Transcript for:
Algebra 2 Key Concepts

in this video we're going to go over a few basic concepts in Algebra 2 so let's begin by solving linear equations so let's start with the basics so if you're given an equation like this 5x - 4 is equal to 11 how would you solve for x feel free to pause the video and work on this example so the first thing that we need to do is get rid of the ne4 the opposite of subtraction is addition so we should add four to both sides -4 + 4 adds up to 0 and 11 + 4 is 15 now the five is Multiplied to the X and to separate the five from the X we need to perform the opposite of multiplication which is division so we need to divide both sides by five in order to solve for x your goal is to get X on one side of the equation 15 / 5 is 3 so X is equal to 3 let's try another example but with variables on both sides let's say that 3x - 7 is equal to 9x plus let's say 17 go ahead and solve for x so now let's subtract both sides by 3x we want to move all the X variables on one side and all of the constants on the other side so I'm going to move the 3x uh to the right side I want to get rid of all of the X variables on the left side 9x - 3x is 6X so now that I only have the X variables on the right side I'm going to try to get rid of the constant on the right side so I'm going to subtract both sides by 17 so now the X variable is on the right the constants are on the left -7 - 17 is -4 now the last thing I need to do is separate the six from the X to do that I can uh divide both sides by six -4 / 6 is uh4 and so this is the answer X is equal to4 here's another one that you can try 5 - 2 * 3x + 1 let's say it's uh equal to 7 * 2x + 5 minus 8 go ahead and try that example now the first thing that we should do is distribute the -2 to uh 3x + 1 and also 7 to 2x + 5 -2 * 3x is equal to -6x X and -2 * 1 is equal to -2 now let's distribute the 7 7 * 2x is 14x and 7 * 5 is 35 now let's combine like terms on the left side we can add 5 and -2 5 + -2 which is the same as 5 - 2 is equal to 3 and on the right side 35 - 8 is 27 now what should we do next at this point the next thing we should do is combine the X variables let's add 6X to both sides so that uh these two will cancel and so we have 3 is equal to 14 + 6 that's 20 20x and + 27 so now we need to take the 27 and move it to this side and we can do that by subtracting both sides by 27 so let's go ahead and get rid of some things on top so these two will cancel 3 - 27 is equal to -4 so the last thing we need to do is uh divide both sides by 20 so X is = to -24 / 20 now we have a fraction but it looks like we can reduce it both numbers 20 and 24 are divisible by four so let's divide each number by four -4 / 4 is -6 20 ID 4 is 5 so the answer in reduced form is -6 over 5 so that's it consider this equation 2 3x - 5 is = 3 how can we solve for x if we have a fraction in the equation what would you do well the first thing that we can do is add five to both sides let's begin by combining like terms so 3 + 5 is simply equal to 8 now what can we do at this point to get rid of the fraction multiply both sides by the denominator of the fraction that is by three if we do so on the left side the threes will cancel 3 ID 3 is 1 so we'll be left over with 2x on the right side it's simply 8 * 3 which is 24 now the last thing that we need to do is divide by two 20 I mean 24 / 2 is 12 so therefore X is equal to 12 and that's it that's all we got to do for that problem let's try another example with fractions so let's say that 12x - 1/3 is equal to 4 what can we do in this example so take a minute and try this one if you have multiple fractions it might be convenient to multiply both sides by the common denominator what if is the common denominator of 2 and three or what is the least common multiple of 2 and three if you're not sure multiply 2 and three 2 * 3 is 6 so let's go of six so everything in this equation we need to multiply by six so what is 6 * a half 6 * a half is the same as 6 / 2 which is three so this is going to be 3x now if we multiply 6 by 1/3 6 ID 3 is 2 so 6 * 1/3 is -2 and finally 6 * 4 is equal to 24 so now this equation is in a form that we can manage so let's add two to both sides 24 + 2 is 26 now the last thing that we need to do is divide by three and that's it we can't really reduce 26 over 3 so we'll leave the answer in this form X is equal to 26 over 3 let's leave it as an improper fraction let's try another example 5x + 2 / 3 is equal to 7 over4 now if you have two fractions separated by an equal sign what can we do to solve for x in a situation like this you can cross multiply 3 * 7 is 21 and we need multiply 4 by 5x + 2 so we need to distribute the 4 to the 5x + 2 4 * 5x is 20x 4 * 2 is 8 now our next step is to subtract both sides by 8 21 - 8 is 13 now the last thing that we need to do is divide by 20 and as we can see X is 13 over 20 and that's the answer now let's go over inequalities and how to graph them on a number line so how would you graph this particular inequality if x is greater than three how would you graph it on a number line so here's zero here's 1 2 3 so if x is greater than three but not equal to three you need to use an open circle if it's greater than and equal to three you need to use a closed Circle greater than three we need to shade towards the right the numbers to the right are larger than three now what about if x is greater than or equal to -2 so here's Z -2 is on the left side of zero this time we need a closed Circle and because X is greater than or equal to it we need to shade towards the right if it was less than we would have to shade towards the left now what if x is less than1 but not equal to it so here's NE 1 relative to zero so we need an open circle if you don't see an underline and because it's less than we're going to shade towards the left and that's how you graph it and let's try one more example what if x is less than or equal to five so five is to the right of zero and because we see the underlined symbol we're going to use a closed Circle as opposed to an open circle and so we're going to shade this way by the way how can we represent the answer using interval notation let's talk about that all the way to the left we have the symbol negative Infinity to the right Infinity so in interval notation you can represent the Shad of region as being from negative Infinity to negative 1 it starts from the left which is here and it stops at negative 1 so you view it from left to right whenever you have an open circle use the parenthesis and for Infinity you should always use parenthesis for closed Circle you can use the bracket so for the next one it starts at negative Infinity but it stops at five so it's negative Infinity to five But it includes five so going to close it with a bracket let's try another problem let's say that 3x - 2 is greater than 10 so solve for x and graph the solution on a number line and also represent the solution using interval notation so let's begin by adding two to both sides 10 + 2 is 12 now our next step is to divide by three so as we can see X is greater than four so now let's plot it on a number line so here's zero let's put four to the right of it so X is greater than four but not equal to 4 so we need to use an open circle we're going to shade to the right since it's greater than all the way to the right we have positive Infinity so it starts from four and it stops at Infinity but because we have an open circle we're going to use a parenthesis so the answer is 4 to Infinity let's try another example go ahead and solve for x so let's begin by subtracting both sides by five 7 - 5 is pos2 now to get X by itself we need to divide by -2 and whenever you multiply or divide by a negative number you need to know that the inequality changes sign it was less than or equal to now it's greater than or equal to 2 / -2 is1 so now let's plot it so1 is to the left of zero and X is equal to or greater than so we need a Clos Circle and we need to shade it towards the right so here we have Infinity so it's going to be from1 to Infinity so this is the answer in indal notation and make sure you use a bracket whenever you have a closed circle now what if we have two inequalities let's say the first one is 2x + 5 and it's less than or equal to1 or we have this expression 3x + 1 is greater than 10 go ahead and solve for the x value and also have graph it on a number line so let's start with the first one let's begin by subtracting both sides by neg five or just by five those two will cancel and we know 1 - 5 is -6 next let's divide by 2 so X is less than or equal to -3 now let's work on the second inequality let's begin by subtracting both sides by one 10 - 1 is 9 and now let's divide by 3 so X is greater than 3 so now let's plot the two inequalities on a number line so the first one X is greater than three we need an open circle and we're going to shade towards the right for the next One X is is less than or equal to-3 so we need a close Circle and we're going to shade towards left so all the way to the left we have negative infinity and to the right we have positive Infinity so here's how we can represent the answer using interval notation so starting from left to right it's going to be negative Infinity comma -3 with a bracket since we have a closed Circle and then Union and now we're going to jump to the next Region 3 comma Infinity so let me rewrite it above so here's the solution in interval notation negative Infinity to3 that's the red section and then Union the blue section three to Infinity now what if we have another compound equality but in this form let's say that 4x + 5 is greater than 1 but less than or equal to 25 so basically you can break this into two separate equations and solve it separately or you can solve it simultaneously in the last example we solved two separate equations so let's Solve IT simultaneously we're going to subtract five to all three sides at the same time so 1 - 5 is4 5 - 5 is z they cancel 25 - 5 is 20 now we're going to divide all three sides by 4 -4 / 4 is -1 4X / 4 is simply x 20 / 4 is 5 so as we can see X is between between -1 and 5 so let's plot the solution on a number line here's Z here's -1 and here's five so X is greater than 1 so we have an open circle shaded towards the right but it's less than or equal to negative I mean to positive 5 so we're going to have a closed Circle shaded to the left so it's between these two values so the answer in interval notation is going to be1 comma five or we could write it like this and anytime you have a closed circle make sure you use a bracket and use a parenthesis for an open circle now let's move on to a different topic let's talk about absolute value Expressions what is the absolute value of five the absolute value of five is five now what about the absolute value of5 this is going to be equal to POS 5 the absolute value expression produces positive numbers or zero the absolute value of zero is zero but you will never get a negative result after using an absolute value expression and it's important for you to understand that now let's solve absolute value equations if the absolute value of x is equal to 4 what is the value of x whenever you take away an absolute value expression you can write two equations 4 and4 the absolute value of four is equal to 4 and the absolute value of4 is four so X can take any one of these two values let's try another example let's say if we have this expression the absolute value of 2x - 3 is equal to 6 what can we do to solve for x so to get rid of the absolute value expression set 2x - 3 equal to the original value 6 and also -6 and then solve for x so here we can add three to both sides and 6 + 3 is 9 and then we could divide by 2 so X is equal to 9 / 2 or 4.5 and and in the other equation let's add three -6 + 3 is -3 and then we could divide by 2 so X is equal to -3/2 or 1.5 now if we plug in any one of these values it will give us six let's try it so let's plug in 9 over two so 2 * 9/ 2 - 3 will that equal 6 well 2 * 9/2 the cancel and so we're going to get 9 and 9 - 3 is indeed 6 the absolute value of 6 is equal to six so that works so now let's try the second answer -3 / two so first we can cancel a two so left with -3 - 3 -3 - 3 is -6 and the absolute value of six is six so this process works it gives us the right answer so now it's your turn try this one now what you don't want to do at this point is write two equations making this 8 and negative 8 before you do that you need to get rid of the numbers on the outside of the absolute value you want to move them towards the right side so first let's subtract one from both sides so 8 - 1 is 7 now let's divide both sides by five when you have the absolute value expression by itself on one side of the equation that's the time when you can separate it into two expressions so now we can write two equations 3x - 2 is = 7 over 5 and 3x - 2 is equal to -7 over 5 so let's get rid of the five first let's multiply both sides by five let's distribute the five 3x * 5 is 15x 5 * -2 is -10 7 over 5 * 5 is just 7 now for the right side if we multiply everything by five it's going to be 15x - 10 but equal to7 so now let's add 10 to both sides so 15x is equal to 17 therefore X is 17 / 15 here let's add 10 both sides -7 + 10 is pos3 and now let's divide by 15 so X is equal to 3 over 15 which you can reduce it if you divide by 3 3 ID 3 is 1 15 ID 3 is 5 so we have two answers 1 over 5 and 17 over five let's try another example but one that contains an absolute value expression along with an inequality so what can we do for this particular problem so like before we can write two expressions 3x - 4 is greater than 7 and the other one we need to change the sign of the inequality and also add a negative to the 7 so 3x - 4 is less than -7 so now let's solve both equations so let's begin by adding four 7 + 4 is 11 and then we could divide by 3 so our first answer is X is greater than 11 over 3 11 over 3 is about 3 over3 I mean 3.3 now for this one let's add four to both sides 7 + 4 is -3 and now let's divide by 3 so X is less than -1 so let's plot the solution on a number line and then we could test the validity of the equation so X is greater than 3.3 here's zero and it's less than 1 so if it's less than Nega 1 we're going to have an open circle shaded towards the left and if it's greater than 3.3 we're going to have an open circle shaded towards the right so an interval notation we can represent the answer as being Infinity to1 Union 3.3 to Infinity so you can write it like this so the equation should work if we plug in any number in these two regions so let's try it let's pick a number that's less than ne1 let's try a -2 if we plug a -2 into the equation let's see if this solution is greater than 7 3 - 3 * -2 is -6 -6 - 4 is -10 the absolute value of -10 is 10 and 10 is greater than seven so it works now let's try an answer in this region let's try Four 3 * 4 is 12 12 - 4 is 8 the absolute value of 8 is 8 which is greater than 7 so that region works as well now let's try a number that is not in the region so if we try anything between Nega 1 and 3.3 it should not work so let's try one one is between 1 and 3.3 one is over here 3 * 1 is 3 and 3 - 4 is1 the absolute value of 1 is POS 1 and 1 is not greater than 7 so it doesn't work any number in this region will make the equation false so therefore the solution is indeed between netive Infinity to 1 and between 3.3 and infinity so here's the last example in terms of solving absolute value equations and inequalities try this one so remember we can't write two two equations yet until we get rid of these numbers until we somehow move it to the other side so let's begin by subtracting both sides by five 5 - 5 cancels that's zero so we have -3 * the absolute value of 6 - 2x -16 - 5 is 21 by the way for each of these examples before I begin solving it you should pause the video and and try to work it yourself it's going to be helpful if you do the problems yourself that's the best way to learn now our next step is to divide both sides by3 anytime you multiply or divide by a negative number the inequality changes sign so instead of being less than or equal to it's now greater than or equal to in this case positive 7 21 divided 3 is seven and the two negative signs will cancel so at this point now that we only have the absolute value expression on one side we can now write two equations so it could be greater than or equal to 7 or less than or equal to -7 so not only must you change the sign of the inequality you must change the postive 7 to7 let's begin by subtracting both sides