Q4 - 5 Mathematics P1 May/June 2026 Grade 12 DBE FUNCTIONS

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[00:00:00]Hey guys, and welcome to the function section of the mathematics paper one for May-June 2026.

[00:00:08]Okay, so question four says, sketched below are the graphs of f of x which is equal to and we have negative a to the power of x plus q. So that is an exponential function and you can see that it is decreasing, okay? And it's below the asymptotes that we have there.

[00:00:24]And then g of x is equal to negative and this is a parabola, okay? It's a sad face or concave down parabola. And then they also tell us that b and c are the x intercepts of g as we can see over there. And then d is the turning point of g and it is also a point on the asymptote of f. Okay, so if you find d then you know what the equation of this asymptote is. Then they tell us that t is a point on f such that dt is parallel to the y axis. So you can see that this point, okay, has an x and a y value.

[00:01:01]This also has an x and a y value. Do you see that these two points that they would share the same x value, okay? So we know that about that. So 4.1 says, calculate the coordinates of d. So d is the turning point of our graph g. How do you find the turning point if they give you the equation of your parabola in standard form? You can use, you know, one of two methods. So I'm going to put down the equation here and you can either use this method where you use the equation, okay? Using the axis of symmetry. So this would be more of a grade 11 method. So that would be, you know that the equation of the axis of symmetry of a parabola or to get the x value of that turning point is x is equal to negative b over 2a. And then you just substitute. So this is negative b, b is 2, so negative 2 over 2 * negative 1. a is negative 1 and so X is equal to That's just -2 over -2, which is one. And then you'll find the X value. Okay? The second way to get that X value would be using the derivative.

[00:02:09]Okay? So, using the function's derivative. And we'll say g prime of X is equal to and then you're going to say the coefficient times the exponent, which is -2x.

[00:02:19]And then exponent -1 is one. You don't have to rewrite that one. Then + 2, and you know the X falls away cuz when you say, you say 1 - 1, the exponent -1, you're just going to get zero, and anything to the power of zero is one.

[00:02:32]So, that stays like that. And then after, you know that you're finding the gradient of this function. To find the turning point, you know that the gradient at the turning point is equal to zero. Okay? Cuz if you had to draw a line all through, you see how the gradients would look all through here.

[00:02:47]For example, at this point over here, the gradient is a positive value because this line is increasing, okay? And so on. So, here, at the turning point, this is where the gradient is equal to zero.

[00:02:58]And so, we say it's equal to -2x + 2.

[00:03:01]And you can simply take that over, so this is 2x is equal to 2. Divide both hand sides of the equation by two, and X is equal to one again. Okay? So, these are just two methods to find that X value of the turning point. To find the Y value, you're just going to use the equation of the graph that you already have. Because it's g of X, it's going to give you the Y value that goes with it.

[00:03:21]So, take that X value of one, and then substitute it into your equation to get its corresponding Y value. Okay? So, we'll just do that. g of one, and put that in the calculator. And our answer is four. Okay? So, it is four. g of one is four. Therefore, the coordinates of that point D are simply just one and four. And that is our answer. Moving on to the next question.

[00:03:46]The next question, 4.2, says, "Write down the values of X where g is increasing. Okay, so tell them the X values where you see G increasing. So, the graph of G, okay, is our parabola.

[00:04:01]And you can see look at this graph. Look what it's doing all through. When you read graphs, you read them from left to right like you read numbers and words, okay? So, look at what this graph is doing from here, from left to right.

[00:04:12]This graph, if you follow it, it is moving up, so it is increasing. And then it reaches this point, the turning point. And then here, this is where it stands still. It no longer increases nor decreases, okay? This is where the gradient is equal to zero.

[00:04:27]And then, look at what it's doing over here, it's decreasing. They want you to tell them where it is increasing, so that would be from here, and then you stop there. Now, you need to tell them the values of X for which this section that has been highlighted exists. To do that, just line it to the X axis so you can understand what these X values are.

[00:04:47]All of these X values, if you line all of these points on the graph to the X axis, do you see how they're all through here, okay?

[00:04:56]So, we know that at the turning point, we have where X is equal to 1, the gradient here is equal to zero. So, the graph is not increasing or decreasing here. But then here, this is where it's increasing, and you can see the X values are values that are less than 1. So, we can simply just say this is where X is less than 1, okay? If you wanted to write that in interval notation, you can simply just say, "Where is it coming from, okay, and where is it heading to?"

[00:05:25]It's coming from negative infinity and it's heading to the 1, okay? It stops at 1. So, it's going to be negative infinity, X is an element of, we say negative infinity to 1, okay? Infinity is not a fixed value, so here you use a round bracket, not a square bracket, because we're not saying it includes the value infinity. It's not a fixed value.

[00:05:48]And for the one, do we include a square bracket? No, it's a round bracket as well because they ask for where this graph is increasing. At one, this is where we have a turning point, okay? At the turning point, the graph is neither increasing nor decreasing. So, you can write it like this or like that, and I'll just also go ahead and put that down here. Okay, so that is our answer.

[00:06:10]And then, moving on to the next question, I probably won't need any of that. Moving on to the next question.

[00:06:16]The next question 3.4 says, "Calculate the x-intercepts of g." Okay, so the x-intercepts, let's see. x-intercepts of g, those would be these, okay? So, B and C cuz this is our graph of g. So, these are the places that it cuts the x-axis.

[00:06:34]So, to find an x-intercept, remember the x-intercept is at the point where the y value is zero, okay? Whatever this is, it'll be negative something. The y value that goes with this is zero, okay? If you try to write that point down. So, we'll take the graph of g and we're going to make y equal to zero. So, let me rewrite that. And so, we're saying zero, okay, in place of g of x cuz that stands for y, is equal to all of this.

[00:07:00]You can simply just multiply or divide both hand sides of this equation by -1.

[00:07:05]So, divide by -1, divide by -1.

[00:07:08]Therefore, we have x squared, okay, it's now positive, -2x -3 is equal to zero.

[00:07:16]Then, we need to factorize this. So, factorize and we're saying x * x gives us x squared. Now, we need factors of -3 that can give us -2. So, one of them need to be negative so that you, you know, multiply and get to the -3. So, that is three and one.

