if I distribute, I'm going to get 60 plus xR equals 180 minus 30 ar, which I can now solve, let's see, that's going to be 40 r is equal to 120. So our, the speed of my current is going to be three miles per hour. This is all that the problem asked for the speed of the current. If I also wanted to solve for the other unknown time, I could do so by plugging in R into one of my equations and solving for T. In this video, I saw the distance rate and time problem by charting out my information for the two situations and my problem using the fact that rates add to fill in some of my boxes, and then using the formula distance equals rate times time to build a system of equations. In this video, I'll do a standard mixture problem in which we have to figure out what quantity of two solutions to mix together. household bleach contains 6% sodium hypochlorite. The other 94% is water. How much household bleach should be combined with 70 liters of a weaker 1% hypochlorite solution in order to form a solution, that's 2.5% sodium hypochlorite. I want to turn this problem into a system of equations. So I'm asking myself what quantities are going to be equal to each other? Well, the total amount of sodium hypochlorite that has symbol and a co o before mixing, it should equal the total amount of sodium hypochlorite after mixing. Also, the total amount of water before mixing should equal the total amount of water after. And finally, there's just the total amount of solution. In my two jugs, sodium hypochlorite together with water should equal the total amount of solution after that gives me a hint for what I'm looking for. But before I start reading out equations, I find it very helpful to chart out my quantities. So I've got the 6% solution. The household leech, I've got the 1% solution. And I've got my Desired Ending 2.5% solution. Now on each of those solutions, I've got a certain volume of sodium hypochlorite. I've also got a volume of water. And I've got a total volume of solution. Let me see which of these boxes I can actually fill in, I know that I'm adding 70 liters of the 1% solution. So I can put a 70 in the total volume of solution here. I don't know what volume of the household bleach, I want to add, that's what I'm trying to find out. So I'm going to just call that volume x. Now since my 2.5% solution is made by combining my other two solutions, I know its volume is going to be the sum of these two volumes, so I'll write 70 plus x in this box. Now, the 6% solution means that whatever the volume of solution is, 6% of that is the sodium hypochlorite. So the volume of the sodium hypochlorite is going to be 0.06 times x, the volume of water and that solution is whatever's left, so that's going to be x minus 0.06x, or point nine four times X with the following the same reasoning for the 1% solution 1% of the 70 liters is a sodium hypochlorite. So that's going to be 0.01 times 70. Or point seven, the volume of water and that solution is going to be 99% or point nine, nine times seven day. That works out to 69.3. Finally, for the 2.5% solution, the volume of the sodium hypochlorite is going to be 0.0 to five times the volume of solution 70 plus x and the volume of water is going to be the remainder. So that's 0.975 times 70 plus x. Now I've already used the fact that the volume of solutions before added up is the volume of solution after in writing a 70 plus x in this box. But I haven't yet used the fact that the volume of the sodium hypochlorite is preserved before and after. So I can write that down as an equation. So that means 0.06x plus 0.7 is equal to 0.025 times 70 plus x. Now I've got an equation with a variable. I'll try to solve it. Since I don't like all these decimals, I'm going to multiply both sides of my equation by let's see, 1000 should get rid of all the decimals. After distributing, I get 60x plus 700 equals 25 times 70 plus x. Distributing some more, I get 60x plus 700 is equal to 1750 plus 25x. So let's see 60 minus 25 is 35x is equal to 1050. Which gives me that x equals 30 liters of the household bleach. Notice that I never actually had to use the fact that the water quantity of water before mixing is equal to the quantity of water after that is I never use the information in this column. In fact, that information is redundant. Once I know that the quantities of sodium hypochlorite add up, and the total volume of solutions add up. The fact that that volume of waters add up is just redundant information. The techniques to use to solve this equation involving solutions can be used to solve many many other equations involving mixtures of items. My favorite method is to first make a chart involving the types of mixtures and the types of items in your mixture. Fill in as many boxes size I can and then use the fact that the quantities add. This video is about rational functions and their graphs. Recall that a rational function is a function that can be written as a ratio or quotient of two power. No Here's an example. The simpler function, f of x equals one over x is also considered a rational function, you can think of one and x as very simple polynomials. The graph of this rational function is shown here. This graph looks