by six so 7 - 6 is one and now let's divide by -2 so anytime you divide by a negative number the inequality changes sign so X is less than or equal to a half now let's solve the other equation -7 - 6 is equal to -3 and now let's divide by -2 so the inequality will change sign and this is going to be positive3 /2 so let's uh write it in terms of decimal values ne5 well negative2 is5 and 13 / 2 is basically positive 6.5 now let's plot the solution on a number line so here's negative a half or5 5 and 6.5 is 132 so X is less than or equal to.5 and it's or it's greater than and equal to 13/2 or 6.5 so to write the answer in interval notation it's negative Infinity to- one2 with a bracket so that includes - one2 and then Union 13 /2 to Infinity so that is the solution to this equation now let's spend some time graphing linear equations how can we graph the equation Y is = 2x + 3 what would you do in order to graph it now you need to realize that this graph or this equation is in slope intercept form that is yal MX Plus m is the slope so the number in front of X is the slope so the slope is 2 B is the Y intercept which is three so the first thing that we're going to do is plot the Y intercept actually let's change the equation to uh 2x - 3 instead of 2x + 3 so in this case b is -3 so plotting the Y intercept we're going to plot the point 03 so the graph starts on -3 on the Y AIS and the slope is two the slope represents the rise over the run so since the slope is two we're going to travel two units up one unit to the right and that will give us the next point so up two over one and then two units up one unit to the right and we can do this a few times and then we could connect these points with a straight line so that's a simple way in which you can graph a function a linear equation that is in slope intercept form try this one go ahead and pause the video and graph the equation -3 over 4x + 2 so this is also in slope intercept form so first identify the slope and the Y intercept so the slope is -3 over 4 the Y intercept is 2 so we're going to start at positive2 on the Y AIS to get the next point we can use the fact that the slope is equal to the rise over the run so the rise is neg3 that means we're going to go down three units and the run is four so we're going to travel four units to the right so the next point is going to have an x value of four but a y value of1 and then you can simply connect those two points with a line and so that's how you can graph this equation now what if we have a linear equation that is in standard form how can we graph it now what would you do to graph it standard form is ax plus b y is equal to C the best way to solve in standard form is by finding the x intercept and the Y intercept to find the x intercept replace y with Z so 3 * 0 is simply 0 so 2x is equal to 6 if we divide both sides by 2 6 / 2 is 3 so the x intercept is 3 comma 0 so we can plot that point on the graph now let's find the y intercept so let's replace x with 0 2 * 0 is 0 so therefore 3 Y is equal to 6 and 6id 3 is 2 so the Y intercept is 0 comma 2 which is here and then all you need to do is really just connect these two points with a line and that's how you can graph a linear equation in standard form now for the sake of practice let's try another example so go ahead and try try this one find the X and Y intercepts plot them on a graph and then connect the two points with a line so let's find the x intercept first this is going to disappear 4 * y I mean 4 * 0 is just Zer so 3x will equal 12 and 12 ID 3 is 4 so the x intercept is 4 comma 0 now to find the Y intercept replace x with Zer so -4 y will equal 12 and 12 /4 is -3 so the Y intercept is 0 -3 and then just connect these two points with a line and that's all you need to do to graph it now how can we graph a linear inequality let's say if y Y is less than 3x + 1 well since it's in slope intercept form Let's uh find the slope and the Y intercept and graph it that way the slope is the number in front of X the slope is three the Y intercept is one so let's plot the Y intercept first so it's at positive one and then we can use the slope to get the next point so the slope is three that means the rise is three and the run is one if we have a slope of three it's the same as 3 over one so as we travel one unit to the right we need to go up three units which will take us to the next Point 1 comma 4 now if you want to you can extend the graph the next point is going to be 2 comma 7 which is somewhere in this region now here's the question for you how should we connect these points with a solid line or with a line made up of dashes if it was less than or equal to if there was an underline here you would graph it with a solid L but because it's just less than but not equal to we need to graph it using a dash line now whenever you have an inequality you need to shade the graph should we shade above or below if Y is less than the function shade below so we're going to shade in this region that is below the curve this region represents above the line and you can use a test point if we pick a number that is outside of the yellow region is not going to work for example if we plug in -1 comma 1 we will not get the right answer the equation will not be true so let's replace y with one and x with Nega 1 3 * -1 is -3 and -3 + 1 is -2 1 is not less than -2 so this region doesn't work this region is not the right solution there's no solution in that region now in the region that's shaded the yellow region any point that we pick in that region will make the equation true so let's pick any point let's try 3 comma 0 so let's replace y with 0 and x with 3 3 * 3 is 9 9 + 1 is 10 0 is less than 10 or 10 is greater than than zero so this statement is true so this is the correct region to shade here's another problem that we could try go ahead and graph this linear equality so let's find the slope and the Y intercept so the slope is always the number in front of x X and the Y intercept in this problem is five so let's plot the Y intercept first so it's at 0a 5 Now using a slope of -2 what's the next point if the slope is -2 which is the same as -2 over one the rise is -2 which means we need to go down to units the run is one so as we go down two units we we need to travel one unit to the right so that's going to take us to the 1 comma 3 and if we repeat the process Down 2 over one we're going to be at 2 1 and if we do it again it's going to take us to 31 which is somewhere in that region and then 4 -3 so notice that it's the function is greater than or equal to Y is greater than or equal to -2x + 5 so in this case because we have the underline we need to use a solid line as opposed to a dash line so let's see if we can connect these points okay my line is not perfect but you get the picture now should we shade above the function or below it so if you focus on a point this is above the function that's below it what would you say now notice that Y is equal to or greater than this expression because Y is greater than the expression we need to shade above it and you can always uh test to see if that's the right solution so let's pick a point that is not in that region let's try 0 if we plug in 0 will it be true let's replace y with 0 and x with 0 -2 * 0 is 0 0 + 5 is 5 Z is not greater than five so 0 0 should not be shaded now let's pick any point in this region so let's pick a random Point let's say 4 comma 3 Let's replace y with three and x with 4 so -2 * 4 is8 8 plus 5 is -3 3 is greater than -3 so this is the appropriate region to shade now let's spend some time talking about how to graph absolute value functions the absolute value of x the parent function function looks like this it's basically a vshape centered at the origin now we're going to talk about Transformations what's going to happen if we put a negative sign outside of the absolute value function what do you think is going to happen to the shape of the graph this graph is going to reflect over the xaxis so it's going to open downward as opposed to Upward now what about this one let's say if we have the absolute value of x + 2 what do you think the shape of the graph looks like and what is the transformation if you set the inside equal to zero and if you solve for x x is -2 what this tells us is that the graph is going to shift two units to the left and it's going to open upward so it's going to look like this likewise if we have a graph that looks like this let's say it's x - 3 with a negative on the outside it's going to be shifted uh three units to the right if you set the inside equal to zero X is equal to 3 so the new origin is at 3 comma 0 and because of the negative sign in front it's going to open uh downward and the slope is going to be one the slope is positive one on the left side negative one on the right side now what about this example the absolute value of X+ one so in this case it's going to shift one unit up so it's going to start at 01 and then it's going to open upward so the graph will look like that and here's another example let's say if we have the absolute value of x - 3 + 2 what's going to happen in this case so the graph is going to shift three units to the right and then it's going to shift up two units so it's going to be at 3 comma 2 and since we have a positive sign it's going to open this way but now let's work on a similar example but we're going to graph it with more care because so far all we've done is just we drew a rough sketch but let's draw a more accurate graph so let's say if we have x + 2 minus uh 3 let's use a table just for this example sometimes you may have a teacher that may want you to graph using a table in a situation like this focus on the x coordinate vertex if you set the inside equal to zero you'll know that X is equal to -2 so that's going to be your Center x value now what I would recommend is plugging two points to the right of X or to the right of -2 which are negative 1 and z and two points to the left and then find the Y value of each of these points the vertex is at -2 comma 3 I mean comma -3 based on this number if you replace x with -2 y will be equal to -3 now if we replace X with1 Y is going to equal -2 the line of symmetry is at xal -2 that's where the vertex is which means that -3 And1 will have the same y value if we replace x with -3 -3 + 2 is1 the absolute value of 1 is 1 1 - 3 is -2 so as you can see these two share the same value the same is true for these two and that's why you want to Center your table around the vertex you only need to find half of the points now let's replace x with 0o 0 + 2 is 2 2 - 3 is 1 so these two points will be negative 1 and then what we can do is plot the graph so the first point the vertex is at -23 which is here and the slope which is the number in front of X is one so because the slope is one we can travel one unit to the right and one unit up that's going to take us to the point1 2 and we can go one unit to the right one unit up that's going to take us to the point 01 and from the vertex we can go one unit to the left up one that will give us the next point which is -3 -2 and then if we go one unit to the left one unit up one it'll give us the next point which is this so if this's a one in front of X once you get the first point just travel with a slope of one to the left side and to the right side and this will give you an accurate function or graph so with that in mind go ahead and graph this function using Transformations you can use a table if you want to whichever method is convenient for you go ahead and use that now the x - one tells us that the graph will shift one unit to the right and then plus three it's going to shift up three units so the vertex is that actually let's change it let's make it minus three I'm not going to have much space up here so the graph shifts three units down now so it's at 1 comm3 now the slope is two so as we travel one unit to the right we need to go up two units so the next point is going to be 2 comma 1 and if we repeat the process one to the right up two it's going to be uh three comma 1 and then the left side is going to be the same as the right side so if we travel one to the left we need to go up two units so this will take us to 01 and then if we repeat the process -1 comma 1 and then we can graph it now we're not quite finished with this topic yet what if we have an absolute value inequality how can we graph this particular function 3 * the absolute value of x -1 - 5 so using what you know go ahead and take a minute and graph this function so what is the vertex the graph shifts one unit to the right and down five so the vertex is 15 actually one neg five so it's right here the slope the number in front of X is three so as we travel one unit to the right we need to go up three units so that will take us to the next Point 2 comma -2 which is over here and then whatever you do to the right side you must also do to the left side so if you travel one unit to the left you need to go up three units so it's going to be here and then another unit to the right up three that's going to take us to the point 3 comma 1 and then one to the left up three1 comma 1 now this is great greater than but not equal to so we need to graph it using dashed lines let's do the same for this side so that's how you can graph this particular function now we need to know where to shade it if it was less than a function if y was less than this expression we would shade below the blue line but because it's greater than we need to shade not below it but above the blue line so we're goingon to shade in this region and keep in mind you can plug in a test Point let's plug in a point inside the region let's try 1 comma 0 if it's true then we know we have the correct region if we plug in one we need to replace y with zero and replace x with one 1 - 1 is z 3 * 0 is 0 0 - 5 is 5 0 is greater than5 on a number line here's zg5 is to the left of zero the numbers to the right are greater than the numbers to the left so this is true which means we have the correct shaded region now before we finish with this problem I have one more question for you what is the axis of symmetry we have the vertex what is the axis of symmetry the axis of symmetry is basically a vertical line and it's a vertical line that passes through the vertex which is this point right here so whenever you need to find the axis of symmetry it's simply the x value of the vertex so it's x equal to 1 and you want to write it like this now there's two other things that you need to know how to graph how can you graph these two equations x = 2 and Y equals 3 go ahead take a minute pause the video and think about it how can we graph those two equations to graph xal 2 simply draw a line a vertical line at xal 2 to draw the line yal 3 it's going to be a horizontal line at a y value of three at any point in this line Y is always three and at any point in its vertical line X is always two next topic how to find the slope between two points so consider the points uh 2 comma 1 and also 4 comma 6 how can we find a slope between these two points here's the formula that you need the slope is basically Y2 - y1 / X2 - X1 it's the change in the Y values divided by the change in the X values now we're going to say that X1 is 2 and X2 is one I take that back that's supposed to be y1 X2 is 4 uh Y2 is 6 so now all we need to do is plug it into the equation so Y2 is 6 y1 is 1 X2 is 4 X1 is 2 so it's going to be 6 - 1 which is 5 4 - 2 which is uh two and that's it so the slope is 5 over 2 let's try another example calculate the slope between these two points -3 comma 2 and also 1 comma 5 go ahead and pause the video and work on this example so let's write the formula first Y2 - y1 / X2 - X1 so this is going to be X1 y1 and X2 and Y 2 so it's -5 minus pos2 / 1 - -35 - 2 is -7 1 - 3 is the same as 1 + 3 which is 4 so it's -7 over 4 whenever you have two negative signs next to each other you can make it into a positive sign a negative times a negative is a positive number now what about if we have two points but with fractions how can we find a slope between the two points so we're going to use the same exact formula so we're going to Define this as X1 y1 X2 and Y2 now this example might require more work but the process is very similar so Y2 is - 1/4 y1 is uh -23 X2 is 12 X1 is 1/3 so we can change these two negative signs into a positive sign so let's go ahead and do that first now what can we do to simplify this process now one thing you can do is you can get common denominators and add the two fractions but personally I think that's more work you'll get the