[00:07:32]But then, which should be negative so that you can get a negative two in the middle?

[00:07:37]That would be the three, okay? Because -3 + 1 is -2. But -3 * 1 is -3. And so we have minus three and plus one.

[00:07:47]Okay. And now we can finally solve. So, therefore X is equal to positive three or X is equal to negative one. So, they asked for just the X intercepts. So, you can leave it like this. They didn't say the coordinates, so these are our answers. Moving on to the next question.

[00:08:06]Question 4.4 says, determine the values of K such that your of X is equal to K has two distinct positive roots.

[00:08:16]Now, this would have been a very simple question where you could have just used nature of roots to solve this.

[00:08:23]But, you see the kind of roots they're asking for. So, they say two distinct and then the roots have to be positive, okay?

[00:08:30]They have to be positive. With that discriminant, okay? When we work with B squared minus 4 AC, that this can manipulate the type of roots. You know, you can show them different types of roots that you can get. You can get, for example, equal roots. So, an idea of equal roots is let's say we have an equation and you have something like X minus two times X minus two is equal to zero. This equation has two roots that are the exact same thing. Cuz if you had to solve for X here, X is just equal to two. Or you could say or X is equal to two, but you don't have to say it twice because it's just X is equal to two. And what that would look like as a graph is let's say we were saying this is equal to zero. So, let's say we are equating it to that X axis. This is where the graph intersects the X axis once, but then it cuts it where X is equal to two.

[00:09:23]And then that's what that is going to look like. This would be one root. This is what it means to have one root, okay?

[00:09:31]So, equal roots just means you essentially just have one root. Now, to have two roots, you see that you're going to have to have a situation where the graph actually cuts twice, okay?

[00:09:44]Cuts whichever line it's equal to. In this case, I'm just using the x-axis.

[00:09:48]Now, when we try to use the discriminant to manipulate this, what we usually say is for us to get equal roots, you need to ensure that delta is equal to zero, okay? So, b squared minus 4ac must be equal to zero. Then, if you want two different roots, so this will be two distinct roots, then you need to ensure that delta is greater than zero. This would be a case of maybe something like x minus two and x plus three, okay? They don't even have to be of opposite signs.

[00:10:17]They can have the same sign, as long as they're two different roots, then you'd be able to solve for x here by simply saying x is equal to two or x is equal to negative three, okay? So, in this case, you have two x values. So, if we're looking at what it looks like as an equation, you'll have your graph and then your graph intersects here at negative three and then here at two.

[00:10:39]Okay, so this would be specifically two distinct roots. But now, do you see the thing with this question? This question does not just say it needs to have two distinct roots. It says two distinct positive roots. The discriminant does not manipulate the kind of roots that you have to this extent, okay? It can't now control whether it's going to be positive or negative. So, that you're going to need the graph to do this. So, we actually need to look at the graph.

[00:11:08]Okay, so let's go ahead and look at our graph, okay? We're looking for where the roots will be positive. Right now, look at what your roots are. You have one here and we already have those values, okay? It's negative one and three. So, this is the negative one, that over there is the three. So, look at the roots you have. We need to ensure that it has two different positive roots. So, the roots need to be positive and you need to make sure that you actually do have two roots. So, if I draw an outline of this graph. Okay, so like this. Let's also just put down we know this, okay?

[00:11:44]This is now one and four. This is actually going to be an important point.

[00:11:47]Okay, so you need to ensure that the parabola has two x-intercepts, but they need to be positive. Another thing is remember how they said g of x is equal to k? If we just put that down, remember g of x was a negative x squared plus 2x and then plus three, right? And if we bring the k over, it just becomes a negative k. Okay, is equal to zero. So, this is the equation that we have.

[00:12:15]Meaning you can really just look at this as what do we subtract this equation by so that that graph will now intersect that x-axis at two points, but then it needs to be at positive values. So, for example, you can't have something like negative one and three anymore. It needs to be positive something and another positive value. Okay. Now, the reason I'm just saying what do we subtract by is just because there's already a minus there. Which is helpful, okay? Just depending on how you need to actually move your graph. Now, also just remember if you're subtracting from an equation, when you look at this on the graph, you're actually shifting so in the parabola, you're actually shifting the graph down, okay? So, let's see. If you look at this graph, how do we move this so that it always has positive roots?

[00:13:04]So, right now it has a negative root and a positive. We want it to have two positives. If you move it up, you see how this does not solve the problem, okay? So, moving it up is not a an option. Okay, it's just going to end up having negative and positive and we don't want that. So, we need to move it down, but how far down? So, a point that is going to be very important, if we keep moving this, okay? Just watch how that graph goes down.

[00:13:26]Do you notice that at this point over here, this is where it has a root of zero and then some positive root over there. Then if you move it beyond that, look at this. There's a positive root here. Here's a positive root, okay?

[00:13:40]We're trying to look for places where it's x-intercept are both positive. So like that. Okay, so let me just move it back to where it was. So with this, what we can really just do is mark where this change happened. There's a point that you need to focus on here. It's this y-intercept. Do you notice that when this graph, okay, when the y-intercept went down to zero, okay? When the y-intercept became zero, then immediately after this, when we move it anything further than this, then we start having two positive x-intercepts and that's what we want. So this is a very important point that we need to look at. How do we get this graph to shift so that it's y-intercept is now zero? How do we do that?

[00:14:25]Well, you have to think about what the y-intercept was here.

[00:14:28]Originally, the y-intercept was, okay, they gave us this in standard form that is very helpful because you know that it's ax squared plus bx plus c and c stands for your y-intercept.

[00:14:40]So that means this is three, okay? We know this is three now. And so you would have to shift this graph exactly three units down and then that's where it has a y-intercept of zero. Then if you move it anything more than three units down, you see what happens? You have positive x-intercept, a positive. Keep moving it, positive and positive until you move it to the point where it's turning point is now sitting on the y-axis and then you will shouldn't move it beyond this. So I'm going to put this here and then let me just go ahead and duplicate this and then I'll put one where the turning point is touching the x-axis. Okay, so we can see that because we know it's moving down, okay? It's minus k. Minus K is a downward shift, okay? Well, we were lucky that it was a minus in this case because we actually had to move the graph down so that you can achieve those positive roots. So, we know it must go down anything more than three units, but no more than what? Okay, so more than three units, but no more than what?

