Just a quick exam review video for the grade 11 functions math course. I'm just going to do a quick overview of the main topics you would do in a functions course and topics that you could probably expect to see on your exam. Uh if you want more in-depth videos with explanations of each individual section, make sure you go to jensummath.ca and check out the videos there for more detailed explanations. So, let's just go a brief overview really quickly. I'm going to try and fit this all in into an hour if I can. Um and I'll break it up into parts for you. Um let's just get right into it. So in the functions course, the first thing you probably would have learned about is what makes a relation a function. So basically each value in the domain uh of a relation can only correspond to one value in the range. So each value of x so each x can only have one corresponding value of y if it's a function. And what you would have done is the vertical line test to check that. You would have checked if a relation passes or fails the vertical line test and that tells you whether it's a function or not. So here's an example of a relation that fails the vertical line test. So if you take a vertical line and you drag it horizontally across a relation, if it ever touches the relation in more than one spot, then it fails the vertical line test and it is not a function. This one is not a function because for an x value of -6, there are two possible y-v values -2 4. So there's more than one y-value that corresponds to a value of x. So it's not a function. fails the vertical line test. This is a relation that passes the vertical line test. If I drag a vertical line across this horizontally, it never touches the relation in more than one spot at a time. Therefore, this relation is a function. This one is a function. Next thing you would have learned is how to state the domain and the range um of a relation. So domain is basically what values of x can the relation take. So that's basically looking at where horizontally if I were to drag this point along the relation. And keep in mind these arrows tell us it extends um forever to the left. So if I were to drag this point along the relation, where horizontally did this point go. So the furthest right it goes is here. It never went further than three. And then it would go forever to the left. So if I were to state the domain of this function, this is how we write it. I would say x is an element of real numbers given that the x value. It never was bigger than three. It was always at three or less. So it was always less than or equal to three. What about the range? So the range is looking at where did the point go vertically. So if you think back to when I was tracing this point along the relation, where did it go vertically? Keep in mind if this graph was extended, it would go forever up and forever down. There's no restrictions on how high or how low that point would go if we continued that graph forever. So, we would just say y is an element of real numbers. There's no restrictions on how high or how low that point can go. Let's look at this relation over here. So, as I'm tracing this point along this relation, the domain, where is this point going horizontally? And keep in mind, if this graph extended forever, if it kept going and going and going and going going, there would be no restrictions on how far left or right that point could go. So it would go forever to the right, forever to the left. So we'd say the domain, there's no restrictions. X is an element of real numbers. It could be any real number. But for the range, notice as I'm tracing this point, where vertically is this point going? It never goes below this point right here. That's the minimum point, negative one. It goes forever up but it never goes below that point. So for our range, y is an element of real numbers but there's a restriction. Y is an element of real numbers given that that's what the vertical line means. It's always at or above one. So greater than or equal to1. So that's how we state domain and range. Uh next you would have looked at what's function notation. So in previous math courses you would write equations like this y equals and then um your relation. What we want to do now is use a new notation. Instead of writing y we write f at x. It means the same thing. This means the y value of the function at a given x value. So these two equations that are written here are equivalent ways of writing the same thing. It's just we use this notation here. We write f at x instead of writing y. This means the y-value of the function at a given value of x. So if I wanted to figure out the y-value when x is -3, we'd write it like this. f at -3, the value of the function when x is -3. And then what you would do is plug into the equation -3 for x and then evaluate. And if we evaluated this -3 + 3 is 0, 0 is 0 * 0.5 is 0 - 1. What we have figured out when x is -3, y is ne1. So on the graph of the function you would find the point -31. Next uh if you remember quadratics uh quadratic the graph of quadratic forms a parabola either one that opens up or one that opens down depending on what your a value is. So this a value bigger than zero. So we know it opens up. So we know we have a parabola that looks like this. which means the vertex, this little point here, is um going to be a minimum point. So, I usually teach my class at least a couple different methods for finding the vertex. Let me review a couple of them with you right now. First of which is completing the square. So, it's basically taking this standard form quadratic, this ax^2 + bx + c quadratic, and changing it into vertex form. because if we change it to vertex form, the h and the k tell us exactly what the vertex of the quadratic is going to be. So to get the standard form to look like the