right answer but it's just going to take more time what I would prefer to do is to multiply the top and the Bottom by the common multiple of two 3 and four so we can clear away every fraction so this expression is known as a complex fraction whenever you have it multiply the top and bottom by the common denominator or the least common multiple of all the denominators that you see here so what number what is the least common multiple of 2 3 and four multiples of two are 2 4 6 8 10 12 14 so forth multiples of three are 3 6 9 12 15 and so forth multiples of four are four 8 12 16 if you notice 12 is the least common multiple it's common between all three numbers 2 three and four and it's the lowest that's found in all three so let's multiply everything by 12 top and bottom so what is 12 * - 1/4 or 12 /4 so that's going to equal -3 now what about 12 * 2/3 there's two ways you can do that you can multiply 12 by 2 and then divide by 3 or you can divide first and then multiply 12 * 2 is 24 24 / 3 is 8 or you can say 12 / 3 is 4 4 * 2 is 8 next we have 12 time a half half of 12 is six and then 12 * - 1/3 or 12 / -3 which is uh -4 -3 + 8 is 5 6 - 4 is 2 and that's it the slope is 5 over 2 there's three common ways to write a linear equation three forms that you need to be familiar with the first one which we mentioned before is the slope intercept form y isal to mx + b m is the slope B is the Y intercept the next one is standard form ax + b y is equal to C and there a third form called the point slope form y - y1 is equal to M * x - X1 so you have the first one which is the slope intercept form m is the slope B is the Y intercept the second one one known as the standard form and the third the point slope form so why do they call it the point slope form it has a slope M and it also carries the point X1 y1 you can write this equation using a point and the slope so now we're going to focus on writing equations let's say that the slope is equal to two and the Y intercept is3 with this information write the equation in slope intercept form and also in standard form so it's easy to write it in slope intercept form all we need to do is replace m and b so it's going to be 2x- 3 so this is the equation in slope intercept form now how can we put it in standard form to put it in standard form we need the X and the Y variables to be on the left side and any constants must be on the right side so we need to move the 2x from the right side to the left side on the right side it's positive 2x but if we move it to the left side it's going to be -2X so the standard form of the equation is -2x + 1 Y which is equal to -3 so to put it in standard form just make sure X and Y are both on the left side of the equation let's try one more example now let's say that the slope is 1/3 and the Y intercept is-5 write the equation in slope intercept form and also in standard form so in slope intercept form it's going to be 1/3 x - 5 now in standard form let's begin by moving the 1/3 x to the left side so it's - 13x + y is equal to5 we're not finished yet we want to get rid of this fraction so to do that let's multiply everything by three 3 * 1/3 is -1 3 * Y is simply just 3 y and5 * 3 is15 so now we have the equation and standard form next you might be given the slope and the point let's choose let's say 0.13 with this information how can you write the equation in point slope form in standard form I mean in slope intercept form and then in standard form in that order so to write it in point slope form we're going to use this particular formula we're going to replace one with X1 and three with uh y1 so y 1 is 3 m is equal to 2 X1 is 1 so this is the answer in point slope form that's how you can write the equation of the line now to write it in slope intercept form we need to get y by itself we need to solve for y to do that let's begin by Distributing the two into X - 1 2 * X is 2X and 2 * -1 is -2 next let's add three to both sides so therefore Y is equal to 2x -2 + 3 is 1 so 2x + 1 this is the equation in slope intercept form so now let's write it in standard form so let's take the 2X and move it to the left side so -2x + 1 Y is equal to 1 so now the equation is in standard form let's try another one so given the slope and the 6 -2 write the equation in point slope form then in slope intercept form and after that in standard form so let's write the equation first y - y1 is equal to M * x - X1 so X1 is 6 y1 is -2 so let's replace y1 with -2 M with- 13 and X1 with 6 y - -2 is the same as y+ 2 so this is the equation in point slope form so now let's convert it to slope intercept form and to do that all we need to do is solve for y so let's begin by Distributing the 1/3 so y + 2 is = to - 13x and - 1/3 * -6 is the same as -6 divid -3 which is equal to pos2 so that is the equation well there's one more thing we need to do we need to subtract two from both sides so they both cancel so in slope intercept form it's 13 3 x you could write plus 0 b is z but we don't need that so this is the equation in slope intercept form in point slope form or I mean standard form we just got to move this to the other side so - 13x + y is equal to Z next let's multiply everything by three just to get rid of the fraction so --1x + 3 Y is equal to Z so this is the equation equ in standard form now what if we're given two points instead of a slope and a point how can we write the equation in all three forms what would you do in this case the first thing we need to do is calculate the slope so let's define each variable so the slope is Y2 - y1 that's 3 - 5 / X2 - X1 that's 2 -1 3 - 5 is -2 2 --1 is 2 + 1 which is 3 so the slope is -2/3 so now we can write the equation in point slope form using a slope and any one of these two points you'll get the same answer but we're going to do it differently let's write the equation in slope intercept form and in standard form you know how to write it in point slope form now so to write it in slope intercept form start with this equation and choose any point I'm going to use -15 so I'm going to replace 5 or Y with 5 M with -2/3 and x with1 and I'm going to solve for b this is another way in which you can do it so 5 is equal to 2/3 plus b the two negative signs cancel now let's get rid of the fraction let's multiply everything by three so 5 * 3 uh that's equal to 3 * 2/3 the 3 cancels and so we're just going to get two and then B * 3 is 3 B now let's subtract two from both sides 15 - 2 is 13 and then we could divide by 3 so B is 13 over 3 so therefore the equation is y is equal to M which is uh -23 * X plus b or Plus 133 so that is the equation in slope intercept form now to put it in standard form let's move this term to the other side so it's going to change from negative to positive next let's get rid of the fractions by multiplying everything by three so 2/3 * 3 is just 2 y * 3 is 3 y 13 over 3 * 3 is 13 so this is the equation in standard form let's try another example so consider the points 31 and -24 this time we're going to write the equation in point slope form first and then in slope intercept form followed by standard form so let's calculate the slope so let's define 3 as X1 -1 as y1 and this will be X2 and Y2 so the slope is Y 2 - y1 4 --1 over X2 - X1 which is -2 - 3 4 -1 is the same as 4 + 1 which is 5 -2 - 3 is5 5 / 5 is 1 I mean negative 1 so that that's the slope so now to write it in point slope form let's begin by using this equation so y1 is1 m is 1 and X1 is 3 so y -1 is y + 1 so this is the equation and point slope for now let's write it in slope intercept form let's solve for y let's begin by Distributing Nega 1-1 * X ISX -1 * -3 is pos3 next let's subtract both sides by 1 so Y is equal tox + 2 so that's the equation in slope intercept form now to convert it to standard form let's take this and move it to the other side so it's going to be positive 1x + 1 Y is equal to two so that's the equation in standard form write the equation of the line that is parallel to the line y = 3x + 5 and that passes through the point 2 comma 1 how can we do that well what you need to know is that parallel lines have the same slope so what is the slope of this line to find the slope look at the number in front of X so the slope is three now it passes through the point in the point 2 comma 1 now we can write the equation of the line in any form slope intercept form point slope form and standard form but the most common way to write it is in slope intercept form so if you're not sure that's probably the default form that you should probably go to so let's write it in slope intercept form so in this case we're going to do it this way we're going to use this equation and we're going to solve for b so Y is equal to 1 x is 2 m is 3 Let's calculate B 3 * 2 is 6 next we need to subtract both sides by six 1 - 6 is5 so B is equal to5 therefore the equation is y is equal to M or 3 * x + b or minus5 so this is the equation of line notice that they have the same slope but the Y intercept is different write the equation of the line that is perpendicular to the line y = -2 5x - 4 and that passes through the point 104 so how should we begin what is the slope of this line the slope of that line is -2 over5 now what is the slope of the line that is perpendicular to it a perpendicular line let's say if this is line one the perpendicular line passes through line one at a right angle that is at a 90° angle but how can we find the slope of the perpendicular line to find it flip the fraction so it's going to be 5 over 2 instead of 2 over 5 and change the sign from negative to positive so that's the slope of the perpendicular line and we know that it passes through the point 104 so let's write the equation in slope and form so let's use this equation and let's solve for b so let's replace y with4 m with POS 5 over2 and x with 10 and let's solve for b so 5 * 10 is 50 50 / 2 is 25 and now let's subtract both sides by 25 so the equation is y is = 5 over 2x minus 29 as you can see4 minus 25 is - 29 so that's our Y intercept value and this is the equation in slope intercept form in 2005 the rabbit population on an island was found to be 1500 in 2015 the rabbit population was determined to be 2400 calculate the average rate of change of the Rabid population so what can we do here to calculate the average rate of change is the same as calculating the slope so we could say the first point is the Year 2005 and the population is 1500 so let's say that the year correlates to X and the population correlates to Y now the second point is 2015 that's our x value and the yvalue is400 so in 2015 there were 2400 Rabbids so this is X2 and Y2 so to to calculate the rate of change we can use the slope formula which is Y2 - y1 or 2400 -500 ID X2 - X1 2015 minus 2005 so in the course of 10 years the rabbit population increased from 1500 or 2400 so it changed by positive 900 so if it changed by 900 in 10 years then if we divide 900 by 10 the robbit population is increasing by 90 per year so every year an average of 90 rabbits is uh being produced to something or that's how much it's increasing you have some rabbits that are dying some that are being born but the net change is 90 so that's the average rate of change the population increases by 90 rabbits per year so in 10 years it increases by 900 in 100 years at that same rate it should increase by 9,000 now let's briefly review some rules on exponents I'm not going to spend too much time on it um because in Algebra 1 you should have learned this uh topic if not you can go to my website video- and look for my algebra playlist and I have a video that focuses on exponents how to simplify them things like that it's a long video but when you watch it it'll cover this topic completely but let's briefly review some rules of exponents what is x s * x to the 3 power whenever you multiply by a common base you can add the exponents 2 + 3 is 5 now when you divide you can subtract it 7 - 3 is 4 and whenever you raise one exponent to another exponent you are allowed to multiply 3 * 4 is 12 anything raised to the zero power is equal to one and if you have a negative exponent you can move it to the bottom to make it positive likewise if you have a negative exponent on the bottom you can move it to the top to make it positive now if you have a fractional exponent you can convert it to a radical so x to the 35ths is the same as the fifth root of x Cub x^ 4 7 is the same as the 7th root of x 4th now how would you simplify 16 raised to the 3/4s let's say if you have a number raised to a fractional exponent what would you do so feel free to take a minute and see if you can come up with the answer go ahead and pause the video one thing you can do is type it in your calculator and it's going to give you the answer another way you can do it is you can separate the four and the three this is equivalent to 16 ra to the 1/4 raised to the 3 power because 1/4 * 3 is 34s so what is the four of 16 what time what time what * what four * is 16 2 * 2 * 2 * 2 is 16 so the four of 16 is 2 and 2 to the 3 power that's 2 * 2 * 2 3 * is equal to 8 so that's how you can simplify fractional exponents let's try another example what is 8 rais to the 4/3 so this is the same as the cube root of 8 raised to the 4th power the cube root of 8 is two and two to the 4 is 16 now what about 81 raised to the 3/4s so this is the four root of 81 raised to the 3 power so what number times itself 4 * is 81 3 * 3 is 9 * 3 is 27 * 3 is 81 so the 4 of 81 is 3 and 3 to the 3 power is 27 here's a question for you what is 5^ 2 * 5 4th this is going to be 5 6 this is not 25 to 6 if the bases are the same you can add the exponents you can't multiply the bases the only time you can multiply the bases is if the exponent is the same for example 5 * 5 is 25 but what it really is is 5 to the 1st power * 5 to the 1st power is 25 to the 1 power so 3 4 * 7 4 is 21 4th so if you choose to multiply the bases the exponent has to remain the same and if you choose to add the exponents the base has to remain the same something has to be the same either the base or the exponent here the exponent is the same so we are allowed to multiply the bases try these examples so in the first example notice that the exponents are the same which means we can multiply the bases 8 * 5 is 40 so it's going to be 40 to the thir power now in the second example the bases are the same so we can keep the base the same and add the exponents 4 + 5 is 9 so this is going to be 7 to the 9th power now in the last example neither the base nor the exponent is the same so we can't multiply it the way it is but however we can change it to a common base for example 2 to the 3 power is 8 2 to the 4th power is 16 so we can replace 8 with 2 to the 3 and we can replace 16 with 2 to the 4th now these numbers were already here just so you know so 2 the 3 raised to the 3 that's 3 * 3 whenever you raise one exponent to another you can multiply 2^ 4 to the 4 4 * 4 is 16 so now we have the same base so we can add the exponents so 9 + 16 is 25 so this answer is 2 to the 25 now the next thing that you need to know how to do is you need to know how to solve a system of equations so here we're going to have a system of two equations and you need to know how to solve for two variables X and Y now there's two methods that we're going to discuss actually three elimination substitution and solving it graphically in this example where X and Y are lined up perfectly the best method to use is the elimination method some textbooks may call it the addition method so let's add the two equations if we add them notice that Y and negative y cancels 2x + 3x is 5x 4 + 1 is 5 so if we divide both sides by five we can see X is equal to 1 now once you have the value of x plug it in to solve for y let's use the first equation so 2 * 1 + y is equal to 4 and if we subtract both sides by two 4 - 2 is 2 so we have the Answer 1 comma 2 and you could check the answer if you plug this point into the second equation it should give you 1 3 * 1 - 2 is equal to 1 so the answer is 1 comma 2 let's try another example try this one let's say that 2x - 3 Y is equal to -6 and 5x + 4 Y is equal to 31 let's use the elimination method now notice that if we add the two equations no variables uh will cancel so we need to modify the equation let's focus on why what is the least common multiple between three and four three and four both go into 12 so let's multiply the first equation by four so we can get -12 Y and the second equation by 3 so we can get posi 12 y now we need to multiply everything in the first equation by 4 so 2x * 4 is 8x x -3 y * 4 is -12 Y and -6 * 4 is -4 now