[00:15:42]Okay, so you see that if you move that graph down, okay? The lowest that you should move it should be to this blue point. Because if it goes anything more than this, then do you see that it has no more It doesn't have any roots anymore. It doesn't even intersect the x-axis anymore, so there are no roots.

[00:15:57]So, we need to stop here where the turning point hits this, okay? The x-axis. So, ask yourself, how would you have moved that original graph so that the turning point hits the x-axis?

[00:16:10]You would have moved it, okay? The turning point here is four. So, you would have moved this graph four units down so that the turning point now hits the x-axis. Exactly four units down will get you to that point. So, move it down anything more than three units, but no more than four units, and then that would be it. Okay, so how do we write this now as an answer cuz we need to tell them the values of [clears throat] K. What should K be? K should be any value that is Remember, move it down anything more than three units, so it can be anything greater than three, but less than four. Okay, and that would be our answer. If you wanted to write this in interval notation, you're simply saying K should just be anything that is greater than three, so from three to four. You can use these numbers.

[00:17:02]Anything between three and four, and then that would work. So, not from three, but between three and four. Okay, so that is our answer there. I hope that made sense. Moving on to the next question.

[00:17:14]The next question 4.5 says, "Write down the range of F." So, remember that the range is it speaks about all the possible Y values of a function. So, all the Y values that this function can ever have, okay? So, we're looking at the graph of F. Yeah, we don't really need these anymore. I hope we won't need them.

[00:17:33]If we do, yeah, we're just going to have to draw them again. So, the range of F F is our exponential function. Now, there's actually something we know about F. Despite the fact that A is not complete yet. You see that this graph Look at the way the graph looks, right?

[00:17:49]And let's look at all the Y values that it can have. This graph has points all over. Okay, so for example, look at that point there. This is its Y value. Look at this point here. Its Y value is here.

[00:18:01]Look at this point. Its Y value is here.

[00:18:03]Do you see how this graph keeps having Y values all through here. Look at all the Y values it has all through here.

[00:18:11]But, if you look above the asymptote, okay, do you see that the graph is no longer here anymore? So, it won't have any Y values that are here.

[00:18:21]And that's because the asymptote is restricting it. So, if you know what value the asymptote passes on the Y axis, then you'll know the range. We can see that this asymptote, okay? They also told us that D, right? That point D lies on the asymptote. So, the equation of this asymptote, it is a horizontal line, okay? It's a line that passes one particular Y value on the Y axis, as we can see. So, its equation is Y is equal to 4. Therefore, from the way this looks, you can clearly see that the range where it has Y values would be all Y values that are simply less than 4. It has Y values all through here. See this point? Look at its Y value. And so on.

[00:19:02]Okay, so we need to tell them Y is less than 4. We can just tell them the range is Y is less than 4. If you want to write it in interval notation, this would just be Y is an element of and you're saying less than four. So, the lower value you're looking at is values less than four, which would be coming from negative infinity and then heading all the way to four. Okay, so you're saying from negative infinity to four, but not including four, okay?

[00:19:30]Cuz the graph will not have a Y value at four since there's an asymptote there.

[00:19:34]Okay, moving on to the next question.

[00:19:36]Question 4.6 says, "If DT is equal to 3 over 2 units, calculate the values of A and Q."

[00:19:47]Okay, let's see. How can DT help us to find A and Q? We don't need this anymore.

[00:19:53]So, DT is the length of this line, okay?

[00:19:57]So, they're just saying from here, yeah, that actually does help, see? From here, this is four, right? This Y value here is four.

[00:20:05]If they tell you that DT is 3 over 2.

[00:20:09]So, if this is 3 over 2, then it means that from D to get to T, since D's lie on the same vertical line, you're just going to move down exactly 3 over 2 units and then you get to T. Now, you have one point, okay? On this graph, on this graph of F. Okay, so if we are at a Y value of four and you move down exactly 3 over 2 units, which is 1.5, then we're going to reach a Y value of 2.5, which is really 5 over 2. Okay, so do you see how this gives us the Y value of T, which is 5 over 2? And we already have the X value. Remember that these lie on the same vertical line. So, here you would have gone one unit to the right and four units up. Well, for this one you'd also go one unit to the right, but now we know we go 5 over 2 units up, okay? So, the X value here is one, the Y value is 5 over 2. So, So info gave us the coordinates of T. So it's 1 and 5 over 2 and we'll just put that down and tell them T is simply 1 and 5 over 2.

[00:21:14]Now, this would actually help us to find A and Q. With Q, we already have it, okay? When they give you the equation of an exponential function, then you see this constant that you have here, okay?

[00:21:28]Whatever does not have that exponent of X, that constant, that is your horizontal asymptote. So that is that 4.

[00:21:38]You had Y is equal to 4, that's Y is equal to Q, okay? So we know that Q is actually 4 and so we can put that down first and tell them that our Q value is 4, Q equals 4, and now you need to find A. Okay, so what is the value of A? A you would find A by simply just taking this equation, updating it, and you're going to use this point, the X and the Y value to find it. Okay, so let me write that equation. That was f of x is equal to -A ^ x and then + Q, but now it's + 4. And then you're going to use this point. So this is x and then so Y value is so we'll just say 5 over 2 is equal to - and then we have A ^ 1 + 4, okay?

[00:22:20]So A is just negative here, okay? 1 still just 1 as an exponent. Just take it over to make it positive and now you have 4 5 over 2 and so A is equal to 3 over 2, okay? Because the 4 - the 3 over 2 gave you the 5 over 2. So either way around it's 3 over 2, okay? So we know what A is and we also found Q and those are answers. Moving on to the next question.