vertex form, there's a bunch of steps we go through. First step is that we put brackets around the first two terms and leave the third term, the y intercept, that plus one off at the end. And then next thing that we do is we factor out what's in front of the x squ. Don't factor out an x, just take out what's in front of the x^2. So take out the four and divide both of the first two terms by the four. So we're factoring the four just from the first two terms. So x^2 - 2x + 1. Next, what we want to do is in the brackets here, we want to create a perfect square tromial by adding and subtracting. strategically half of that number squared. So half of two is 1 and 1 squared is 1. So we add and subtract one. We don't actually want this negative one here. I mean we're allowed to do this, right? Because + 1 minus 1 is essentially zero. So we're not changing the value of our equation at all. But we don't really want that minus one. So what we do is we take it out by multiplying it by the coefficient, the four that's out front. And then if we do that, what we're left with in the brackets is just our perfect square tromial x^2 - 2x + 1. And outside the brackets, 4 *1 is -4. And there's still that plus one off at the end. And our last step is to factor that perfect square tromial. Factor it by finding the numbers that multiply to 1, add to -2. And the numbers are 1 and 1. 1 * 1 is 1.1 + 1 is -2. So it factors to x -1 * x -1 which we could just rewrite as x -1^ 2 and -4 + 1 is -3. Notice how this is now in vertex form and I told you that the vertex of this is equal to the h and the k from the vertex form equation. So our vertex is 1 -3. There's another method for finding the vertex uh called partial factoring. So if you remember our standard form equation here, uh this C value right here, this tells us the y intercept of the function. So the y intercept of the function is one. It means it crosses the y ais uh around here somewhere and we know it opens up. So roughly we would know the parabola looks something like this. We know it has a y value of one uh at on the y axis, right? the y intercept uh is at one. So that point there we know is 01. What we're going to do is we're going to use the method of partial factoring to find another point point something um one. We're going to find another point that has the same y-value and then use the property that parabas are symmetrical. Um add the zero plus this unknown x value to figure out where the vertex is. So this is how it works. We set the equation equal to So we set the original equation equal to the y intercept. So the y intercept is one. We set it equal to one. 4x^2 - 8 x + 1. So we set it equal to one and then solve. So we're finding for what x values have a y-coordinate of one. So we know one of our answers should be zero. We should get zero. And then we're going to find the other answer, average them, and that'll tell us where our vertex is. So we can now move this plus one to the other side and we get zero. 4x^2 - 8x. So a lot of people will skip to this step and essentially just think about like erasing your c value and then solving for x. And to solve for x now we can common factor this. So we can common factor out a four and an x. Divide both of these terms by 4x and we get x - 2 for this product to be zero. Um either factor could be zero. So we set both factors to zero and then solve each separate equation. Divide the four. 0 divided 4 is zero. Add the two to the other side. X could also be two. So we have two possible values for x. So we figured out this other point here is 21. So our function has a value of one at zero and at two. So we know the vertex is in between those. So we can find the x coordinate of the vertex just by adding those just by averaging those two x values. So 0 + 2 over 2. 2 over2 is 1. And we can find the y-coordinate of the vertex just by plugging that original value of 1 back into our original equation. So 4 * 1^ 2 - 8 * 1 + 1. So 4 - 8 is4 + 1 -3. So we found once again our vertex is 1 -3. So that's just an alternate method to find the same answer of the vertex being at 1 -3. So two ways of getting the same answer. Uh next thing you would have done with quadratics is solve by factoring. So one where the a value is one and one where it is not one and can't be factored out. So two different methods of how you would factor these. So the first one all you need to do is find two numbers of a product of your c value 12 and a sum of your b value seven. And if we find those two numbers, those two numbers are three and four. We can go right to our factors. Since the a value is one, we can go right to our factors just by adding three and four to x in separate brackets. So our two factors are x + 3 and x + 4. And if we want to actually solve this equation, that means find the x intercepts. We know at each x intercept the y value is zero. So we set the equation to zero. And now solve this. So for this product to be zero that means either this factor has to be zero or the other factor has to be zero to get our product to be zero. Solve each equation your first x intercept is -3 your second x intercept is -4. If we want to solve this one so remember solving means find the x intercept. We know along the x-axis the ycoordinate is always zero. No matter where it crosses we set it to zero. And now we want to solve for what value of x makes this happen. Since uh since our a value is not one, it's three. We check if we can factor it out, but three doesn't go into four. So we can't common factor it out. So we're going to have to factor this the long way by grouping. So we would find two numbers who have a product of not -15, but of 3 * -15. So of - 455 and a sum of our b value, uh -4. So the two numbers that work for this one are -9 and five. Now we can't go right to our factors for this