for the next equation we need to multiply everything by three so 5x * 3 is 15x 4 y * 3 is 12 y 31 * 3 is 93 now let's add the two equations so the Y variables cancel 8 + 15 is 23 and 93 - 24 is 69 so now we can divide both sides by 23 69 / 23 is 3 so now that we have the value of x let's find the value of y using the original equation before we multipli it by four so let's replace x with three so 2 * 3 is 6 next let's subtract six from both sides -6 - 6 is -12 and -12 / -3 is pos4 so the answer is 3 comma 4 x is 3 and Y is equal to 4 now what if you have an equation in this form let's say Y is equal to 2X + 3 and Y is = 7 x - 7 if you see it in this form the best method to use is the substitution method so we can replace this y with 7 x - 7 since Y is equal to 7 x - 7 so we could say that 7 x - 7 is = to 2x + 3 and now we just got to solve for x so let's subtract both sides by 2X and simultaneously let's add seven to both Sid sides so the X variables will cancel and the sevens will cancel on the left 7 - 2 is 5 3 + 7 is 10 so now we could divide both sides by five 10 / 5 is 2 so X is equal to 2 so now we can use any one of the two equations to find the Y value let's use the first one let's replace x with 2 2 * 2 is 4 + + 3 is 7 if we use the second equation we should get the same answer 7 * 2 is 14 14 - 7 is 7 so Y is equal to 7 so the answer is 2 comma 7 now let's try another example 2x + 3 Y is equal to 14 and also Y is equal to 5x -1 so if you see two equations written like this do you think we should use the elimination method or the substitution method typically If X and Y are not aligned it might be best to use the substitution method since we have this equation which is equal to Y we can replace y with 5x - 1 so therefore 2x + 3 * 5x - 1 is equal 14 now we can solve for x let's begin by Distributing the three 3 * 5x is 15x and 3 * -1 is -3 next let's add like terms 2x + 15x is 17x and now let's add three to both sides so 17x is equal to 17 and if we divide both sides by 17 we could see that X is equal to 1 now once we have that value we could take it and plug it into this so Y is going to be 5 * 1 minus 1 so that's 5 minus 1 which is four so we have the point 1 comma 4 to make sure that this answer is correct let's plug in those values into this equation so let's replace x with one and let's replace y with four and let's see if it's equal to 14 2 * 1 is 2 3 * 4 is 12 and 2 + 12 is indeed equal to 14 so this is the right answer consider these two equations now you can solve it using the substitution method but what if you're told to solve it graphic what would you do for a situation like this you can graph the two equations and you want to find out where they intersect the point of intersection is the solution so let's start start with the first graph the Y intercept is5 and it has a slope of three so we need to go up three over one to get the next point so it's going to be at 1 comma two and if we go one to the right up two it's going to be 2 one one to the right up two and then 34 so we can just draw a straight line now let's graph the second equation so the Y intercept is POS 5 the slope is -2 so we need to go down two over one to get to the next point so that's going to be 13 the next one if we go down two over one we could see that it's going to be one two I mean 21 so that's the point of intersection so the answer is 2 comma 1 and we could check it let's replace x with 2 and let's make sure that Y is equal to 1 for both equations 3 * 2 is 6 6 - 5 is equal to 1 now let's check the other equation -2 * 2 + 5 -2 * 2 is4 -4 + 5 adds up to one so this is indeed the right answer so that's how you can solve a system of equations uh graphically graph both equations and simply find a point of intersection as you can see X is 2 and Y is one now let's move on to quadratic equations the first thing that you need to be able to do is you need to be able to solve quadratic equations by factoring so let's say if we have the equation x^2 - 25 and it's equal to zero how can we Factor this equation and then solve for x so this equation is in the form a 2us b^ 2 also known as the difference of perfect squares a factor it it's going to be a + b * a minus B so all you got to do is take the square root of these two values put it in uh two parentheses make sure one is positive and the other is negative and that's it so let's work on this example the square root of x^2 is X the square root of 25 is five one is going to be positive and the other will be negative and then set it equal to zero next set each factor equal to zero and solve for x so if x + 5 equal 0 that means X is equal to neg 5 if you subtract both sides by five here we have to add five to both sides so X is five so these are the two answers now let's work on some more examples try this one let's say if x^2 let's say if we have X2 - 36 is equal to0 go ahead and solve for x so take a moment pause the video and work on this example so the square root of X2 is x the square OT of 36 is 6 one is going to be positive and the other will be negative and then let's solve for x let's write two equations the reason why you can do this is because if this is equal to Z the whole thing is zero this is also known as the zero product property here x is equal to -6 and for the other one X is equal to posi 6 now let's try some other examples that have more steps try this one so what is the square root of 4x^2 the square root of 4 is 2 the square root of x^2 is X so the sare otk of 4x^2 is 2x the square root of 81 is 9 so this is going to be 2x + 9 2x - 9 and then set each one equal to zero so first let's subtract both sides by 9 so 2x is equal to9 and then let's divide by two so the first one is going to be 9 over2 the other one has to be positive 9 /2 and so these are the two solutions now what about this one 16 x^2 - 64 now the first thing that we can do in this problem is we can take out the GCF the greatest common factor the greatest common factor between 16 and 64 is 16 16 x^2 / 16 is x^2 64 4 / 16 is 4 and so now we can Factor it so it's going to be x + 2 * x - 2 so x + 2 is equal to 0 x - 2 is equal to 0 so X is equal to positive2 and X I mean for that one should be Nega -2 I take that back and X is equal to pos2 so that's it for that problem try this one 3x^2 - 75 is equal to zero so right now it's not really advisable to square tk3 or 75 because it's not a perfect square in this case always check to see if you can remove the greatest common factor the greatest common factor is 3 3x^2 / 3 is x^2 -75 ID 3 is -25 so now we can Factor by um perfect squares the difference at perfect square method so we know it's going to be x + 5 * x - 5 and so X is equal to5 and positive five so always check to see if you can take out the GCF now what if we have a trinomial where the leading coefficient is one how can we Factor this expression in order to solve for x whenever you have a problem like this look at the constant term what two numbers multiply to 15 but add to the middle coefficient 8 so let's make a list we have 1 and 15 and 3 and 5 1 + 15 is 16 so that doesn't work but 3 + 5 adds up to eight so to factor it it's just going to be x + 3 * x + 5 and you could check your work if you foil it it should give us the original expression x * X is x^2 x * 5 is 5 x 3 * X is 3x and 3 * 5 is 15 as you can see 5x + 3x adds up to 8X so it gives us the original expression so now let's goad ahead and solve for x so let's set each factor equal to zero and so we can see that X is equal to -3 and X is equal to5 take a minute and try this one x^2 + 4x - 21 is equal to Z so go ahead and work on this example now what two numbers multiply to 21 but that add to the middle coefficient four it's not going to be 1 and 21 however 3 and 7 can work 3 * -7 is -21 also -3 * POS 7 is -21 but -3 + 7 adds to pos4 3 and -7 adds to4 so the answer is going to be x - 3 * x + 7 in its factored form so now we can set each factor equal to 0 and so X is equal to 3 and X is equal to -7 now let's work on a few more examples let's try x^2 - 4x - 45 so what two numbers multiply to - 45 but add to4 so 1 and 45 is not going to work 2 doesn't go into 45 but 3 does 3 * 15 is- 45 but 3 and5 add to -12 so that's not going to work but 5 and9 Works 5 * 9 is 45 but 5 +9 adds up to4 so this is going to be x + 5 * x - 9 therefore X will equal -5 and positive 9 try this example so what two numbers multiply to 35 but add to -12 well we know 5 * 7 is 37 and also 5 * -7 is POS 35 I think I said 37 but I meant to say 35 so it's x - 5 * x - 7 so X is equal to 5 and 7 now if you're not sure how to factor it you can get the answer using the quadratic equation so let's get the same answer using that method now this particular equation is in standard form that is ax^2 + BX plus c so you need to realize that a is equal to 1 B is -12 and C is 35 the quadratic formula helps you to solve for x and it's equal to B plus or minus < TK b 2 - 4 a c / 2 a by the way the quantity that's inside the square root b^2 - 4 a c is the discriminant if that value b^ s - 4 AC if it's greater than zero that means that you have two real solutions if it's equal to zero that means the equation has one real solution and if it's less than zero there are two imaginary Solutions so if you ever need to calculate the discriminant it's simply equal to B ^2 - 4 a c without the square root so now let's finish this problem so B is -12 b^2 -12 * -12 is POS 144 a is 1 and C is 35 ID 2 a or 2 * 1 which is 2 so we have pos2 and what's 4 * 35 4 * 35 is 140 so we have 144 - 140 so 144 - 140 is 4 and the square root of four is two so at this point we can separate the equation into two expressions so we have 12 + 2 / 2 and 12 - 2 / 2 12 + 2 is 14 14 / 2 is 7 12 - 2 is 10 10 / 2 is 5 so notice that we have the same answers as we did before POS 7 and POS 5 now what if we have a quadratic expression where the leading coefficient is not one in this case it's two how can we Factor this expression and let's confirm our answer using the quadratic formula to factor it multiply the leading coefficient by the constant term so 2 * -2 is equal to4 now once you get this number find two numbers that multiply to4 but that add to the middle coefficient -3 this is4 and positive 1-4 + 1 adds up to3 now what we're going to do at this point is we're going to replace the middle term with -4x + 1X because -4x + 1X adds up to -3x and now let's get rid of this so we have four terms we're going to factor this expression using a technique called factoring by grouping in the first two terms we're going to take out the GCF the greatest common factor which is 2x 2x^2 / 2x is X -4x / 2x is -2 in the last two terms all we can do is take out a one and so we're going to be left with just x - 2 if these two factors are the same then you're on the right track so it's going to be x - 2 times the stuff on the outside that is the 2X and the plus one so that's how you can factor that particular expression now let's solve for x so let's set each factor equal to zero so X is equal to 2 and for the other one let's move the one to the other side so it's going to be negative 1 and then divide by two so I'm going to write the answers here x is equal to 2 and also -2 so now let's confirm the answer using the quadratic formula so keep in mind this is a that's B and this is C so X is equal to B plus orus the sare < TK of b^ 2 -3 * -3 is POS 9 - 4 a * C all divid 2 a which is 2 * 2 that's 4 so * -3 is POS 3 and on the inside -4 * 2 is 8 * this other -2 that's POS 16 so it's 9 + 16 on the inside 9 + 16 is 25 and the square root of 25 is five so now let's separate into two expressions so we have 3 + 5 / 4 and 3 - 5 ID 4 3 + 5 is 8 8 / 4 gives us 2 which is this answer 3 - 5 is -2 -2 / 4 is2 so you can also get the answers using the quadratic formula now let's work on some more examples go ahead and try this one factor the quadratic expression and solve the equation so we got to multiply 6 by15 so 6 * -15 is equal to9 90 which is a large number so now let's make a list whenever you have a large number it's good to make a list let's start with one9 ID 1 is 90 if we divide it by two it's - 45 if we divide it by three it's uh -30 to find the factors that go into 90 look at 6 and 15 factors of six are 3 and two and factors of 15 are three and five so I'm going to show you something so the next number we're going to choose is 5 and if you multiply the remaining numbers 3 3 and 2 3 * 2 is 6 * 3 is 18 so 18 * 5 is 90 if you take 90 / 5 you're going to get 8 you can try that in your calculator now so what should be the next Factor the next factor is 2 and three which is six and the remaining two numbers are three and 5 which is 15 so 90 / 6 is15 now the next one we could choose is 3 * 3 3 * 3 is 9 what we have left over is 2 and 5 so 9 * 10 is 90 notice that these two they differ by one but we need to change a sign 9 + -10 is 1 but 9 and posi 10 adds up to positive 1 so let's replace 1X with with -9x + 10 x and now let's Factor by grouping so in the first two terms let's take out the GCF which is 3x 6x^2 / 3x is 2x -9x / 3x is -3 and the last two terms let's take out five 10 x / 5 is 2x -15 / 5 is -3 and since d these two factors are the same we want to write track so it's 2x - 3 times the stuff on the outside the 3x + 5 now let's set each factor equal to zero so the first one let's add three to both sides so 2x is equal to 3 and then divide by 2 so x equal 3 2 for the next one let's subtract both sides by 5 and then divide by 3 so it's 5/3 and 3 over2 now let's try a much harder expression to factor how would you factor this expression 15 x^2 minus 167 x minus 156 so this is a challenge problem feel free to pause the video and try this example use what you know to factor it so what do you think we should do here well we can multiply 15 * -56 and that's going to give us a big number so I'm going to use the calculator 15 * 156 is- 2340 what two numbers multiply to 2340 that add to 167 now that's going to take some time we can make a list starting from one just as we did in the last example but that can take a very very long time in a situation like this there's another way in which you can Factor it you can basically reverse engineer the problem you want to use the quadratic formula to help you to factor it in fact you can Factor any factorable expression using the quadratic formula let me show you so this is a b and c so X is = to B plus or minus b^2 so - 167 * - 167 is 27889 - 4 * a * C / 2 a or 2 * 15 which is 30 so we have 167 4 * 15 * 156 that's positive 9,360 now let's add the numbers inside the square root so 2788 9 + 9360 is 37249 if I type that in correctly so now let's take the square root of that number the square root of 37249 is 193 167 Plus + 193 is 360 and 360 / 30 is 12 so that's the first answer 167 - 193 is -26 / 30 so -26 over 30 we can divide by two to reduce it so it becomes -3 over 15 so that's how you can solve for x now how can we use the answer to factor the expression so here's what we need to do write two equations X is equal to 12 and X isal to -3 over 15 your goal is to get a zero on one side to put it back in its factored form so let's subtract both sides by 12 so x - 12 is equal to 0 and for this one let's add 13 over 15 to this side to the other side so we have this now because we have a fraction we want to get rid of it so let's multiply that equation by 15 x * 15 is 15x 13 over 15 * 15 the 15s will cancel you get 13 0 * 15 is 0 so therefore we can Factor the expression like this we can write it as 15x + 13 * x - 12 now just to see if it's correct Let's uh foil the expression 15 x * X is 15 x^2 15x * -12 I'm going to type that in that's uh - 180x 13 * X is just 13 x and 13 * -12 is -156 so the two numbers that we needed that multiplied to uh to the product of these two which was um what's 15 * 156 that was - 2340 the two numbers that we needed that multipli to - 2340 and that added to - 167 are these two numbers - 180 and 13 - 180 * 13 is - 2340 and it adds up to- 167 so that's how you can find those two numbers and once we combine these two terms we're going to get the original expression so now you know how to factor it so the factored answer is 15x + 13 * x -2 so any expression that is factorable if you don't know how to factor it use the quadratic equation and work backwards the next technique that you need to be familiar with with is the complete in the Square method so let's say if you're given an equation that looks like this complete the square and solve for x and use the quadratic formula to confirm your answer to complete the square look at the number that is in front of X the B