[00:22:45]And then question 4.7 says, "Why is F a function?" Okay, so in grade 12 you would have learned about functions and when you're working with inverse functions, okay? So if you have the equation of some graph, right? To determine whether it is a function or not, you need to ensure that this graph, okay? All its X values map to a single Y value, okay? One Y value for an X value.

[00:23:16]It doesn't have to be one X value, okay?

[00:23:18]For example, a parabola is a function.

[00:23:20]You see that if I draw this line here, I can see that this There are two points that have this Y value, okay? This Y value may be something like two, okay?

[00:23:30]And so, this point has a Y value of two, but then this point would also have a Y value of two, okay? But then look at its X values. It has this X value and this X value. You can have multiple X values, but they need to map to one single Y value. And you can see with this graph, okay? Each X value maps to a single Y value. See? This one is a one-to-one function. There's one X value, and then its Y value is whatever that is, okay?

[00:23:57]You can just take a point that you have, okay? One and five over two. One X value to a single Y value, one Y value. Okay, so I'll just tell them it's because we can see that each X value maps to a single Y value, okay? And that's my answer there. Moving on to the next question. Okay, the next question 4.8 says, "It is further given that f of x is equal to this, okay? And then we have and h of x is equal to negative f of x and then plus four, okay? Determine the equation of the inverse of h in the form of y is equal to." So, to find the equation of the inverse of h, you need the equation of h. They told you that h of x is equal to negative f of x and then plus four. So, wherever you see f of x, you're going to put a minus in front of it, and it's going to multiply in. So, we have h of x is equal to minus and then this is f of x. It is negative, and then 3 over 2 raised to the power of x, then plus 4, and then you're going to have a plus 4 on the outside because they said negative f of x plus 4. Okay, so just simplify this, find the equation of h, and then you can find its inverse.

[00:25:14]So, when we're multiplying, do you see that inside the bracket, how many terms do you have? You have one and two. Okay, terms are separated by plus or minus signs. So, this multiplies here, and then it also multiplies to the 4. So, you're going to get negative times a negative gives you a positive. So, now we have, okay, positive 3 over 2 raised to the power of x, and then that multiplies to the positive 4 to give you a negative 4. Then outside we have a plus 4. Yeah, it's like one of those cases like why couldn't have just given us a simple equation, but okay. h of x is equal to, and then we have 3 over 2 raised to the power of x. Negative 4 plus 4 gives you 0. This is h of x. And we need to find the inverse of this.

[00:25:57]Okay, so to find the inverse, you're just going to call it y is equal to 3 over 2 raised to the power of x. And step one, swap x and y. So, we replace this with x is equal to 3 over 2, and then the x becomes y. Then make y the subject. To do that, we'll say y is equal to you're making an exponent the subject. So, you're going to use log.

[00:26:21]You say log, and what is our base? Our base is whatever number is carrying the exponent, so that is the 3 over 2. So, our base is 3 over 2, and then you just say x, okay? So, there we go. That is the inverse of h. Moving on to the next question. Question 4.9, the last question for question 4, says we have t prime is the image of t on h. Okay, t We had a t, okay? We had this point, which was 1 and 5 over 2. So, let me just keep track of this. So, this point T, which was 1 and 5/2, this was on the graph of That was F, okay? So, G was the parabola. So, this was on the graph of F. So, they're saying now that T' is the image of T on H, okay? So, this one is on graph H. We can actually get its coordinates. So, H is a translation of the graph of F, okay? So, do you see what they did to the graph of F? They said -f(x) + 4. When you say -f(x), remember that f(x) stands for Y, okay?

[00:27:30]So, if you're saying negative and then Y, you're just multiplying -1 to all the Y values, okay? Of that graph. And so, what that means is if you take a point, and let's say we have A 1 and 2, and you only decide to multiply its Y value, okay? By -1. So, that means that Y value becomes -2. So, if you plot this point down, it's just going to be 1 and -2. Do you see what happened? Only the Y value changed, and if you look at what happened with this point and this point, this point was simply just reflected through the X axis. So, that's what -f(x) means. It means that the graph was reflected through the X axis. So, we know that we're simply going to take the Y value of that point and multiply by -1. But then they still say this, after multiplying by -1, you say +4. So, what would that point become? You had this before, so it would have become 1 and -5/2, then you must still add 4 to it. So, I multiplied by -1 the Y value by -1, and then add 4. So, what is that going to be? Well, I can just say 8/2.

[00:28:43]That is 4, okay? And then 8 - 5, okay?

[00:28:47]And then 8 - 5 is 3, so that's just 3 over 2. Okay, so this just becomes 1 and 3 over 2. Okay, so that is what that point would look like on H. That's T prime. Now, we need to sketch the graph.

[00:28:58]We need to sketch the graph of the inverse. Okay, so this is really what we're doing. Sketch that inverse of H.

[00:29:05]And they say indicate clearly on your graph the intercepts with the axes, the coordinates of T double prime, the image of this. Okay. So, this that they're referring to, it is going to be the image of whatever this is on the inverse, okay, of H. So, you know how to get the inverse function, right? To get the equation of the inverse function, we first swap X and Y, and then you make Y the subject. You saw how we did that.

[00:29:36]So, the whole point is, if you have a graph, all the points on the inverse would simply just be a swap of X and Y.

[00:29:44]So, if you had a point that it was 1 and 2, if you get the inverse graph, it's just going to be 2 and 1. So, it means that this point that they're referring to, okay, it would just be 3 over 2 and 1. Okay, so this is the point they're referring to on the inverse. We're just going to put that down. You can also just test if it is there. So, 3 over 2 and 1, right? If you look at the inverse function, it's log 3 over 2 and then X. So, if we did that, we say log and we're saying 3 over 2, and here what I'm saying is my X value is this. Okay, 3 over 2. So, let's see the Y value that goes with this. If it goes with 1, then yeah, it it is that point. Okay, there we go. It goes with 1. So, this is the point that they want us to put on the graph as well. Okay.