different from the graph of a polynomial. For one thing, its end behavior is different. The end behavior of a function is the way the graph looks when x goes through really large positive, or really large negative numbers, we've seen that the end behavior of a polynomial always looks like one of these cases. That is why marches off to infinity or maybe negative infinity, as x gets really big or really negative. But this rational function has a different type of end behavior. Notice, as x gets really big, the y values are leveling off at about a y value of three. And similarly, as x values get really negative, our graph is leveling off near the line y equals three, I'll draw that line, y equals three on my graph, that line is called a horizontal asymptote. A horizontal asymptote is a horizontal line that our graph gets closer and closer to as x goes to infinity, or as X goes to negative infinity, or both. There's something else that's different about this graph from a polynomial graph, look at what happens as x gets close to negative five. As we approach negative five with x values on the right, our Y values are going down towards negative infinity. And as we approach the x value of negative five from the left, our Y values are going up towards positive infinity. We say that this graph has a vertical asymptote at x equals negative five. A vertical asymptote is a vertical line that the graph gets closer and closer to. Finally, there's something really weird going on at x equals two, there's a little open circle there, like the value at x equals two is dug out. That's called a hole. A hole is a place along the curve of the graph where the function doesn't exist. Now that we've identified some of the features of our rational functions graph, I want to look back at the equation and see how we could have predicted those features just by looking at the equation. To find horizontal asymptotes, we need to look at what our function is doing when x goes through really big positive or really big negative numbers. Looking at our equation for our function, the numerator is going to be dominated by the 3x squared term when x is really big, right, because three times x squared is going to be absolutely enormous compared to this negative 12. If x is a big positive or negative number, in the denominator, the denominator will be dominated by the x squared term. Again, if x is a really big positive or negative number, like a million, a million squared will be much, much bigger than three times a million or negative 10. For that reason, to find the end behavior, or the horizontal asymptote, for our function, we just need to look at the term on the numerator and the term on the denominator that have the highest exponent, those are the ones that dominate the expression in size. So as x gets really big, our functions y values are going to be approximately 3x squared over x squared, which is three. That's why we have a horizontal asymptote at y equals three. Now our vertical asymptotes, those tend to occur where our denominator of our function is zero. That's because the function doesn't exist when our denominator is zero. And when we get close to that place where our denominator is zero, we're going to be dividing by tiny, tiny numbers, which will make our Y values really big in magnitude. So to check where our denominators zero, let's factor our function. In fact, I'm going to go ahead and factor the numerator and the denominator. So the numerator factors, let's see, pull out the three, I get x squared minus four, factor in the denominator, that factors into X plus five times x minus two, I can factor a little the numerator a little further, that's three times x minus two times x plus two over x plus 5x minus two. Now, when x is equal to negative five, my denominator will be zero, but my numerator will not be zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when x equals two, the denominator is zero, but the numerator is also zero. In fact, if I cancelled the x minus two factor from the numerator, and in nominator, I get a simplified form for my function that agrees with my original function as long as x is not equal to two. That's because when x equals two, the simplified function exists, but the original function does not, it's zero over zero, it's undefined. But for every other x value, including x values near x equals to our original functions just the same as this function. And that's why our function only has a vertical asymptote at x equals negative five, not one at x equals two, because the x minus two factor is no longer in the function after simplifying, it does have a hole at x equals two, because the original function is not defined there, even though the simplified version is if we want to find the y value of our hole, we can just plug in x equals two into our simplified version of our function, that gives a y value of three times two plus two over two plus seven, or 12 ninths, which simplifies to four thirds. So our whole is that to four thirds. Now that we've been through one example in detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking where the denominator is zero. The holes happen where the denominator and numerator are