value take half of it and square it half of four is two so we're going to add two square to both sides now on the left side I'm I'm going to leave it as 2^2 and you can just look at it and see how to factor it to factor this expression on the left it's simply going to be whatever this variable is X whatever this symbol is plus this number before you square it squared x^2 + 4x + 4 is x + 2 * x + 2 which can be written as x + 2^ 2 here we have 12 + 4 which is 16 so now let's solve for x so let's take the square root of both sides the square root will cancel the square and so you're just going to get what's inside x + 2 the square root of 16 is plus or minus 4 so there's two equations x + 2 = 4 and x + 2 is equal to4 so X is equal to 2 and -6 now let's see if we can get the same answer using the quadratic formula so first we need to move the 12 to the left side it has to be in standard form so a is 1 B is 4 C is -12 so x equal to B plus or minus theun of b^ 2 4 is 16 - 4 * a * C / 2 a 2 * 1 is 2 so -4 * -12 is POS 48 and 16 + 48 is 64 the square root of 64 is 8 so now let's separate it into two equations -4 + 8 / 2 and4 - 8 / 2 -4 + 8 is 4 and 4 / 2 is 2 so we get the first answer -4 - 8 is -12 / 2 is equal to -6 let's try this one x^2 - 6X is equal to 24 so go ahead and try this example complete the square solve for x and confirm your answer using the quadratic formula so half of six is three you could ignore the negative sign so we're going to add 3^ squ to both sides now to factor the expression on the left you can just see everything it's going to be x minus 3 squared 3 squ is 9 24 + 9 is 33 so let's square root both sides so we have uh x - 3 is equal to plus orus < tk33 so now all we need to do is add three to both sides so the final answer is three plus or minus otk 33 we can't really simplify this expression so we're going to leave it like this now let's start from the beginning and let's use the quadratic formula to get the same answer so the first thing we need to do is move the 24 from the right side to the left side so it's going to be x^2 - 6 x - 24 is equal to Z so a is 1 B is -6 C is -4 so X is equal a B plus orus b^ 2 -6 * -6 is 36 - 4 * a * C / 2 a so this is 6 plus or minus < tk36 -4 * -4 is positive 96 now thir 36 + 96 is 132 what is the square root of 132 what can we do here first we need to find out what numbers multiply to 132 because it's even we know it's divisible by 2 2 132 / 2 is 66 and 66 is divisible by 2 it's 2 * 33 33 is divisible by 3 so these are the prime factors so the square otk of 132 is the square < TK of 4 * 33 2 * 2 * 33 is 132 now the reason why I chose to write it like that is because four is the perfect square you know that the square root of four is two so therefore the square root of 132 is 2un 13 I mean 2 < tk33 so let's replace root 132 with 2 < tk33 so now the next thing we need to do is divide each of these numbers by two 6 ID two is three two ID two is 1 so this is the answer now let's spend a few minutes talking about how to graph quadratic equations so let's start with this one the parent function X2 and let's draw a rough sketch this is basically a parabola that opens upward it looks like a u-shape negative x^2 is a parabola that opens downward so it looks like this now what about these two based on your knowledge of Transformations go ahead and draw a rough sketch of these two graphs so the first one is going to shift one unit to the left and it's going to open upward the second one let's put a negative in front it's going to shift two units to the right and it's going to open in a downward Direction now what about these two go ahead and draw a rough sketch so the first one it's going to shift three units up and it's going to open upward the next one is going to shift up two units but there's a negative in front of the X squ so it's going to open in a downward Direction now what about this one x - 2^ 2 minus 3 so this particular graph is going to shift two units to the right down three but it's going to open upward if you need to find the axis of symmetry is basically a vertical line that passes through the vertex the axis of symmetry is X is equal to this number positive2 that's the line of symmetry for the graph now what if we need to draw an accurate sketch so first let's use the table method because some teachers may want you to do this identify the vertex the vertex is going to be positive 3 POS 1 so this sign you need to change this sign you should not change once you identify the vertex make that point the center point in your table so the x value of the ver vertex is three we're going to choose two points to the right of three and two points to the left so to the left two and one to the right four and five now let's find the Y value that correspond to these two points if you Center it based on the vertex the Y value for these two x values will be the same the yv value for these two x values will be the same so let's replace x with 4 so 4 - 3^ 2 + 1 4 - 3 is 1 1 2 is 1 1 + 1 is 2 if we plug in two we're going to get the same answer 2 - 3^ 2 2 - 3 is 1 - 1 2 is POS 1 1 + 1 is 2 so you could see it's the same now let's replace x with five so 5 - 3 is 2 2^ 2 is 4 4 + 1 is 5 so now we can graph it so let's start with the vertex 31 the next point is 42 and we also have 2 two and then after that it's 55 and then 15 now the other technique involves finding the vertex and then you can find the next point to find the second Point travel one unit to the right and up one to find the next Point travel two to the right up four this works if there's a one in front of the x value and the reason why it works is because 1 squ is 1 2 squ is 4 3 squ is 9 and so forth so because 1 squ is 1 as you travel one to the right you need to go up one because 2^2 is four as you travel two to the right you need to go up four so we're going to use that technique in the next example but for now let's go ahead and graph it and that's it by the way what is the axis of symmetry the axis of symmetry is the x coordinate of the vertex so it's X is equal to 3 now does this function have a maximum value or a minimum value whenever the graph opens upward it has a minimum value if it opens in a downward Direction it has a maximum value so the location of the minimum value is the x coordinate of the vertex the actual minimum value is the yvalue of the vertex so this function has a minimum value of one now let's try this one feel free to pause the video so first what is the vertex the vertex is going to be positive one change the sign on the inside and don't change this side so it's going to be POS 1 comm4 the Vex form of a quadratic expression is in the form a x - h^ 2 + K and the vertex is H comma K just in case you were wondering so let's begin by plotting the vertex so it's at 1-4 now notice that we have a one in front of the x - 1^ 2 so the parent function is y is equal to x^2 so when X changes by one y should change by one when X Change by two y should change by four when X changes by three y should change by n based on that we can find the other points so as we travel one unit to the right we need to go up one because 1 squ is one so here's an X point and as we travel one to the left we need to go up one as well but the left side and the right side should always be the same they should have the same y value now as we travel two to the right we need to go up four so notice that this point is on the x axis that is the x intercept as we travel two to the left we also need to go up four so the other x intercept is 1 0 now if we choose to travel three units to the right we need to go up 9 units so -4 + 9 is 5 so we should have the point 4 comma 5 and -2 comma 5 and now we can graph it at this point let's identify the axis of symmetry and the vertex so the axis of symmetry is X is equal to 1 the vertex well we have that but the minimum value that's what I meant to say the minimum value is equal to4 so here's the last example of graphing a quadratic expression in vertex form so the vertex is -2 comma 7 so reverse this sign but keep that the same the axis of symmetry is X is equal to -2 so let's plot the vertex so -27 is over here now we don't have y = x instead we have y = 2x^2 so if we plug in one y is going to be 2 and if x changes by two 2^ 2 is 4 * 2 is 8 Y is going to change by 8 so as we travel one unit to the right we need to go down two units so the next point is going to be one comma five and the reason why we're going down two units is because there's a negative sign anytime there's a negative in front the graph is going to open downward now as we travel one unit to the left we need to also go down two units as we travel two units to the right we need to go down8 units so 7 - 8 is 1 so it's going to be 01 which is the Y intercept and also uh -41 so now we can draw a decent shape graph now do we have a minimum or a maximum what would you say anytime it opens downward we have a maximum the maximum value is 7 and as you mentioned before the axis of symmetry is x equal to -2 the maximum value is located at an x value of -2 but the value itself is seven now how can we graph the equation if it's in standard form what would you do first see if it's factorable if it's factorable go ahead and find the X intercepts to find the X intercepts replace y with zero two numbers that multiply to three but add to4 is -1 and -3 so you can see that the X intercepts are 1 and three it turns out that the vertex is usually located in between the two x intercepts so if you take the average of 1 and three if you add them up and divide by two this is equal it's going to equal to two and it turns out that the x coordinate of the vertex will be equal to two another way in which you could find the x coordinate of the vertex is using this formula B over 2 a so we can see that b is -4 and a is 1 * -4 is 4 4 / 2 is 2 which gives us the same x coordinate value so now you have two ways in which you could find the x coordinate the vertex now let's go ahead and find the Y intercept to find the Y intercept replace x with uh zero so we have the point0 comma 3 now let's organize our data into a table because with the table is easy to find any missing points let's put the vertex to the middle and then we have the X intercepts and we also have the point 03 which means that at four it should have the same yv value since this is the center value these two must be symmetrical to each other now all we need to do is find the y-coordinate of the vertex so let's plug in two 2^ 2 is 4 4 * 2 is 8 4 - 8 is4 -4 + 3 is1 now we can graph it so the vertex is at 21 the X intercepts are 1 0 3 0 the Y intercept is 03 and we have another point at 43 and so that's it now what if it's in standard form and if we cannot Factor it what can we do there's two things you can do you can put it in vertex form or you could just find the vertex using that equation I showed you earlier and graph it that way let's find the vertex it's B / 2 a which is two once again now let's find the y coordinate of the vertex so it's 2^ 2 - 4 * 2 + 2 so it's 4 - 8 + 2 -4 + 2 so it's -2 now notice that there's a one in front of the X2 so we can find the other points based on that so the first Point's at 2 -2 and since 1^2 equal 1 as we travel one to the right from the vertex we need to go up one one to the left and up one as well now 2^ s is four as we travel two to the right we can go up four so you can graph it using that technique as well two to the left up four so that's the Y intercept which is 02 and now you can graph it so that's what you can do if you have it in standard form now let's go ahead and put it in vertex form to do that you need to use the complete the square method in this case we need to take half of this number and square half of four is two so we're going to add two squared to both sides now since we added two squared to the right side we have a choice we can add two squar to the left side or we can subtract two squ from the right side let's do it this way now to factor this expression it's going to be x minus whatever sign you see there this number before you square it squared and then we have two -4 so it's going to be x - 2^ 2 - 2 so now you can see the coordinates of the vertex it's pos2 and -2 which is what we have now let's spend a few minutes talking about domain and range so let's say if we have a linear function 2x - one what is the domain and the range for this function we know the Y intercept is1 and the slope is two so the first point is going to be at 01 the next point is going to be at 1 two and let's just connect these two points with a line so here's a rough sketch of the graph the domain represents all of the allowed X values in a function the range is associated with the Y values for a linear function X can be anything so the domain is from negative Infinity to Infinity also y can be anything it's negative Infinity to Infinity when you're analyzing the domain of a graph look at the graph from left to right the lowest x value is negative infinity and the highest is infinity now we considering the range look at the Y values from the bottom to the top the lowest y value is negative infinity and the highest is infinity this graph will continue to go up forever and it can continue going down forever so X and Y can be anything for a linear function now what about an absolute value function what is the domain for this particular function so we know it's going to shift three units to the right and up one so it starts here and it's going to open in both directions we're going to get the v-shaped graph for an absolute value function X could be anything if you extend a graph the lowest x value is negative infinity and the highest is infinity therefore that's the domain X can be anything now what about the range what is the lowest y value the lowest y value is one the highest is infinity because it can keep going up forever so the range is restricted it starts from one and it includes one but it goes to Infinity let's try another example four minus the absolute value of xus 3 so this graph shifts three units to the right and it's going to shift four units up if you want to you can rewrite it like this it's x - 3 + 4 so the vertex is at 3 comma 4 but there's a negative sign in front of it so it's going to open in a downward Direction so this is just a rough sketch the domain is going to be the same for any absolute value function it's negative Infinity to Infinity however the range will vary so now what if we have a fraction like this how can we simp simpy the imaginary expression and put it in standard form so just like the last example we need to multiply the top and the Bottom by the conjugate of the denominator so on top we need to foil 4 * 2 that's going to be 8 and then 4 * I is 4 I on the bottom I * 2 is 2 I and finally we have I * I which is just I 2 on the bot bottom 2 * 2 is 4 and then it's 2 * I which is 2 I and then I * 2 that's -2 I and finally I * I is i^ 2 so let's combine like terms we can combine these two so that's going to be 6 I and we could change i^ 2 toga 1 on the bottom these two terms will cancel which will always happen whenever you multiply by the conjugate and negative i^2 is POS 1 so whenever you multiply by the conjugate the imaginary numbers will disappear which is what we want to happen on the bottom 8 - 1 is 7 so we have 7 + 6 I / 5 now to put it in standard form separate the fraction into two smaller fractions so it's going to be 7 over 5 + 6 over 5 * I as we can see the lowest y value is to negative Infinity the graph will continue to go all the way to the bottom the highest y value is four so from low to high the range is from negative infinity and it stops at Four and since it includes four we need to use the brackets you should always use parentheses when dealing with infinity symbols now let's graph a quadratic function x - 1^ 2 minus 3 so this graph shifts one unit to the right and down three units and there a positive sign in front of it so it's going to open any upward Direction so that's the general shape of the graph now let's analyze the domain X can be anything the graph starts from negative Infinity to Infinity as it goes up it continually moves towards the left and towards the right so there's no restrictions on the value of x that we can plug in now the range has restrictions the lowest yvalue is-3 the highest is infinity so the range is from -3 to Infinity so that's how you can find the domain of quadratic functions linear functions and absolute value functions consider these two expressions go ahead and Factor each one and solve for x on the left side to factor it it's x + 3 * x - 3 so X is equal to -3 and pos3 on the right side if you have the sum of perfect squares as opposed to the difference of perfect squares to factor it you need to use imaginary numbers so it's going to be x + 3 I and x - 3 I so X is going to be -3 I and POS 3 I the imaginary number I is equal to the square root of 1 i^ 2 is1 I Cub is I and I to 4th is equal to 1 now if you're wondering why I Cub is to the negative I I Cub is I * i^ 2 and i^ 2 is1 so 1 * I is negative I so make sure you know those four values now what about this one let's say if x^2 + 10 is equal to Z what is the value of x so you can Factor it or you can move the 10 to the other side and if you take the square root X is going to be plus or minus < TK -10 which is plus or minus < TK 10 I so anytime have a negative number inside a square root function it's going to give you the imagin number I now let's talk about how to perform operations with imagin numbers go ahead and simplify this expression in Algebra 2 you'll see problems like this which is not too difficult to do let's distribute the four 4 * 5 is 20 4 * 2 I is 8 I and now let's distribute the -3 -3 * 2 is -6 -3 * 7 I is - 21 I our goal is to put it in standard form which is in a plus b i form a is a real number the term bi is an imaginary number so let's combine like terms the two real numbers are 20 and -6 which adds to 14 the two imaginary numbers are 8 I and -21 I which adds to -3 I so this is the answer in a plus bi I form go ahead and simplify this expression 2 - 3 I * 4 + 2 I so let's go ahead and foiler 2 * 4 is 8 2 * 2 I is 4 I and -3 I * 4 is -12 I and finally -3 I * 2 I is -6 i^ 2 so we could combine the two middle terms to 4 i - 12 I is8 I i^2 is 1 so -6 * 1 is + 6 so so now we can add 8 + 6 which is 14 and so we have this 14 - 8 I what if you were to see a problem like this what's 5 - 2 I SAR go ahead and simplify this example so the first thing you should do is expand it we can write it like this and then once again we're going to foil 5 * 5 is 25 5 * -2 I is -10 I and then -2 I * 5 is also -10 I and finally -2 I * -2 I is postive 4 I 2 so -10 I + -1 I is -2 I i^ 2 is1 so 4 i^ 2 is4 25 - 4 is 21 so this is the answer now what about this term let's say if we have a fraction how can we put this in a plus bi form how can we write it in standard form if you have a fraction that looks like this all you need to do is multiply the top and bottom by I so it's going to be 7 I over 3 i^ 2 your goal is to get rid of the imaginary number on the bottom 3 I 2 is basically -3 so your final answer is -7 over3 * I so B is-7 over3 the a value doesn't exist a is zero now how can we put this expression in standard form if you see this multiply the top and the Bottom by the conjugate of the denominator the conjugate is going to be the same thing but all you need to do is change the negative sign into a positive sign on top we're going to distribute the five 5 * 3 is 15 5 * I is 5 I on the bottom we need to foil whenever you multiply by the conjugate the two middle terms will cancel 3 * 3 is 9 3 * I is 3 I and then I * 3 is3 I and I * I is i^ 2 so as you can see the two middle terms cancel and so we have 15 + 5 I over 9 i^2 is POS 1 so 9 + 1 is 10 now what we can do is separate it into two smaller fractions and we can reduce each fraction each of these numbers are divisible by by 5 15 / 5 is 3 10 / 5 is 2 5 / 5 is one and then we could reduce this into a two so this is the answer in a plus bi form now what about graphing imaginary numbers how can we graph this expression you need to realize that the X axis is associated with the real axis the Y AIS is the imaginary axis so the real number is string so we need to travel three units to the right on the x axis the imaginary part of this expression is four so we need to go up four units so 3 + 4 I is located at the point 3 comma 4 now how can we find the absolute value of this expression what is the ABS value of 3 + 4 I to do that draw a triangle the absolute value of that expression it turns out it's equal to the hypotenuse of the triangle whenever you have a triangle a right triangle that is C is equal to the hypotenuse C is equal to the square root of a 2 + B it comes from the pyan theorem formula A + B is equal to c^2 so to find the magnitude of the absolute value it's going to be the < TK of 3^2 plus 4 I 2 well not technically 4 i^ S but just 4 squ you could ignore the I in the equation everything inside the square root has to be positive 3^ s is 9 4^2 is 16 9 + 16 is 25 so the absolute value of this expression is simply equal to five so knowing that calculate the absolute value of 5 - 12 I so all you got to do is just equals to 5^2 + 12 S 5 S is 25 12 s is uh 144 and if you add them it's 169 which is equal to 13 now how can we simplify the expression I raised to the 17 power sometimes you'll see things like this you want to first break it down into two numbers one of which is a multiple of four the highest multiple of four just under 17 is 16 so you want to write it like this 16 + 1 is 17 now I to 16 is the same as I 4th raised the 4th power because 4 * 4 is 16 and if you recall I to the 1 power is the < TK of 1 which we're not going to use i^ 2 is1 which we can use I to the 3 power is I and I to the 4th power is one so we can replace I to 4th with one 1 to 4th is simply 1 so the final answer is equal to I so I to the 17th simplifies to I let's try a few more examples so let's try I to the 27th the highest multiple of four just under 27 is 24 so 27 is 24 + 3 and 24 / 4 is 6 and I to the 4 power is 1 I the 3 we know it's I 1 6 power is 1 1 * I is I so this is the answer let's try one more example what about I raised to the 34 feel free to try this one so this is equal to I raised to the 32 * I raised to the second power since 32 is the highest multiple of 4 just under 34 now if you're not sure how to find this number here's what you can do take your calculator and divide 34 by 4 34 ID 4 is uh 8.5 so what you do is you separate 8.5 into 8 and .5 if you multiply 8 by 4 it will give you the 32 if you multiply the 05 by 4 it will give you the two and so you're going to use these numbers as the new exponents this might be useful when dealing with large numbers let's say if you want to find I to the 236 or something that technique might be helpful now 32 ID 4 is 8 and i^2 is 1 I to the 4th power is 1 and 1 to the E is simply 1 so 1 * 1 is 1 so that's the final answer now let's say if we have a polom function of degree 3 and we wish to find a zeros of this function how can we do so if you wish to find the zeros of the function set it equal to zero you're basically looking for the X intercepts of the graph the best way to do that is to factor the expression how can we Factor this cubic polom notice that the ratio of the first two terms is the same as the ratio of the last two terms for instance if we divide 2 by 1 it's equal to pos2 if we divide 8 it's also equal to pos2 when you see that that tells you that you can Factor by grouping so let's take out the GCF in the first two terms which is X2 X Cub / x^2 is X 2x^2 ID x^2 is 2 and the last two terms let's take out the GCF which is4 -4x /4 is X8 /4 is 2 so since we have the common factor x + 2 we can write it as x + 2 times the stuff on the outside which is x^2 - 4 and x^2 - 4 is a perfect square we can factor that as x + 2 * x - 2 so we do have a repeat answer which are these two so therefore X can be equal to -2 or two in this problem so that's one way in which you can find the zeros of a polinomial function is by factoring by grouping that's that's only going to work if the first two coefficients has the same ratio as the last two coefficients when it's placed in standard form when the exponents are placed in decreasing order let's find the zeros of this function so can we Factor by grouping 6 / is not the same as -2 ID one so we can't Factor by grouping so what can we do in this problem in a situation like this you want to list the possible zeros to do that write out the possible factors of the constant term six which are plus or minus one plus - 2 plus - 3 and plus - 6 and the factors of the leading coefficient which is just Plus or -1 so the zeros of the function can be any one of these numbers some of them are the right answers the others are just there so these are the possible zeros so let's start with the smallest number one we want to find out which x value will make the function equal to zero once we have that then then we can use synthetic division to find the other zeros so let's begin by plugging in one into the equation 1 Cub is 1 1^ 2 is 1 * 2 that's -25 * 1 is5 and then plus 6 so we have 1 - 2 which is15 + 6 is POS 1 -1 + 1 is z so one is a factor so now we're going to use synthetic division to find the other uh zeros so let's place the coefficients in this area so it's going to be 1 -25 and 6 so let's bring down to one 1 * 1 is equal to 1 and now let's add 2 + 1 is 1 and then multiply 1 * 1 is1 and then add5 +1 adds up to -6 and then multiply 1 * -6 is -6 so you should get a zero when you divide X Cub by an x value you should get x2 so the first one is going to be 1 x^2 - 1X - 6 so therefore to factor this particular expression we need to convert the zero into its factored form the zero that we found was X is equal to 1 in its factored form it looks like this x - one if you set x -1 equal to Z you'll get X is equal to 1 now the remaining part that we got from the synthetic division that belongs here if you foil X X - 1 by x^2 - x - 6 you will get the original expression so now we have a trinomial with a leading coefficient of one so we can Factor it using old techniques so what two numbers multiply to the constant term of -6 but add to the middle term of 1 this is going to be -3 and 2 so the other two factors are x - 3 x + 2 this so therefore X is equal to 1 x is equal to 3 and X isal to -2 so we can check it let's plug in -2 into the equation it should give us zero -2 to the 3 power is8 -22 is 45 * -2 is 10 -2 * 4 is 8 10 + 6 is 16 8 - 8 is 16 I mean -6 and6 + 16 is zero so the other zero works and if you plug in three it should work as well now let's try another example like that go ahead and find the zeros of this particular polinomial function so let's make a list of the possible zeros the factors of 24 are 1 2 3 4 6 8 12 and 24 and here the factors of one is just one so the answer can be any one of these so let's begin with one let's see if one is a solution 1 3 is 1 --3 * 1^ 2 is -3 -10 * 1 is -10 + 24 -2 - 10 is -12 + 24 is pos2 so 1 is not a zero let's try negative one so this is going to be -1 - 3 + 10 + 24 just by looking at this we know that's not going to work now let's try positive two 2 the 3 power is 8 2^ 2 is 4 10 * 2 is 20 3 * 4 is 12 -20 + 24 is is pos4 8 - 12 is -44 + 4 is 0 so 2 is one of the zeros so if x is equal to 2 as a factor it's xus 2 which I'm going to write it here for now now let's use synthetic division so make sure you plug in this value positive two for this to work now let's write the coefficients that we have here in descendant order so it's 1 - 3 -10 and 24 so let's bring down to one 2 * 1 is 2 now let's add -3 + 2 adds up to 1 now let's multiply 2 * -1 is -2 and add -10 + -2 is -12 and then multiply 2 * -12 is -4 so we can write this as 1 x^2 - 1X - 12 so that's the trinomial that we can Factor two numbers that multiply to -12 but add to1 are4 and 3 so to completely Factor the expression it's going to be x - 2 * x - 4 * x + 3 so therefore the solutions are 2 4 and-3 if you have a ACC to a graphing calculator you can simply graph the function and find the X intercepts these numbers will be the X intercepts or you could just go to Google type in the online graphing calculator and then you could just go to the first website that you see and literally just graph this function and you could find these intercepts now let's say if F ofx is X 4th - 3xb + 7 x^2 - 24 and if you're asked to calculate the value of f of four what would you do there's two ways in which you can do this you can just plug in four or you can use synthetic division let's do it both ways so let's replace x with four now this might require some work hopefully you have a calculator with you 4 to the 4th power is 256 4 to the 3 power is 64 but times 3 that's 192 4 2 is 16 * 7 that's 112 256 - 192 is 64 112 - 24 is 88 and 64 + 88 is 52 so now let's get that same answer using synthetic division so I'm going to write it here so that's the answer that we need to get so using synthetic division here's what you need to do first you need to realize that we're missing a term the X term so we're going to add 0x I mean 0x to the first Power you need to incorporate that number when using in synthetic division so there's a one in front of x 4 a -3 in front of X cub and then a 7 and then a Zer and then -4 so we wish to calculate the value of f of four it turns out that the remainder is equal to the Y value when you plug in four so let's bring down the one 4 * 1 is 4 -3 + 4 is POS 1 4 * 1 is 4 and 7 + 4 is 11 4 * 11 is 44 + 0 it's still 44 and 4 * 44 is 176 -4 + 176 is 152 so as you can see this method is probably a lot easier to get 152 then doing it the other way so that's how you can evaluate functions using synthetic division now let's say if we have this expression 4x^2 I mean cub + 7 x - 5 / x - 2 so divide this function using long division and synthetic division so let's start with synthetic division so we have a four in front of the X Cub now there is no x s so we need to put a zero and then we have a seven and then A5 now if we're dividing by x - 2 that means that X is equal to two so let's put a two here don't use netive -2 so let's bring down the four 2 * 4 is 8 0 + 8 is 8 2 * 8 is 16 7 + 16 is 23 2 * * 23 is 46 -5 + 46 is 41 so the remainder is 41 so using synthetic division the answer can be written this way by the way X Cub / X will give you X2 so the first number is associated with x^2 so it's going to be 4x^2 + 8 x + 23 plus the remainder divided by what you try to divide by which was x - 2 so this whole thing is the answer so let's get the same thing uh using long division so let's put the numerator on the inside so we have 4X Cub + 0 x^2 + 7 x - 5 and we're going to put xus 2 on the outside so first let's divide 4X Cub by x 4x Cub ID X is 4x2 now after you divide multiply 4x^2 * X is 4X cub and 4x^2 * -2 is 8 x^2 so now let's subtract 4xb - 4xb is 0 0x 2-x 8x2 that's going to be positive 8 x^2 7x minus nothing or zero is just 7x so you can bring down to 7x and you can bring down to ne5 so now after you subtract divide 8x^2 / X is positive 8X now let's put a negative sign and then multiply 8X * X is 8 x^2 8X * -2 is -6x so now let's subtract 8x^2 - 8 x^2 is z which we don't have to write you could just cross those two out 7 xus -6x is the same as 7 x + 16x which is 23x and you can bring down 5 so now let's divide 23x / X is 23 now let's multiply 23 * X is 23x 23 * -2 is -46 and then subtract 23x - 23x is z they cancel -5 minus - 46 or5 + 46 is positive 41 now we can't divide 41 by X so we need to stop here so the answer is going to be what we see on top plus the remainder which is 41 / what we tried to divide it by which is xus 2 so that's how you can divide that's the answer to the problem now let's go over composite functions let's say if f ofx is = to x^2 + 4 and G of X is 2x - Str what is f of G of X so to find the value of this expression we need to replace x with G of X so we got to take a g and insert it into F since G is inside of f so let let's write the function for f but let's not write the x value so it's going to be x^ 2 + 4 so instead of writing x^2 + 4 in the case of f replace x with 2x - 3 so it's going to be 2x - 3^ 2 + 4 so that's F of G of X and if we want to we can foil 2x - 3^ 2 2x - 3 * 2x - 3 is going to be 4x^2 - 6x - 6X + 9 so we can combine like terms -6x + -6x that's -12x and 9 and 4 is 13 so this is f of G of X now what about G of f ofx so this time we need to take F and insert it into G so this is going to be 2 * x^2 + 4 - 3 so all we did is we replace x with X2 + 4 so now let's distribute the two so it's 2x^2 + 8 and then let's combine like terms so it's 2x^2 + 5 so that's how you can find the value of the composite functions now how can we evaluate F of G of three so first let's find the value of G of3 that's 2 * 3 - 3 which is 6 - 3 that's 3 so G of 3 is equal to 3 so we can replace G of 3 with 3 so now let's find the value of f of 3 let's plug it into here this is going to be 3^2 + 4 which is 9 + 4 that's 13 so let me give you another problem evaluate G of f of one so first let's find the value of f of one using this