[00:30:28]So, at the end of the day, they're just saying you need to sketch the graph of the inverse of H. So, sketch the inverse of H, make sure to include any X or Y intercepts, and then also that point And the asymptotes, of course. Okay. So, let's get our graph. Since the original h, right? This is the original. That's the final form of it. Okay, so since the original graph of h, okay? This is the final form of h. Since it was an exponential function, then it means that the inverse is going to be a logarithmic function. Okay, so we can just put that on our Cartesian plane. Well, we can look at what the this graph would look like first. So, h of x is equal to 3/2 to the power of x. Do you see this number 3/2? You can just quickly find out whether the graph is going to be increasing or decreasing. So, there's just certain things that I want to find out about this, then it will make sketching the inverse easier. So, this is the equation of h. A graph is always an exponential function is always written in the form of y = a * b raised to the power of x, and then - p + q.

[00:31:38]Sometimes there's a shift, okay?

[00:31:39]Sometimes there is not. So, it means that the value of a, the number in front of this, okay? Base and exponent, it's an invisible one. Here's the thing. If a is a positive number in an exponential function, what it means is that your graph is going to lie above the asymptote, okay? So, it could look like that, increasing, or it could look like this, decreasing. One of these two. And then, to determine whether it's increasing or decreasing, you actually just look at the value of this and this at the same time. There's the whole uh thing that you would study where you say, "If a is greater than zero and then b is a certain thing, it's going to be increasing." I don't really use that method because there's a simpler way to just do this. You can simply take this, right? And then on your calculator, you're going to put 3/2 raised to power of x, and you're going to substitute two x values. Take two x values that come right after each other. Like x is equal to 1 and x is equal to 2. And then And going to check the y values that you get here. Okay? If you substitute one, you just get 3/2. Remember, that is 1.5. If you substitute two, you see that you're going to get 9/4, which is if you take 9/4, that is just 2.25. So, this is 2.25.

[00:32:56]Then you just look at from where X was equal to one, okay? You got a Y value of 1.5 and then here 2.25.

[00:33:04]Did this increase or decrease?

[00:33:07]Okay? From 1.5 to 2.25, this is an increase. And so, you know now that your graph is going to be an increasing function that lies above the asymptote.

[00:33:17]And so, it looks like this. That is the graph you have. Okay, so I don't really need any of these. I just wanted that so that I can understand the shape of the inverse. Okay, so if we know what a rough sketch of the graph of H would look like, then you would know that the inverse would do this. You see how the X and the Y values just swap? This is exactly what will happen here as well.

[00:33:39]So, here if you had to put down a Y axis, it could be somewhere like here.

[00:33:42]Do you see how your graph, right? If you look at the X axis, so whatever the graph does on the X axis, it's now going to do on the Y axis. Okay, so let's put down the Cartesian plane. So, you can look at it like this. So, look at this part, the negative part of the X axis, and then look at the positive part of the X axis. Do you see how the graph, okay, comes from being really close to that X axis, the negative part of it, and then it starts to move away from it.

[00:34:10]Now, the inverse is going to do exactly that. It's just going to swap the X and Y values. So now with the inverse, you're going to have to imagine this is your Y axis. Okay? So, here's a Y axis.

[00:34:23]It means that if you look at the negative parts of your Y axis towards the positive part, your graph must also be moving away from this axis. So, it's going to look like this. It's going to come from being really close to it and then it's going to move away. And that is what your inverse function is going to end up looking like. Okay, so we can just draw a rough sketch of it. Okay, so like that and now we need to put down those points they asked for. So, they said put down the x-intercept and then put down this point. Okay, so to find the x-intercept, this is the inverse, okay? This is the inverse function. So, the x-intercept on the original would not be an x-intercept. This x-intercept, if you looked at the original graph, it would be a y-intercept because remember that the x and the y values swap. So, whatever your y-intercept is here, that would serve as your x-intercept here.

[00:35:16]So, what is the y-intercept of this function? You simply just say make x equal to zero. To find y-intercept, you make x equal to zero. And so, if you say 3/2 raised to the power of zero, you get one. Okay, so we can see that that's just going to be one. And so, here the x-intercept would also be one. So, it means that it cuts the x-axis at one.

[00:35:41]Okay, you can say one and zero or just one like that. And then, we also have to put down that other point they spoke about. That was the 3/2 and one. 3/2 is 1.5, so it's like somewhere here and then one. So, let's just say this is one. And there we go. So, this is T, I'll call that double prime. So, 3/2 and one. And this is our graph. Okay, so from this you see that you don't even need this part. But I know that they would have given you guys like graph paper to work on this. So, that is it for that question. And that is the last question for question four and moving on to question five. We are now on question five and question five says, "The graphs of f of x, which is you can see this is a hyperbolic function and then we have g of x is equal to a half x plus four. So that's a straight line. Okay, so there's your hyperbola and your straight line.

[00:36:36]So they are drawn below and then a and b are the x and y intercepts of f respectively and then we have a is also a point of intersection of f and g.

[00:36:48]Okay, right there. So they say write down the domain of f. So the domain of a function refers to all its x values. So all the x values that this function will have. So a hyperbola will have almost every x values. You can see where that graph looks. So look at the x axis and just see all the points that it maps to.

[00:37:10]So it has x values here, okay, here, here. If you map this to the x axis, there it has an x value over here. It also has here. It will have x values at almost every point, okay? It will have x values except for one particular case.

[00:37:27]And that would be wherever your vertical asymptote is. So if your vertical asymptote passes a particular number, that's where the graph will just never have an x value there because it never reaches that point, okay? It never has this y value. It's undefined at that point. So to find that you can simply use the equation of your function.

[00:37:47]Remember this is a hyperbola. It's written in the form of y equals a over x minus p plus q, okay? And so that value of p, that's what tells you your vertical asymptote. Right now we have x plus two. And so it means that the value of p would simply be you just look at this as x minus negative two. That's what gives you x plus two. So p is actually negative two. As you can see that fits that perfectly. Okay, so that means that this is the x values will never have negative two. And so how we write its domain would be like this. You can simply just say X is an element of all real numbers, but X is just not going to be equal to -2. And that's our answer. Moving on to the next question.