both zero and those factors cancel out. The vertical asymptotes are all other x values where the denominator is zero, we find the horizontal asymptotes. By considering the highest power term on the numerator and the denominator, I'll explain this process in more detail in three examples. In the first example, if we circle the highest power terms, that simplifies to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going to be going very close to zero. And therefore we have a horizontal asymptote at y equals zero. In the second example, the highest power terms, 2x cubed, over 3x cubed simplifies to two thirds. So as x gets really big, we're going to be heading towards two thirds, and we have a horizontal asymptote at y equals two thirds. In the third example, the highest power terms, x squared over 2x simplifies to x over two. As x gets really big, x over two is getting really big. And therefore, we don't have a horizontal asymptote at all. This is going to infinity, when x gets through go through big positive numbers, and is going to negative infinity when x goes through a big negative numbers. So in this case, the end behavior is kind of like that of a polynomial, and there's no horizontal asymptote. In general, when the degree of the numerator is smaller than the degree of the denominator, we're in this first case where the denominator gets really big compared to the numerator and we go to zero. In the second case, where the degree of the numerator and the degree of the dominant are equal, things cancel out, and so we get a horizontal asymptote at the y value, that's equal to the ratio of the leading coefficients. Finally, in the third case, when the degree of the numerator is bigger than the degree of the denominator, then the numerator is getting really big compared to the denominator, so we end up with no horizontal asymptote. Final Finally, let's apply all these observations to one more example. Please pause the video and take a moment to find the vertical asymptotes, horizontal asymptotes and holes for this rational function. To find the vertical asymptotes and holes, we need to look at where the denominator is zero. In fact, it's going to be handy to factor both the numerator and the denominator. Since there if there are any common factors, we might have a whole instead of a vertical asymptote. The numerator is pretty easy to factor. Let's see that's 3x times x plus one for the denominator or first factor out an x. And then I'll factor some more using a guess and check method. I know that I'll need a 2x and an X to multiply together to the 2x squared and I'll need a three and a minus one or alpha minus three and a one. Let's see if that works. If I multiply out 2x minus one times x plus three that does get me back to x squared plus 5x minus three, so that checks out. Now I noticed that I have a common factor of x in both the numerator and the denominator. So that's telling me I'm going to have a hole at x equals zero. In fact, I could rewrite my rational function by cancelling out that common factor. And that's equivalent, as long as x is not equal to zero. So the y value of my whole is what I get when I plug zero into my simplified version, that would be three times zero plus one over two times zero minus one times zero plus three, which is three over negative three or minus one. So my whole is at zero minus one. Now all the remaining places in my denominator that make my denominator zero will get me vertical asymptotes. So I'll have a vertical asymptote, when 2x minus one times x plus three equals zero, that is, when 2x minus one is zero, or x plus three is zero. In other words, when x is one half, or x equals negative three. Finally, to find my horizontal asymptotes, I just need to consider the highest power term in the numerator and the denominator. That simplifies to three over 2x, which is bottom heavy, right? When x gets really big, this expression is going to zero. And that means that we have a horizontal asymptote at y equals zero. So we found the major features of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this would give us a framework for what the graph of our function looks like. horizontal asymptote at y equals zero, vertical asymptotes at x equals one half, and x equals minus three at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator of graphing program, we can see that our actual function will look something like this. Notice that the x intercept when x is negative one, corresponds to where the numerator of our rational function or reduced rational function is equal to zero. That's because a zero on the numerator that doesn't make the denominator zero makes the whole function zero. And an X intercept is where the y value of the whole function is zero. In this video, we learned how to find horizontal asymptotes have rational functions. By looking at the highest power terms, we learned to find the vertical asymptotes and holes. By looking at the factored version of the functions. The holes correspond to the x values that make the numerator and denominator zero, his corresponding factors cancel. The vertical asymptotes correspond to the x values that make the denominator zero, even after