expression so it's 1^ 2 + 4 F of 1 is 5 so replace F of 1 with 5 so now we're looking for G of 5 which is 2 * 5 - 3 that's 10 - 3 so the final answer is 7 so I just want to give you a brief overview into composite functions now let's spend some time talking about inverse functions so let's say if we have the function f ofx and let's say it's x^2 - 5x what is the inverse function of F ofx how can we find it so find the inverse function here's what you needs to do replace f ofx with Y that's the first step in the next step switch X and Y and then after that solve for y so the first thing we need to do is add fives to both sides so this is going to be x + 5 is equal to y^2 and then take the square root of both sides so the square root of x + 5 is equal to Y and it's plus or minus so then the inverse function is plus orus root x + 5 let's try another example so let's say if f ofx is equal to 3x - 7 over 4 go ahead and find the inverse function let's begin by replacing f ofx with Y next let's switch X and Y and then solve for y so let's multiply by four so 4X = 3 y - 7 next let's add 7 to both sides and then let's divide both sides by three so we have 4x + 7 3 isal y so we could say that the inverse function of F ofx is equal to 4x + 7 / 3 so let's say if F ofx has the point 2 comma 3 the inverse function is going to have the point 3 comma 2 if F ofx has the point4 comma 8 the inverse function is going to be 84 so basically all of the X and Y values are switched two examples of inverse functions is the e to the X function and the natural log function here's a rough sketch of e the X exponential functions they increase at an increas in rate logarithmic functions increase at a decreasing rate if you plot them notice that they're symmetrical about the line Y that's a property of inverse functions they reflects over the line Y X so these two are inverses of each other now how can we prove if two functions are inverses of each other what should we do to find the answer now we know that 3x - 7 over4 and 4x + 7 over3 are inverses of each other because we found this function from that but how can we prove that they're inverses of each other if f and g of X are inverses of each other then the composite function f of G ofx should be equal to X also the composite function G of f ofx should also be equal to X if you can prove that this is true then f and g are inverses of each other so let's find F of G ofx so let's insert g into F so we're going to replace x with the value of G that 4x + 7 ID 3 and let's see if the whole thing equals x so the first thing that we can do is we can divide by 3 the 3es cancel 3 IDE 3 is 1 and so that leaves us with 4x + 7 and then - 7 over 4 7 - 7 is 0 and 4x / 4 is X so far so good so now we need to test the other one let's prove that g of f ofx is equal to X so this time we're going to take F and insert it into G so let's replace x with 3x - 7 / 4 so as you can see we can cancel the force and so we have 3x - 7 + 7 left over -7 + 7 is zero so we now have 3x ID 3 which is X so that's how you can prove if two functions are inverses of each other now let's move on into logarithms if you see an expression like this what is log base 2 of 8 what is that equal to evaluate this expression there's two things you can do you can use the change of Base formula and type in your calculator if you type in log 8 / log 2 you will get the answer now to do it without a calculator ask yourself how many twos do I need to multiply to get to eight 2 raised to the what power is equal to 8 it turns out that 2 the 3 power is equal to 8 you need to multiply three twos to get to 8 so log base 2 of 8 is three so let's work on some examples what is log base 3 of 81 how many threes do I need to multiply to get to 81 3 * 3 is 9 * 3 is 27 * 3 is 81 so I need to multiply four 3es to get to 81 to convert this to exponential form this is equivalent to 3 to the 4th is equal to 81 so if you see the expression log base a of B is equal to C you can convert it to exponential form by doing this a to the C is equal to B go ahead and try these examples log base 4 of 16 log base 2 of 32 and log base 5 of 1 so 4 raised to the what power is 16 it turns out that 4 S is 16 you need to multiply 2 4S to get 16 now 2 raised to the what power is 32 2 * 2 is 4 * 2 is 8 * 2 is 16 * 2 is 32 so it takes 5 twos to get to 32 so this is equal to five 5 to the what power is 1 anything raised to the zero power is one five to the Zer is one so log one is always equal to zero so now what about log 10 what is log 10 equal to if there is no base it's always assumed to be base 10 10 raised to the 10 I mean 10 to the first power is 10 so log 10 is 1 log 100 is two log a th000 is three whenever you have a log with a base 10 you can simply count as zeros so knowing that what is log of 100,000 how many zeros do you see log 100,000 is five log1 is1 log1 is equal to -2 log1 is equal to-3 so hopefully you see a pattern here let's evaluate a few more Expressions try these four problems so what is log base 3 of 9 well this one is straightforward 3^ 2 is 9 3 * 3 is 9 it takes 2 3 to get to 9 so if that's the case what is log Bas 3 of 1 over9 now think about it now if you recall x to the -2 is the same as 1x^2 so by changing the sign of the exponent we can move the X variable to the bottom so if you want to move n to the bottom we need to change the exponent instead of being two it's going to be -2 if you think about it 3^2 is 9 so 3 to the -2 is 1 over 9 so anytime you have a fraction here it means that this answer is going to be negative now notice that we have a nine and three in all of these problems so this going to be a two somewhere either it's going to be on the top or on the bottom either it's going to be positive or negative so what is log base 9 of 3 if log base 3 of 9 is 2 log base 9 of 3 is 1 / two if you think about it 9 raised to the half is three anything raised to the half is the same as the square root 9 to the half is basically the square root of 9 which is three now what about the last one base 9 to the 1/3 so let's review anytime this number is bigger than the base then this is going to be greater than one if it's a whole number it won't be a fraction if you reverse it if the base is bigger chances are this is going to be a fraction and if this is a fraction this is going to be negative so this is a fraction that tells us that it's going to be negative and then 9 is bigger than the three which tells that this should be a fraction so the final answer is -2 9 raised to the2 is 1/3 let's prove it 9 to the2 is the same as 1 over 9 to the positive2 and anything raised to the 1/2 is the same as the square root the square root of 9 is three so that's why it's one over three try these log base 4 of 64 log base 4 of 1 over 64 log base 64 4 and log base 64 1/4 so if you could find the first one you can easily find the other ones 4 to the what power is 64 4 * 4 * 4 is 64 so 4 the 3 power is 64 so this one here is reversed we don't have a fraction so the answer is going to be positive if this number is larger then you simply have to flip three over one so we're going to have a fraction it's going to be 1 over three now looking at this one this number is larger but it's a fraction so the answer is going to be negative but it's going to be3 it's still going to be like a an integer since 64 is bigger than four now here we have a fraction and 64 the base is larger than the value that we have here 64 is greater than four so it's going to be negative and it's going to be a fraction it's going to be negative 1/3 hopefully you see a pattern but if you can find the first one you can easily find the other ones so anytime you have a fraction here the answer will be negative and if this number is larger then you shouldn't have a fraction as in the case of these two your answer won't be a fraction however if this number is larger then your answer will be in the form of a fraction it's going to be the reciprocal of this answer now how can we simplify the expression log base 5 of 125 raised to the 7th power what can we do a property of logs allows us to take the exponent and move it to the front so log a raised to the N power is n * log a so what we can do do is move the seven to the front so this is equivalent to S log base 5 125 now what is log base 5 of 125 how many fives do we need to multiply to get to 125 it turns out that 5 to the 3 power is 125 so log base 5 of 125 is 3 and 3 * 7 is 21 so that's the final answer let's try another problem like that try this one log base 3 of 27 raised to the 4th power so let's move the four to the front so this is 4 log base 3 of 27 3^ the 3 power is 27 so log base 3 of 27 is simply equal to 3 and 4 * 3 is 12 now there are some other properties of logs that you need to know you can condense two logs into a single log by multiplication log a plus log B is the same as log ab and log a minus log B is log a / log B so whenever you wish to condense multiple log Expressions into a single log go this way if you wish to expand a log go the other way so using all the logarithmic rules that youve learned so far how can we solve this particular equation let's say that log base 2 of 5x - 6 is equal to 3 what can we do in order to solve for x now if you recall log base a of B is equal to C and we could change it to its exponential form a to the C is equal to B so for instance we know that the log base 2 of 8 is equal to 3 to write it in its exponential form 2^ the 3 power is equal to 8 so let's apply that technique to this expression so therefore we could say 2 raised to the 3 power is equal to what's inside so 2 3 is equal to 5x - 6 and 2 the 3r is 8 so now we just got to solve for x so let's add six to both sides so 14 is equal to 5x and then let's divide by five so X is 14 over 5 try this problem let's say that the log base 4 of 2x - 8 is equal to log base 4 5x -2 how can we solve for x so go ahead and take a minute to try this problem so notice that the bases are the same and the log is the same so therefore what's inside of the log expression must be equal to each other so 2x - 8 must equal 5x - 12 so let's solve let's subtract 2x from both sides and let's add 12 to both sides 8 + 12 is 4 5x - 2x is 3x and so in this problem X is equal to 43 so let's say if we have two logs on the left side and a constant on the right side what can we do in order to solve for x in a situation like this condense the two logs into a single log since the logs are added we can turn it into multiplication for instance we have this expression we know that that log A+ log B is equal to log a so we can condense these two expressions into a single log by multiplying x and x - 2 so therefore we can say that log base 2 * x * x - 2 is the same as x^2 - 2x and this is equal to three so now that we have a single log on one side and a constant on the other side we can convert it into its exponential form so two raised to the thir power is equal to what's inside and 2 the 3 power is 8 so notice that we have a quadratic expression and the only way to solve it is to move the a to the other side so 0 is equal to x^2 - 2x - 8 so now let's see if we can Factor the this particular expression what two numbers multiplies to8 but add to -2 this is going to be -4 and pos2 so it's x - 4 * x + 24 + 2 adds up to -2 but multiplies to8 so we have two potential answers x equal to 4 and X is equal to -2 now you have to check to make sure both answers work you can never have a negative number inside a log so log -2 doesn't exist so this is an extraneous solution so now let's check out four log base 2 of 4 plus log base 2 of 4 - 2 is it equal to 3 so let's find out 2^ 2 is 4 so log base 2 of 4 is simply = to 2 4 - 2 is 2 so we have log base 2 of 2 and 2 to the first power is two so log base 2 of two is just one and 2 + 1 is three so it does work let's try another example like that let's try this one log base 3 5x + 7us log base 3 x + 5 let's say that is equal to 1 so we have a minus sign which means we can condense the two logs into a single log by means of division we have to divide these two so the equation that we're going to use is this log a minus log B is equal to log a / log V so it's going to be log 3 5 that x + 7 over x + 5 and that's equal to 1 so now let's convert it into its exponential form 3 to the first power is equal to what's inside let's write it as 3 over 1 and Let's cross multiply 1 * 5x + 7 is 5x + 7 and 3 * x + 5 is 3x + 15 so now let's solve for x so let's subtract both sides by 3x and let's subtract both sides by seven 5x - 3x is 2x 15 - 7 is 8 and if we divide both sides by two x is 4 so now let's check to see if that's correct so let's plug in four 5 * 4 is 20 + 7 that's 27 4 + 5 is 9 what is log base 3 of 27 now 3 thir power is 27 so this is equal to 3 and 3^ 2 is 9 so log base 3 of 9 is 2 and 3 - 2 is 1 so it works now let's spend some time talking about how to simplify radical expressions how can we simplify the square root of 18 to do this you need to know the common perfect squares for instance 1 2 is 1 2 2 is 4 3 2 is 9 4 squ is um 16 5 S is 25 6 s is 36 and so forth so which of these perfect squares can go into 18 the highest perfect square that can go into 18 is nine so 18 is 9 * 2 and the square root of 9 is three so Square < TK 18 is also 3 < tk2 so that's how you can simplify a radical expression try this one go ahead and simplify the square root of 75 so notice that 25 goes into 75 75 ID 25 is three and the square root of 25 is 5 so that's how you can simplify what about the square root of 80 80 is divisible by four and by 16 but choose the perfect square with the higher value 80 ID 16 is 5 and the square root of 16 is four so it's 4un 5 try this one go ahead and simplify this particular expression so take a minute and work on this problem the square root of 8 is theun of 4 time the < TK of two so four is a perfect square that goes into eight a perfect square that goes into 32 is 16 and a perfect square that goes into 50 is 25 2 * 25 is 50 now the square root of 4 is 2 and the square root of 16 is 4 and the square root of 25 is five so now we can multiply 5 * 2 which is 10 3 * 4 is 12 and 7 * 5 is 35 each term has a square root of two attached to it so they're like terms so therefore we can add the coefficients 10 min-2 is -2 and -2 + 35 5 if we add these two that's equal to POS 33 so the final answer is POS 33 squ < tk2 so now it's your turn try this one a perfect square that goes into 27 is 9 9 * 3 is 27 12 is 4 * 3 and a perfect square that goes into 48 is 16 16 * 3 is 48 the square root of 9 is three and the square root of 4 is two and the square root of 16 is 4 4 * 3 is 12 5 * 2 is 10 9 * 4 is 36 12 - 10 is 2 and 2 + 36 is 38 so the final answer is 38 < tk3 so now you know how to add and subtract rational expressions I mean radical expressions if you see a test question that looks like this and it tells you to simplify the expression what would you do to simplify you need to rationalize the denominator that is to get rid of the square root in the bottom to do that multiply the top and the Bottom by theare otk of 3 theare < TK of 3 * the < TK of 3 is the < TK of 9 and theun of 9 is 3 so this is your answer that's how you can rationalize the denominator of a fraction so let's try another example so for this one we need to multiply the top and the Bottom by the square root of five 5 * 5 is 25 and the square root of 25 is 5 so the goal is to get rid of a radical in the bottom of the fraction so now what if we have a problem that looks like this what can we do in this situation multiply the top and the bottom bom by the conjugate of the denominator so on top let's distribute the 12 12 * 5 is 60 and 12 * < TK 2 is simply 12 < tk2 on the bottom we need to foil 5 * 5 is 25 5 * < tk2 aun2 * 5 those two terms will cancel the last term is < tk2 * < tk2 2 which is < TK 4 and the < TK of 4 is 2 so we have 25 - 2 so the final answer is 60 + 12 < tk2 / 23 and you could separate into two fractions if you want to now what if we want to multiply two radical expressions how can you simplify this entire expression what would you do in a problem like this the best thing to do is to simplify the radicals 20 is 4 * 5 and 28 is 7 * 4 the square OT of 4 is 2 so we have 6 * 2 which is 12 and over here we have 8 * 2 and that's uh 16 so now what we can do is multiply 12 and 16 12 * 16 is 192 and now we can also multiply the five and the 7 5 * 7 is 35 so this is the answer now what about dividing two numbers inside a radical how can we simplify an expression that looks like that so what you want to do is break it down into smaller numbers this is equivalent to