one. What we have to do is split up the middle term. Uh since -9 and five satisfy our product and sum, we split up the middle term into 9x + 5x. And we leave our 3x^2 and our -5. And keep in mind this is set to zero. We're allowed to do this, right? Because 9x + 5x is -4x. And now what you do is you group the first two terms together. and you put an addition sign. And then you group the last two terms together. And then you common factor each group. And always separate with an addition sign here. Uh common factor each group. So I can take a 3x out from this first group. I would get x - 3. When I divide both of those by 3x, common factor this group, take out a five and I would get xus 3. And what you'll notice is what's in brackets with both of these should be the same. So, um, since they're both the same, we can common factor out that xus 3. So, if I take that out from both terms, what I'm left with is 3x + 5 as my second factor. Now, it's factored. And now what you can do is set each factor to zero. And then solve each separate equation. And those are your x intercepts. So, the first one is three. The second one, if I move the plus five over becomes neg five, then divide by three, I get5 over 3. Uh, last thing before we move on to the next section or actually two more things. Um, if you want to solve a quadratic that's not factorable like this one here, like we want to solve it, right? Set it to zero, right? Because the y-coordinate, solve means find the x intercepts and at the x intercept the y coordinate is zero. So set it to zero. x^2 + 6 x + 4. If you were trying to solve it by factoring, you'd find numbers that multiply to four, add to six, and the numbers don't exist. So, what you have to do is use the quadratic formula. And hopefully you remember the quadratic formula.b plus or minus the square root of b ^ 2 - 4 a c all over 2 a. And keep in mind that uh the coefficient of the x squar in this case is one. So our a value is one, our b value is six, our c value is 4. Let's plug all of that into the quadratic formula and get our x intercepts. And it's good to know that the quadratic formula always works. It works for quadratics that are factorable as well. But factoring doesn't always work. Factoring only works if you can find numbers to satisfy your product and sum. So let's go ahead and use quadratic formula. So b so -6 plus or minus the square t of b^2. So 6^ 2 - 4 a c. And this all needs to be divided by 2 a. So if I I'm going to start by evaluating the discriminant. So that's the part underneath the square root. We call that the radicand. If I evaluate that 36 - 16 that would give me 20. So I don't need this whole part here. And then this is all over two. And what we want to do is the key thing here. We want to give exact answers. That means um we're not going to just evaluate roo 20 and get an approximate decimal answer that has to be rounded. We want to give an exact answer. So you're going to have to simplify um this uh this radical expression here root 20. So root 20 what we want to do is we want to find are there any perfect square numbers right 1 square is 1 2 is 4 3 squar is 9 and so on 16 25 are there any perfect square numbers that divide evenly into root 20 yeah this one right here four goes into 20 so I'm going to break up roo 20 into roo4 * 5 and hopefully you'll see why in a second so I'll break it up -6 plus or minus I'll break it up into roo4 4 * 5 all over two. Why did I do that? Well, because four is a perfect square number. I can take the square root of four. So I get -6 plus or minus the square root of 4 is two. So 2 5 over two. And lastly, I can simplify this. I notice I have two terms in the numerator here. Uh they're both even. So I could common factor out a two from both of these. So uh -3 plus or - 1 5 over 2. And now that this two is a factor of the entire numerator, I could simplify it with the denominator. And what I'm left with is uh my first x intercept is -3 + 5. Second x intercept -3 minus 5. And those are exact answers. Last thing before um we move on to the next section is solving a linear quadratic system. So that a linear quadratic system means you have a parabola and you have a line and to solve it means to find where they intersect. If you get two points of intersection we would call that the line we call it a secant line. There's other possibilities. You could have it so your line only touches the parabola in one spot and we call that line a tangent line. So you can only get you could get two solutions if you have a seeant line. You could get one solution if you have a tangent line or it's possible for your line and your parabola to never cross and you could get no you could get no solutions. So, keep in mind those are always possibilities when you're solving a linear quadratic system. And it'll all depend on what you get um for your discriminant when you're using your quadratic formula or I or um you don't have to use quadratic formula. Uh you could uh factor as well if it's factoable, but let's see how it works. So, basically, you're going to use substitution to solve a linear quadratic system. So you're going to make the y values equal and then solve for what value of x makes that happen. So I'm just going to substitute 5x + 9 into the other equation for y. So I'll get 5x + 9 = x^2 + 4x + 3. And then you'll go ahead and solve this equation. So set it to zero. Move everything to the right. x^2 + 4x - 5x + 3 - 9. So I moved the 5x and the nine over. Simplify as much as you can by collecting like terms and then solve this equation. So you could use quadratic formula like I said, but I noticed this one is easily factorable. Find two numbers with a product of -6 and a sum of -1. So those two numbers are -3 and two. They multiply to -6 add to 1. So it factors to x - 3 and x + 2. Set each