factoring any any common and in common factors in the numerator denominator. This video is about combining functions by adding them subtracting them multiplying and dividing them. Suppose we have two functions, f of x equals x plus one and g of x equals x squared. One way to combine them is by adding them together. This notation, f plus g of x means the function defined by taking f of x and adding it to g of x. So for our functions, that means we take x plus one and add x squared, I can rearrange that as the function x squared plus x plus one. So f plus g evaluated on x means x squared plus x plus one. And if I wanted to evaluate f plus g, on the number two, that would be two squared plus two plus one, or seven. Similarly, the notation f minus g of x means the function we get by taking f of x and subtracting g of x. So that would be x plus one minus x squared. And if I wanted to take f minus g evaluated at one, that would be one plus one minus one squared, or one, the notation F dot g of x, which is sometimes also written just as f g of x. That means we take f of x times g of x. In other words, x plus one times x squared, which could be simplified as x cubed plus x squared. The notation f divided by g of x means I take f of x and divided by g of x. So that would be x plus one divided by x squared. In this figure, the blue graph represents h of x. And the red graph represents the function p of x, we're asked to find h minus p of zero. We don't have any equations to work with, but that's okay. We know that for any x, h minus p of x is defined as h of x minus p of x. So for x equals zero, H minus p of zero is going to be h of zero minus p of zero. Using the graph, we can find h of zero by finding the value of zero on the x axis, and finding the corresponding y value for the function h of x. So that's about 1.8. Now P of zero, we can find similarly by looking for zero on the x axis, and finding the corresponding y value for the function p of x, and that's a y value of one 1.8 minus one is 0.8. So that's our approximate value for H minus p of zero. If we want to find P times h of negative three, again, we can rewrite that as P of negative three times h of negative three. And using the graphs, we see that for an x value of negative three, the y value for P is two. And the x value of negative three corresponds to a y value of negative two for H. Two times negative two is negative four. So that's our value for P times h of negative three. In this video, we saw how to add two functions, subtract two functions, multiply two functions, and divide two functions in the following way. When you compose two functions, you apply the first function, and then you apply the second function to the output of the first function. For example, the first function might compute population size from time in years. So its input would be time in years, since a certain date, as output would be number of people in the population. The second function g, might compute health care costs as a function of population size. So it will take population size as input, and its output will be healthcare costs. If you put these functions together, that is compose them, then you'll go all the way from time in years to healthcare costs. This is your composition, g composed with F. The composition of two functions, written g with a little circle, f of x is defined as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically and diagram f x on a number x and produces a number f of x, then g takes that output f of x and produces a new number, g of f of x. Our composition of functions g composed with F is the function that goes all the way from X to g of f of x. Let's work out some examples where our functions are defined by tables of values. If we want to find g composed with F of four, by definition, this means g of f of four. To evaluate this expression, we always work from the inside out. So we start with the x value of four, and we find f of four, using the table of values for f of x, when x equals four, f of x is seven, so we can replace F of four with the number seven. Now we need to evaluate g of seven, seven becomes our new x value in our table of values for G, the x value of seven corresponds to the G of X value of 10. So g of seven is equal to 10. We found that g composed with F of four is equal to 10. If instead we want to find f composed with g of four, well, we can rewrite that as f of g of four Again work from the inside out. Now we're trying to find g of four. So four is our x value. And we use our table of values for G to see that g of four is one. So we replaced you a four by one. And now we need to evaluate f of one. Using our table for F values, f of one is eight. Notice that when we've computed g of f of four, we got a different answer than when we computed f of g of four. And in general, g composed with F is not the same thing as f composed with g. Please pause the video and take a moment to compute the next two examples. We can replace f composed with F of two by the equivalent expression, f of f of two. Working from the inside out, we know that f of two is three, and f of three is six. If we want to find f composed with g of six, rewrite that as f of g