the square root of 160 / the square root of 45 so 160 is basically 16 * 10 45 is 9 * 5 so theare root of 16 is 4 r 10 we can make it radal 5 * radical 2 5 * 2 is 10 and the < TK of 9 is 3 so notice that we can cancel the < TK 5 so the final answer is 4 < tk2 / 3 you just have to break down the large numbers into small numbers and cancel how can we solve this particular radical equation what would you do in order to solve it we need to take the square of both sides this will cause the radical to disappear for instance the square of the square root of 7 is the same Asun 7 * 7 2 * 7 * 7 is 49 and theun of 49 is 7 so as you can see you just get the number on the inside so the square of theare root of 2x + 7 is just 2X + 7 5^ 2 is 25 so now let's subtract both sides by S 25 - 7 is 18 and if we divide by two we can see that X is equal to 9 let's try another one let's say that the cube root of 3x - 4 is equal to two solve for x in this case we need to raise both sides to the third power so that these will cancel and so we're just going to have what's on the inside 3x - 4 2 to the 3 is 8 and if we add four to both sides 3x + 3x is equal to 12 8 + 4 is 12 and then 12 ID 3 is 4 so that's the answer now the last thing that we need to cover in this section is graph in radical functions the graph of the square root of x looks like this so knowing that what is the graph of negative < TK X if you put a negative in the front it's going to reflect over the x axis so it's going to look like that now what if we put a negative on the inside if you do that it's going to reflect over the Y AIS so it's going to look like that and if you have a negative on the outside and on the inside it's going to reflect over the origin so it's going to go that way let's draw a rough sketch of this graph using your knowledge of Transformations and also let's write the domain and range of the function so in this case there's no reflection over the X and Y AIS inside the radical there's a positive sign in front of the X and outside of it there's a positive sign so it should be going this way but we need to find a new origin so it's been shifted two units to the right and up three units so the graph is going to go like that so what is the domain of the function to write the domain look at the X values the lowest x value that's where the graph starts is two and it will continue to go on forever into Infinity so it's 2 to Infinity now to write the range consider the Y values the lowest y- value is three as this graph moves to the right it's going to keep going up but slowly so it will keep going up to positive Infinity so the range is three to Infinity let's try another example let's graph this function 5 minus the < TK x + 3 so the graph shifts three units to the left and up five units so the new origin is here now which way is the graph going to go is it going to go towards quadrant one towards quadrant 2 quadrant 3 or towards Quadrant 4 now notice that we have a negative outside of the radical so what that means is that it's going to reflect over the x axis so instead of it going like this it's going to go like this and so that's a rough sketch of the graph so now what is is the domain of the function if you look at the X values the lowest x value is three the highest is infinity this will keep on going forever but it's going to decrease as it goes to the right it just decreases slowly so the domain is three to Infinity now what about the range this is going to keep going down slowly but forever so the lowest y value is negative Infinity the highest is five so the range is from negative Infinity to 5 now let's move on to our last topic of the day that is simplifying rational expressions or expressions with fractions in this example we're going to multiply two rational expressions go ahead and simplify this example so in this problem I'm going to expand everything 24 we can write it as 8 * 3 X2 is x * X Y Cub is y * y * y 15 we could break it down into five and three now focusing on the 18 we can make that 6 * 3 and X to 5 means that we have five x values and Y Cub 3 y values 48 is 8 * 6 and X 4th means that we have four x values we have a y value so we can cancel an eight we can cancel a six and we can cancel a three so we have a three on top and a five on the bottom now we can cancel an X and we can cancel a y on the other side we can cancel four x values and we can cancel a y value so on top what we have left is we have two x values combined that's X2 and we also have four y values so the final answer is 3x^2 y 4 divid 5 now let's multiply two rational expressions if you see a problem that looks like this what do you think we should do how can we simplify this expression the best thing you can do is Factor everything and then cancel so in the first expression on top x^2 - 4x all we can do is take out the GCF the greatest common factor which is X and so we're going to be left with x - 4 on the inside now we can Factor this trinomial two numbers that multiply to -20 but add to the the middle coefficient of 1 is going to be5 actually posi 5 and4 5 * -4 is -20 but 5 +4 is POS 1 here we have the difference of two perfect squares the square root of X2 is x the square otk of 25 is 5 and it's going to be plus and minus and for the last term or the last uh expression all we could do is take out the GCF which is 7x and so we're going to have x + 2 left over so now let's see what we can cancel we can cancel an x + 5 we can cancel an x - 4 and we can cancel an x value so the final answer is x - 5 / 7 overx + 2 now what about dividing two rational expressions so let's put a division symbol so how can we simplify this perhaps you heard of the expression Keep Change flip so this expression is equivalent to the following expression we're going to keep the first fraction the same way we're not going to change it we're just going to rewrite it we're going to flip the symbol or we're going to flip division into multiplication I mean change division into multiplication and then we're going to flip the second fraction so keep change flip so now that we wrote Our equivalent expression let's Factor everything so we can cancel so in this expression let's begin by taking out a four so we're going to have X2 - 4 and we can Factor x^2 - 4 as x + 2 * x - 2 since we have the difference of perfect squares now for this particular trinomial what two numbers multiply to 12 but add to 7 this is going to be x + 3 and x + 4 3 * 4 is 12 3 + 4 is 7 here we can take out the GCF which is five and so we're going to have x - 4 left over and for this one two numbers that multiply to5 but add to -2 is5 and 3 so it's going to be x - 5 and x + 3 so the only thing that really cancels is the X+ 3 and that's it so this is just 4 * x + 2 x - 2 x - 5 / 5 x + 4 x - 4 simplify the following expression 5 / x + 2 + 3 / x -1 so if you wish to add or subtract rational expressions you need to find the common denominator which is simply a combination of these two so the fraction on the right multiply top and bottom by the other denominator and for the fraction on the left let's multiply the top and the Bottom by the other denominator that is x - one so let's distribute the 5 to the x - one so it's going to be 5x - 5 and let's distribute 3 to x + 2 so it's going to be 3x + 6 now since the denominator is the same for both fractions we can add the numerators so 5x + 3x is 8x -5 + 6 is positive 1 and we can write this as a single fraction so this is how you can simplify uh this problem is by combining it into a single fraction let's try another example so go ahead and simplify this rational expression so let's multiply this fraction by x - 2 let's multiply the other fraction by x + 3 so let's distribute the 8X to x + 3 so it's going to be 8x^2 + 24x / the common denominator and on the right side we need to foil 7 x * X is 7 x^2 7 x * -2 is -4x -3 * X -3x -3 * -2 is POS 6 and on the bottom let's just keep it the way it is so now let's combine like terms 8 x^2 + 7 x^2 is simply 15 that does not look like a five 15 x^2 14x - 3x is -7x and 24x + -7x is POS 7x and the last term is a six and then we can divide everything by the common denominator so this is the answer solve for x in this rational expression or rational equation so what should we do in order to solve for x what would you do in this problem the best thing to do is to get rid of the fractions let's multiply by the common denominator so if you're not sure what it is multiply everything by every denominator that you see so 3 * 4 * 7 so what is 5 over 3 * these three numbers if we multiply those two the threes will cancel and we're going to have 5 * 4 * 7 5 * 4 is 20 20 * 7 2 * 7 is 14 so 20 * 7 must be 140 so now let's multiply these two together so the fours will cancel 3 * 7 is 21 * X is just 21x and then these two so this time we can cancel a seven 3 * 3 is 9 * 4 is 36 so it's 36x and as you can see it's a lot easier to solve now so now let's subtract both sides by 21x 36 - 21 is 15 and 140 / 15 is not a whole number so let's uh reduce that so X is 140 ID 15 both numbers are divisible by five 15 ID 5 is 3 140 / 5 is 28 so the answer is 28 over 3 Let's solve another rational equation so in this case we're going to multiply everything by the two denominators x + one and X+ 2 so these two will cancel we're going to have 12 * x + 2 if we take this fraction and multiply it by these two terms the x + 2 term will cancel and we're just going to have 8 * x + one left over but don't forget about the negative sign in front now this two we need to multiply by both terms so let's distribute the 12 to the x + 2 that's going to be 12x + 24 now let's distribute the8 so it's -8x - 8 let's foil x + 1 * x + 2 when you do it it's going to be x^2 + 2x + x which adds up to 3x + 2 so let's combine like terms 12x - 8 x is 4X 24 - 8 is 16 and let's distribute this two so it's going to be 2x^2 + 6 x + 4 so what we're going to do now is subtract both sides by 4X and by 16 so on the left side it's zero on the right side 6x - 4x is 2x 4 - 16 is -12 so let's divide everything by two 0 ID 2 is 0 this is going to be x^2 + 1 x - 6 two numbers that multiply to -6 but add to 1 are 3 and -2 so we can Factor this expression and write it as x + 3 x - 2 which means that X is equal to -3 and pos2 so let's check the work to see which one is correct let's plug in3 first -3 + 1 is -2 a -3 + 2 is 1 let's see if this is equal to two 12 / -2 is -6 and 8 /1 is just8 and 6 -6 -8 is the same is -6 + 8 which is indeed equal to two so the first answer is not an extraneous solution it works now let's try the second answer 2 + 1 is 3 and 2 + 2 is 4 12 / 3 is 4 8 / 4 is 2 4 - 2 is 2 so that works now let's talk about how to graph rational equations let's start with the graph y is equal to 1x so this graph has a vertical ASM toote at xal 0 because you can't have a zero in the bottom of the fraction it's undefined there and also it has a horizontal ASM toote at y equal Z so this graph is symmetric about the origin it looks like this so that's the parad function of one /x so from this parent function we could find the graphs of the other functions so let's say if we wish to graph y is equal to 1x - 2 the vertical ASM toote is going to shift two units to the right to find a vertical ASM toote set the inside or the bottom equal to zero so the vertical ASM toote is xal 2 so it's right here anytime you have a rational function where the degree of the denominator is higher than the degree of the numerator the horizontal ASM toote will always be Y is equal to Z so the horizontal ASM toote is the x-axis so this graph has been shifted two units to the right so it looks like that now what is the domain and range for these two functions so let's start with the function on the left all the way to the left we have negative Infinity all the way to the right we have Infinity the only thing is X cannot equal zero so therefore we have to remove Zero from the domain so it can be negative Infinity to 0 Union 0 to Infinity so for rational functions remove the vertical ASM toote from the domain and the range remove the horizontal ASM toote notice that y cannot be zero because that's the horizontal ASM toote so the range is also negative Infinity to 0 Union 0 to Infinity now what about the function on the right the range is the same y can't be zero but the domain is different the vertical ASM toote is now xal 2 so we have to take that out so the domain is going to be negative Infinity to 2 Union 2 to infin now let's try another one go ahead and sketch a rough graph so where is the vertical ASM toote now if you set x - 3 equal to 0 x is equal to 3 that's the vertical ASM toote so it shifted three units to the right so it's now over here here now what about the horizontal ASM toote is it still y equal 0 this part would be y equal 0 but notice that we added four to it so it's going to go up four units so keep this in mind this value here affects the vertical ASM toote it's xal 3 and this number on the outside affects the horizontal asmt which is uh y = to 4 so the graph is going to be the same it's going to look like this now if there was a negative sign in front the graph would look like this it would be reflected over the horizontal asmt just so you know you can also view it as being reflected over the vertical ASM it's the same so now that we have the graph what is the domain and a range so for this for any rational function just remove the vertical ASM toote out of the domain so it's negative Infinity to 3 Union 3 to infinity and for the range remove the horizontal asmt so y cannot be four it could be everything else but four so the range is negative Infinity to four Union four to infinity and don't use a bracket because that means that you're including four when you shouldn't now let's try one more example this is going to be the last example of this uh very long video go ahead and graph the function so the first thing that we can realize is that we can cancel xus 2 and the term that can be cancelled is a hole it's a a point of discontinuity so the whole is that X is equal to two I'm not going to focus on the Y value I'm just going to draw a rough sketch of the graph so it's not a a graph that's drawn to scale just a simple rough sketch now you see that we have a surviving term in the bottom that factor in the bottom that was not cancelled that is going to give us the vertical ASM toote if you set x + 1 equal to0 x is it's going to be- 1 so that's the vertical ASM toote let's draw it here now if we write the surviving function it's 3/x + 1 + 2 so now it looks like the last example so the horizontal ASM toote is 0 Plus two so it's over here and there's a positive in front of the three so it's going to be shaded or graphed in the upper quadrant and in the third quadrant so that's how you can sketch the graph now we do need to put the hole in the graph so at xals 2 we need to have an open circle and that's important make sure you put that in your graph to write the range of the function all we need to do actually we need the Y value of the hole now that I think about it to find a y value plug in two into the surviving equation so it's 3 / 2 + 1 + 2 2 + 1 is three 3 / 3 is 1 1 + 2 is 3 so the whole has the point 2 comma 3 now the reason why this is important is because you need to remove the whole from the domain and the range so let's start with the domain in this domain remove the vertical ASM toote and the x coordinate of the hole from left to right so it's going to be negative Infinity Nega 1 comes before two so1 union1 and then for the whole 2 Union 2 to Infinity so you have to remove the vertical ASM toote and the hole the x coordinate of the hole now for the range you need to remove the horizontal ASM toote and the y coordinate of the hole because as you can see y can't be three so it's going to be negative Infinity to two two comes before three that's for the horizontal asmt and then three for the whole Union 3 to Infinity so that's the range so that is it for this video by the way if you want to watch more videos you can go to my website video.net where you can find all my playlists or you can click that little red button that says subscribe it'll take you to my channel where you have the option to subscribe but you don't have to and you can see my playlists on algebra trigonometry pre-calculus calculus uh General chemistry organic chemistry and physics so you can find a lot of these algebra topics that we went over in the algebra playlist you can find more examples practice problems so you can uh do well in your algebra course so that's it for this video thanks for watching and have a great day