factor to zero and solve. So my first uh point of intersection is going to have an x coordinate of three. My second point of intersection is going to have an x coordinate of -2. So let's look at our first point of So we're going to get two answers, right? It's going to be a secant line. our point of intersection number one. If we want the y-coordinate, all we have to do is take the x value of 3 and plug it back into either of the original equations. It should give us the same answer because they they're crossing at this point. Uh the linear equation is going to be the easier evaluation. So 5 x + 9. So I'll do 5 * 3 + 9. That gives me 24. So my first point of intersection is the point 324. my second point of intersection. I'll just plug in my second x value -2. So y = 5 * -2 + 9. So -10 + 9 is -1. So my second point of intersection is the point -21. What we're going to do next uh is rational expressions. How to simplify rational expressions. So I'll do this pretty quickly. Uh, when dividing two rational expressions, the first thing you're going to want to do is to change this to a multiplication question. So, when dividing fractions, what you have to do is keep the first fraction the same, and then flip the second fraction. So, change it to multiplication, flip the second fraction. Okay, now that it's a multiplication question, what you're going to want to do is factor as much as possible. So the numerator of the first fraction multiplies to 10 adds to -7 are -2 and5. So that would go to x -2 * x - 5. The denominator is a difference of squares. It's an x^2 - a 2^ 2. So that goes to x - 2 * x + 2. This fraction over here common factor the top divide a 3 from both terms. We get 3 * x + 2. Denominator uh multiplies to -5 adds to -4. Our 5 and 1. So it goes to x - 5 x + 1. And when we have two fractions being multiplied, we are allowed to reduce within the same fraction, but we're also allowed to cross reduce. So like these x -2's, I have an x -2 / an x -2 within the same fraction. You're always allowed to reduce within the same fraction as long as your numerator and denominator are both fully factored. So x -2 over x -2 is one. So I can essentially cross those off. But because we're multiplying, we are also allowed to cross reduce. I can reduce this x - 5 with this x - 5 to be one as well. And then I can cross reduce this x+2 with this this x plus2. Um, and that's all that reduces. And please only reduce when um your numerators and denominators are fully factored. And what I'm left with is just 3 over x + one. And you'll have to state your restrictions. And that just means um what values of x would make the denominator at any point be zero? And keep in mind since it was a division question, this whole thing was a denominator at once. So we have to check the numerator and the denominator of this one. So uh right because this um x^2 - 4x - 5 or sorry this 3x + 6 used to be in the denominator of this fraction um but then was flipped. We still have to check it. So we check um here x can't be -2. x can't be negative 1. Can't be 5. Can't be -2. Can't be two. The only one we wouldn't check is here. That's the only one we wouldn't have to worry about checking for a division question. But check the rest. So x can't be 2 -2 5gative 1 -2. And we'll write them in ascending order. So -21 2 5. So restrictions are what values of x would make the denominator at any point be zero. So check this, check this. And it's easiest to check when it's all been factored. Let's do a subtraction question with rational expressions. So, when subtracting, you're going to need a common denominator, but it's probably going to be easiest to find a common denominator once we make sure this is fully factored. So, this numerator, uh, keep in mind, it's not a difference of squares. It's adding. So, we can't use a special product for this. It's not a difference of squares. We just have to leave it. There's nothing we can do with x^2 + 4. But the denominator multiplies to 6, adds to -7. That's -6 and -1. So, we can factor it to x - 6. x -1 - 8 over x - 6. We want a common denominator. They both have an x - 6. So I just have to multiply this one top and bottom by x -1. And now they have a common denominator. So I can write them as a single fraction over that common denominator. So I can put the x^2 + 4 - 8 * x -1 all over our common denominator of x - 6 * x -1. And then what we do is we simplify the numerator as much as possible and then see if there's anything uh that can be reduced further. So please make sure you know you can do not do this. You can't cancel these. This xus one in the numerator is not a factor of everything in the numerator. The numerator is not fully factored. So you cannot reduce anything right now. Start by expanding the numerator. So x^2 + 4 distribute the negative8 so I get - 8x + 8 leave the denominator alone. Leave it fully factored and simplify the numerator. Collect your like terms. So I have x^2 - 8 x and 4 + 8 is 12. And then we would check uh can the numerator be factored? Is there anything that multiplies 12 adds to8? Yep. -6 and -2. So I can factor it to x - 6 * x - 2. Now the entire numerator and the entire denominator are factored. So now I can reduce if there's anything that reduces. So what I can do is I can reduce this x - 6 over x - 6 to be equal to 1. And what we're left with is x - 2 over x -1. and we just have to state our restrictions. So even though this x - 6 has been canceled out throughout the process, you still have to take that into consideration when stating your restrictions. At any point throughout the process, if x was 6 or 1, our denominator would have gone to zero and we would have had an undefined equation. So x can't be one, it can't be six. So look at the denominator all the way throughout. The next section you would have done