of six, using the table for g, g of six is eight. But F of eight, eight is not on the table as an x value for the for the f function. And so there is no F of eight, this does not exist, we can say that six is not in the domain, for F composed with g. Even though it was in the domain of g, we couldn't follow all the way through and get a value for F composed with g of six. Next, let's turn our attention to the composition of functions that are given by equations. p of x is x squared plus x and q of x as negative 2x. We want to find q composed with P of one. As usual, I can rewrite this as Q of P of one and work from the inside out. P of one is one squared plus one, so that's two. So this is the same thing as Q of two. But Q of two is negative two times two or negative four. So this evaluates to negative four. In this next example, we want to find q composed with P of some arbitrary x, or rewrite it as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula for that. That's x squared plus x. So I can replace my P of x with that expression. Now, I'm stuck with evaluating q on x squared plus x. Well, Q of anything is negative two times that thing. So q of x squared plus x is going to be negative two times the quantity x squared plus x, what I've done is I've substituted in the whole expression x squared plus x, where I saw the X in this formula for q of x, it's important to use the parentheses here. So that will be multiplying negative two by the whole expression and not just by the first piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my expression for Q composed with p of x. Notice that if I wanted to compute q composed with P of one, which I already did in the first problem, I could just use this expression now, negative two times one squared minus two and I get negative four, just like I did before. Let's try another one. Let's try p composed with q of x. First I read write this P of q of x. Working from the inside out, I can replace q of x with negative 2x. So I need to compute P of negative 2x. Here's my formula for P. to compute P of this expression, I need to plug in this expression everywhere I see an x in the formula for P. So that means negative 2x squared plus negative 2x. Again, being careful to use parentheses to make sure I plug in the entire expression in forex. let me simplify. This is 4x squared minus 2x. Notice that I got different expressions for Q of p of x, and for P of q of x. Once again, we see that q composed with P is not necessarily equal to P composed with Q. Please pause the video and try this last example yourself. rewriting. And working from the inside out, we're going to replace p of x with its expression x squared plus x. And then we need to evaluate p on x squared plus x. That means we plug in x squared plus x, everywhere we see an x in this formula, so that's x squared plus x quantity squared plus x squared plus x. Once again, I can simplify by distributing out, that gives me x to the fourth plus 2x cubed plus x squared plus x squared plus x, or x to the fourth plus 2x cubed plus 2x squared plus x. In this last set of examples, we're asked to go backwards, we're given a formula for a function of h of x. But we're supposed to rewrite h of x as a composition of two functions, F and G. Let's think for a minute, which of these two functions gets applied first, f composed with g of x, let's see, that means f of g of x. And since we evaluate these expressions from the inside out, we must be applying g first, and then F. In order to figure out what what f and g could be, I like to draw a box around some thing inside my expression for H, so I'm going to draw a box around x squared plus seven, then whatever's inside the box, that'll be my function, g of x, the first function that gets applied, whatever happens to the box, in this case, taking the square root sign, that becomes my outside function, my second function f. So here, we're gonna say g of x is equal to x squared plus seven, and f of x is equal to the square root of x, let's just check and make sure that this works. So I need to check that when I take the composition, f composed with g, I need to get the same thing as my original h. So let's see, if I do f composed with g of x, well, by definition, that's f of g of x, working from the inside out, I can replace g of x with its formula x squared plus seven. So I need to evaluate f of x squared plus seven. That means I plug in x squared plus seven, into the formula for for F. So that becomes the square root of x squared plus seven to the it works because it matches my original equation. So we found a correct answer a correct way of breaking h down as a composition of two functions. But I do want to point out, this is not the only correct answer. I'll write down my formula for H of X again, and this time, I'll put the box in a different place, I'll just box the x squared. If I did that, then my inside function of my first function, g of x would be x squared. And my second function is what happens to the box. So my f of x is what happens to the box, and the box gets added seven to it, and taking the square root. So in other words, f of x is going to be the square root of x plus seven. Again, I