[00:38:32]The next question we have to find the coordinates, calculate the coordinates of B. Okay, so B, what would be the coordinates of B? Okay, so do you see that B, okay, is on the Y axis and it is on this graph, okay? This is where the graph cuts the Y axis. So this is the Y intercept. How do you find the Y intercept? You simply make X equal to zero. Okay, so this is our graph of F.

[00:38:54]We're going to take that graph, I'll rewrite it here, and we're finding the Y intercept, so we simply make X equal to zero. Okay, so f of zero equals -6 over and you're saying zero plus two minus one. So f of zero is equal to that's just going to be -3 minus one, which is -4. And so the coordinates of B, the X value, remember we made that equal to zero. Okay, the X value is zero, the Y value is -4, and that is our answer.

[00:39:22]Moving on to the next question.

[00:39:24]The next question asks us 5.2.2 for the coordinates of A. So, do you remember how they told you that A, right? A and B are the X intercepts of F respectively.

[00:39:35]Okay, so this is A and then they also said that it's a point of intersection between F and G. So if this point lies on the X axis and it lies on both graphs, then just use the easier graph to find it. Okay, so the easier graph would be the straight line. So just use that. It is the X intercept of our straight line graph and of course our hyperbola, but we're just going to use the straight line cuz it's easier to work with that.

[00:40:00]Okay, so we're just going to do that. So let's go ahead and take the graph of G and we're looking for an X intercept, so you need to make Y equal to zero. So g of X, okay, we make that equal to zero.

[00:40:12]A half X plus four. And then you're going to take the four over, so we have -4 = 1/2 x. And so how do you solve x?

[00:40:20]You divide both hand sides by 1/2, or you can multiply by two, okay? Cuz that's going to cancel that two out in the denominator. So finally, x is simply equal to -8, okay? So as a coordinate, that point is simply just going to be A is -8, and the y value, remember, we made that equal to zero. So, -8 and zero, because this is an x-intercept. Moving on to the next question.

[00:40:47]Next question 5.3 says, "H is the axis of symmetry of F that has a positive gradient. Determine the equation of H in the form of h of x is equal to." So, remember that a hyperbola, there are two axes of symmetry you can have, okay? You can have this one where the equation is y = and you just say x - p + q, okay? If you remember the general form of your hyperbola, then you remember this, because it's just the x - p, and then plus the q. And then there's the one with the negative gradient, but you see that right now we're focused on this one where the gradient is positive, okay?

[00:41:22]The gradient is the number in front of x, so y = mx + c, so that number. So, this is all we need. y = x - p + q. So, you need to know your p and your q value. And we actually have that here.

[00:41:39]So, do you remember how I wrote this like this, just so that you can easily see the value of p? Okay, that that is -2. Then what is the value of q?

[00:41:48]The value of q is -1, okay? So, we have those. We know that p is -2, and q is simply -1. So, the equation of this, which we must write as h of x is equal to, it is just x minus the -2, and then minus the one, okay? Plus -1, which is just minus one. So, finally, we'll say h of x is equal to and we're just going to have x and this is becomes plus two minus one and so finally h of x is equal to x plus one and that is our equation. Moving on to the next question.

[00:42:27]Question 5.4 says, calculate the coordinates of the point of intersection of g and h. Okay, so this is the yeah, the axis of symmetry with the positive gradient and the graph of g. So, the graph of g is a straight line. The axis of symmetry with a positive gradient also has a, you know, positive gradient like this one. So, we just need to find out where they intersect. If you're trying to find a point of intersection, this is the point where the y values of the graph are exactly the same and so we're just going to make the y values equal. So, you're going to take h of x, you're going to equate them. Okay, that stands for the y values of h or the y values of h and then it's equal to g of x. And so we have x plus one is equal to and there this one, the graph of g, its equation was this, okay, 1/2 x plus four. So, we'll just say a half x plus four. And then just solve for x and y, okay? So, we'll solve for x here and then we'll find the y value. So, x and then you're going to take that over so it's negative a half x is equal to and you have four, bring the one over, becomes negative. So, this becomes a half x is equal to three.

[00:43:40]Then you can divide both hand sides by a half or we can multiply both hand sides by two to get rid of that two in the denominator. So, finally here, x is simply equal to six.

[00:43:51]Then you need to find the y value. Now, remember, it's a point of intersection between two graphs. For example, if we're looking at this, okay, the point of intersection between these two graphs is going to be whatever that y value is.

[00:44:03]Do you see that because they lie on the the graph, you can substitute that X value into either one of the graphs and it will work out. So, substitute this X value into H or G and you'd find your Y value. So, I'll just use the easier one.

[00:44:16]I'll use H, so H of X is finally just H of 6. I'm substituting 6 as the X value is equal to 6 + 1, which is 7.

[00:44:27]And so, that coordinate is 6 and 7 and that is our answer. We'll be on to the next question.

[00:44:34]Okay, question 5.5 says, "Use the graphs to determine the values of X for which f of x + g of x is greater than or equal to 0."

[00:44:44]The minute you see them adding or subtracting things, just take the one graph and transpose it. So, this equation is actually f of x is greater than or equal to when that goes over, it will become a negative g of x.

[00:45:00]So, we need to understand what this will look like and if this should really be the graph that should be negative, is it going to help? So, the graph of g. Okay, that's our straight line. You see, if we put a negative in front of g of x, this is what is going to happen to it and you should have seen this from the previous question. If you say negative g of x, you are taking the Y values of g and you're multiplying them by -1. That would mean you are reflecting this graph, okay? You're reflecting it across the x-axis. So, what would this look like? You know that all the Y values just become negative. So, because this is a straight line, so yeah, we're working with the right graph. It's good to actually make this one negative instead of the graph of f. So, what we're going to do here is to make sure that we can reflect it properly, you can take two points cuz it's a straight line. You only need two points to join the line. So, you can take two points that would be important here. I'll take the x-intercept and the y-intercept.

[00:45:59]So, this is a straight line. It's written in standard form. You can see the Y intercept. It's always going to be that constant value, that plus C. Okay?

[00:46:07]Y equals MX plus C. And so, this value of four, that is what we have here.

[00:46:14]Remember how this means we are reflecting across the X axis. And so, it makes all the Y values to become negative. Okay? They get multiplied by negative one. So, this as a coordinate, it would have been zero and four.