can check that this works. If I do f composed with g of x, that's f of g of x. So now g of x is x squared. So I'm taking f of x squared. When I plug in x squared for x, I do in fact, get the square root of x squared plus seven. So this is that alternative, correct solution. In this video, we learn to evaluate the composition of functions. by rewriting it and working from the inside out. We also learn to break apart a complicated function into a composition of two functions by boxing one piece of the function and letting the first function applied in the composition. Let that be the inside of the box, and the second function applied in the composition be whatever happens to the box. The inverse of a function undoes what the function does. So the inverse of tying your shoes would be to untie them. And the inverse of the function that adds two to a number would be the function that subtracts two from a number. This video introduces inverses and their properties. Suppose f of x is a function defined by this chart. In other words, Have two is three, f of three is five, f of four is six, and f of five is one, the inverse function for F written f superscript. Negative 1x undoes what f does. Since f takes two to three, F inverse takes three, back to two. So we write this f superscript, negative one of three is to. Similarly, since f takes three to five, F inverse takes five to three. And since f takes four to six, f inverse of six is four. And since f takes five to one, f inverse of one is five. I'll use these numbers to fill in the chart. Notice that the chart of values when y equals f of x and the chart of values when y equals f inverse of x are closely related. They share the same numbers, but the x values for f of x correspond to the y values for f inverse of x, and the y values for f of x correspond to the x values for f inverse of x. That leads us to the first key fact inverse functions reverse the roles of y and x. I'm going to plot the points for y equals f of x in blue. Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a moment and see what kind of symmetry you observe in this graph. How are the blue points related to the red points, you might have noticed that the blue points and the red points are mirror images over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse of x can be obtained from the graph of y equals f of x by reflecting over the line y equals x. This makes sense, because inverses, reverse the roles of war annex. In the same example, let's compute f inverse of f of two, this open circle means composition. In other words, we're computing f inverse of f of two, we compute this from the inside out. So that's f inverse of three. Since F of two is three, and f inverse of three, we see as to similarly, we can compute f of f inverse of three. And that means we take f of f inverse of three. Since f inverse of three is two, that's the same thing as computing F of two, which is three. Please pause the video for a moment and compute these other compositions. You should have found that in every case, if you take f inverse of f of a number, you get back to the very same number you started with. And similarly, if you take f of f inverse of any number, you get back to the same number you started with. So in general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x. This is the mathematical way of saying that F and n f inverse undo each other. Let's look at a different example. Suppose that f of x is x cubed. Pause the video for a moment, and guess what the inverse of f should be. Remember, F inverse undoes the work that F does. You might have guessed that f inverse of x is going to be the cube root function, we can check that this is true by looking at f of f inverse of x, that's F of the cube root of function, which means the cube root function cubed, which gets us back to x. Similarly, if we compute f inverse of f of x, that's the cube root of x cubed. And we get back to excellence again. So the cube root function really is the inverse of the cubing function. When we compose the two functions, we get back to the number that we started with. It'd be nice to have a more systematic way of finding inverses of functions besides guessing and checking. One method uses the fact that inverses reverse the roles of y and x. So if we want to find the inverse of the function, f of x equals five minus x over 3x. We can write it as y equals five minus x over 3x. Reverse the roles of y and x To get x equals five minus y over three y, and then solve for y. To solve for y, let's multiply both sides by three y. Bring all terms with y's in them to the left side, and alternate without wizened them to the right side, factor out the why. And divide to isolate why this gives us f inverse of x as five over 3x plus one. Notice that our original function f and our inverse function, f inverse are both rational functions, but they're not the reciprocals of each other. And in general, f inverse of x is not usually equal to one over f of x. This can be confusing, because when we write two to the minus one, that does mean one of our two, but f to the minus one of x means the inverse function and not the reciprocal. It's natural to ask us all functions