[00:46:30]What is going to be the new points on the graph that has been reflected across the X axis?

[00:46:36]That would be zero and negative four.

[00:46:39]Now, we already had We actually had a negative four. Do you see that? B was zero and negative four. So, that means that point is going to be here. Okay?

[00:46:47]It's going to intersect with the graph of F. And then, this point, this is an X intercept, which we already had. Okay?

[00:46:54]So, A was negative eight and zero. So, if this is negative eight and zero, remember all the Y values change. So, if zero gets multiplied by negative one, it stays zero. So, that point stays that.

[00:47:06]And so, your graph is just going to look like this. It's just going to be from there to there, and then you have your graph.

[00:47:13]Let me just draw it properly. Okay? So, this is what we have, and that's our graph. Okay? We just join them. So, I know that this is minus four.

[00:47:21]Okay.

[00:47:22]So, the question here is if we take that graph, and then we reflect it across the X axis, we need to find the X values for which f of X is now greater than or equal to negative g of X. Okay? So, you're looking for all the places where your graph of F has greater Y values, bigger Y values than this pink graph, this new graph that has been reflected.

[00:47:49]Now, simply put, that would just be all the places where that graph lies above the graph of this pink graph. Okay? The graph of F must be above the graph of this reflected G. Okay, so we don't really need the vertical line here. You can just do this. We're going to look all through and just look and look for where the graph of F is above the graph of G. So starting here, do you see what's going on here? Look where the graph of F is. It's below this graph, this pink graph. So we can't take any of these [clears throat] points, okay?

[00:48:19]Because if you line the two to the Y axis, do you see that this this has a lower Y value than whatever this Y value is. So we can't take that.

[00:48:30]The graph of F must be above the graph of that reflected one. Okay, so G is reflection across the X axis. So it wouldn't be there.

[00:48:40]But then here's the point where they intersect. So remember they asked for where the graph of F, right? F of X must be greater than or equal to negative G of X. So we can also take where they're equal. They are equal at A cuz this is where they intersect. Okay? So they equal at A. They also equal here at this point, okay? Where we have zero and negative four. So at this point. And so the graph of F is greater than the pink graph all through here. So look at all the places. See? This is where it is above that pink graph. And then we have an asymptote after this. And then if you keep going here, then look at what happens here. See, this is the graph of F. Unfortunately, it is currently below the pink graph. We don't want that. And then we're going to take where they're equal. And then again the graph of F must be above the pink graph. And this is what we're looking for. So these highlighted sections, we're going to find out what exactly their X values are. So if you take these and you line it to the X axis, so I understand. So it starts at negative eight. And then do you see that if you line them to the X axis, these are all values that are simply greater than negative eight. They're moving to the right hand of -8. Here's -8. Values greater than -8, but then here we have an asymptote, so it stops here, cuts off here. So, this would be where x is greater than or equal to -8 because there's a point of intersection here, but less than -2. You can't say equal to -2, there's an asymptote here.

[00:50:12]Okay, the graph of f doesn't even have a y value at that x value. So, we're done with that side. Now, this side, okay?

[00:50:18]So, look at the x values again. We're trying to line it to the x-axis and then stand. Okay, so look at the x values.

[00:50:24]What are these x values? So, it starts here at this point. The x value that goes with this is an x value of 0. And then if you keep lining all the highlighted sections, do you see where it's moving? Okay? It's moving to the right. And do you see that this graph, these graphs they'll continue like this forever. This graph is going to shoot down like that forever, and this one is going to move closer and closer to this asymptote, but never touch it. So, do you see that all through here the graph of f is always going to be above this graph, okay, this pink graph. So, we can see that it starts at where x is equal to 0 and it's moving towards all these values that are greater than 0. So, you can say here x is greater than 0. Okay, in interval notation, you're just saying all the parts we've selected go from -8 all the way to -2, so you'd say x is an element of and you'll say -8 to -2. But by -8, do you see that here -8 is an included x value. X is greater than or equal to -8, so you're going to use a square bracket here, then here a round bracket. And then you can say or you just say union, okay, to just include this solution. This is where x is greater than 0. So, it starts at 0 and it's heading to positive infinity, okay?

[00:51:43]So, we're just going to say starting at 0, and yes, 0 is an included value here. So I can't just say greater than zero. Okay, I have to say greater than or equal to zero because at this point we have a point of intersection. I almost missed that. Okay, so then you're going to say it's from zero all the way to positive infinity. So by zero use a square bracket because the X value of zero is included but infinity is not a fixed value so use a round bracket. So these are the two ways you can write it.

[00:52:14]So like this and you say or and of course put that one as well and then you can also write it like this and that's fine. Okay, so these are answers you can write it in set builder notation or interval notation and that's fine.

[00:52:27]Moving on to the next question.

[00:52:29]The next question says 5.6 the last question for functions. It just says a right angle triangle is created having A as one of its vertices. Okay, let me just get that diagram first and then we can work on this. Okay, so they tell us that a right angle triangle is created having A as one of its vertices and then we have D which is K and zero as another vertex and G as one of its sides. Okay, so A is going to be one of the points that we have on this triangle and then there is this point K, okay, which is going to be another point on the triangle. So a triangle has it's made up of three vertices but now do you see that one of its sides is the graph of G. So this graph of G. So looking at this graph, okay, one of its sides is this graph. They want us to calculate the values of K when the area of the right angle triangle is 64 units squared. Okay, so 64 square units. With this we have to now draw this triangle. We know that it has one point at A, then it uses this, okay? This line G, and then it has a point D, which is K and 0. Okay, that actually helps a lot. So, K and 0. Do you see that the Y value is 0? Which point has a Y value of 0? That would be an X intercept. Okay, this is where we make Y equal to 0. So, that is an X intercept, meaning it looks something like this. So, there's a point over here, it goes to G, and then down, okay?

[00:54:10]I don't know where. Doesn't go down here, but maybe extended like this. It just depends on where K is, whether K is a positive or negative value. Okay, but let's say it looked like that. So, we would have this, and then that point D they were referring to, that would be down here. So, this would be D, which is K and 0. Okay, so, this is the triangle they're creating. It is a right-angled triangle, like they said. And now we have to use this info to calculate the values of K when the area of the triangle, the right-angled triangle, is 64 units. Okay, it's a right-angled triangle, so it's very simple to find the area of this triangle. To find the area of a right-angled triangle, you just say area of a triangle is equal to a half multiplied by whatever its base is, and then multiplied by whatever its height is. So, we need the base and the height. Those are the two things we need.

[00:55:08]Looking at this triangle, the base is going to go from -8 all the way to K.

[00:55:15]So, if you actually had this value here, to find the length of this base AD, technically, that's what it's going to be called, AD. To find AD, we would simply just say the higher X value minus the lower X value. Okay, so if like this was two, you would know that, okay, I just went two units to to right, and then eight units to the left, And then that's a total of 10 units. But then an easier way to find that is just to say two minus -8, okay? Cuz the coordinate is -8. So, we would just say, in the same way, this x value of K, okay? Minus -8. And so AD, our base, is K + 8. That's what we have there. Okay, we have the base. What else do we need?

[00:56:01]We need the height. So, we've got the base. We need the height. Okay, so now how do we get the height? So, the height is whatever this length is, which we cannot see right now, but we can actually find an expression that will serve as the height. So, here's the thing. You see, wherever this point is, I don't know where it is. I'm just like saying, maybe it's here, okay? It could be on this side. This triangle may be way smaller than this. We'll see. It just depends on the value that K is, okay? If K is positive or negative. So, to find that height, you would have to find a y value here. Because you see that this is a vertical line. To find the length of a vertical line, okay? If you knew what this was, you take whatever its y value is, say it was four, okay? You'd say four minus whatever this point's y value is. In this case, this point's y value is zero.

[00:56:53]So, you just say, "Oh, 4 - 0, the height is four." But in this case, we know that it's not four. They didn't tell us anything like that. So, we need something to stand as the y value of this point.

[00:57:04]Use the fact that this point lies on the graph of G.

[00:57:08]Okay? Look at the equation of G. It says g of x is equal to this, okay? g of x stands for y. That means this 1/2 x + 4.

[00:57:19]It stands for all the y values of G, depending on what their x value is.

[00:57:26]Right now, let's look at that point, okay? Since this is a right angle triangle, it would mean that you see this point and the point D, they would have the same x value. That is the only way this can be a right angle triangle.

[00:57:40]Okay, if they share the same x value.

[00:57:42]And so, right now we can call this point K and then it's y value would just be this. Okay, a half x plus four. But then, to find its y value, you're just going to say, "Okay, I know the x value is K, so I'm going to substitute that in this equation in place of x, okay?" To find its y value. So, it'll be a half and then in place of x you substitute K plus four. So, this is what it will be, a half K plus four. And then now, how do you find that height? So, you find the height of a vertical line by taking the higher y value minus the lower y value.

[00:58:22]And that would be it. So, we're just going to take this minus zero, which as you can see, it's just going to stay this. So, we can finally just conclude that the height of this triangle is simply just a half K plus four units.

[00:58:37]Okay, that's what it is. Okay, so we've got the base, we've got the height, we can go ahead and work with this. So, they said the area of this triangle, so area of the triangle is equal to and we're saying a half multiplied by its base, which is K plus eight, so we'll say times K plus eight, multiplied by its height, which is a half K plus four.

[00:58:59]Then this ends up being equal to, they said that that area is 64 square units.

[00:59:07]So, we're going to just say, 64 is equal to and then all of this. Okay, I just wanted to write that like that, but now I'm just seeing, you see this? You can do some factorizing here. You can take out a half as a common factor. So, I'll take a half out as a common factor just in this bracket, so I'll leave this K plus eight. If you take a half out as a common factor, you're going to ask yourself, what do I multiply to a half to get back to a half? Okay? Or you can just say this divided by a half. That's just going to give you K. And then, if you say 4 divided by a half, that gives you 8. So, see how you have the same thing there? Okay, so this is 64 is equal to a quarter, and then this is just same thing times the same thing. In the same way x times x is x squared.

[00:59:54]Then, if you have the same thing times the same thing, it is that thing all squared. So, this is what we have, and we can just get rid of the 1/4. So, divide, or really multiply both hand sides by 4. That will get rid of this 4 in the denominator, because this one is in the numerator. And then, we have K plus 8 squared is equal to, and we know that 64 times, um, two. Okay, 64 plus 64 is 128, and then times two is 256. So, this is 256 that we have here. Now, we can finally solve for K. So, we need to get rid of the square to get to the K.

[01:00:30]So, you would simply square root both hand sides of the equation. Okay, what is the square root of 256? It is 16. So, I'm just going to say this is K plus 8.

[01:00:43]Now, remember, when you're square rooting both hand sides in an equation, okay, you need to make sure that you remember the positive and the negative solution. Because if I say x squared is equal to 4, x is not just equal to 2.

[01:00:56]Okay? This is an equation, meaning you have, here, this is, uh, quadratic equation. There are two possible solutions. So, when you square root, you need to consider the positive and the negative solution. So, x is equal to positive 2, or x is equal to negative 2.

[01:01:11]Okay? Cuz you're saying something squared gives you 4. 2 squared gives you 4. Yes, but negative 2 squared also gives you 4. Okay? So, don't forget that plus minus. And so, this is equal to positive or negative 16. And finally, K is equal to we'll take that eight over becomes negative. So, it is negative eight plus minus 16. Therefore, K is either equal to negative eight plus 16 or K equals negative eight minus 16. And then you just get your answers. Okay, so here K would be equal to eight or here K would be equal to negative 24.

[01:01:52]And these are our answers. Okay, K equals eight or K equals negative 24.

[01:01:57]They said calculate the values of K.

[01:02:01]And those are the values of K. Okay. So, that is a definite one. I won't say that they have never given something like this, but yeah, it's not typical that they will give like a shape. Okay, so that is our answer. That is the last question for question five. And that concludes it for this video, the function section. And I will see you guys in the financial math section.