have inverse functions. That is for any function you might encounter. Is there always a function that its is its inverse? In fact, the answer is no. See, if you can come up with an example of a function that does not have an inverse function. The word function here is key. Remember that a function is a relationship between x values and y values, such that for each x value in the domain, there's only one corresponding y value. One example of a function that does not have an inverse function is the function f of x equals x squared. To see that, the inverse of this function is not a function. Note that for the x squared function, the number two and the number negative two, both go to number four. So if I had an inverse, you would have to send four to both two and negative two, the inverse would not be a function, it might be easier to understand the problem, when you look at a graph of y equals x squared. Recall that inverse functions reverse the roles of y and x and flip the graph over the line y equals x. But when I flipped the green graph over the line y equals x, I get this red graph. This red graph is not the graph of a function because it violates the vertical line test. The reason that violates the vertical line test is because the original green function violates the horizontal line test, and has 2x values with the same y value. In general, a function f has an inverse function if and only if the graph of f satisfies the horizontal line test, ie every horizontal line intersects the graph. In it most one point, pause the video for a moment and see which of these four graphs satisfy the horizontal line test. In other words, which of the four corresponding functions would have an inverse function? You may have found that graphs A and B violate the horizontal line test. So their functions would not have inverse functions. But graph C and D satisfy the horizontal line test. So these graphs represent functions that do have inverses. functions that satisfy the horizontal line test are sometimes called One to One functions. Equivalently a function is one to one, if for any two different x values, x one and x two, the y value is f of x one and f of x two are different numbers. Sometimes, as I said, f is one to one, if, whenever f of x one is equal to f of x two, then x one has to equal x two. As our last example, let's try to find P inverse of x, where p of x is the square root of x minus two drawn here. If we graph P inverse on the same axis as p of x, we get the following graph simply by flipping over the line y equals x. If we try to solve the problem algebraically we can write y equal to a squared of x minus two, reverse the roles of y and x and solve for y by squaring both sides adding two. Now if we were to graph y equals x squared plus two, that would look like a parabola, it would look like the red graph we've already drawn together with another arm on the left side. But we know that our actual inverse function consists only of this right arm, we can specify this algebraically by making the restriction that x has to be bigger than or equal to zero. This corresponds to the fact that on the original graph for the square root of x, y was only greater than or equal to zero. Looking more closely at the domain and range of P and P inverse, we know that the domain of P is all values of x such that x minus two is greater than or equal to zero. Since we can't take the square root of a negative number. This corresponds to x values being greater than or equal to two, or an interval notation, the interval from two to infinity. The range of P, we can see from the graph is all y value is greater than or equal to zero, or the interval from zero to infinity. Similarly, based on the graph, we see the domain of P inverse is x values greater than or equal to zero, the interval from zero to infinity. And the range of P inverse is Y values greater than or equal to two, or the interval from two to infinity. If you look closely at these domains and ranges, you'll notice that the domain of P corresponds exactly to the range of P inverse, and the range of P corresponds to the domain of P inverse. This makes sense, because inverse functions reverse the roles of y and x. The domain of f inverse of x is the x values for F inverse, which corresponds to the y values or the range of F. The range of f inverse is the y values for F inverse, which correspond to the x values or the domain of f. In this video, we discussed five key properties of inverse functions. inverse functions, reverse the roles of y and x. The graph of y equals f inverse of x is the graph of y equals f of x reflected over the line y equals x. When we compose F with F inverse, we get the identity function y equals x. And similarly, when we compose f inverse with F, that brings x to x. In other words, F and F inverse undo each other. The function f of x has an inverse function if and only if the graph of y equals f of x satisfies the horizontal line test. And finally, the domain of f is the range of f inverse and the range of f is the domain of f inverse. These properties of inverse functions will be important when we study exponential functions and their inverses logarithmic functions.