# Video: AQA GCSE Mathematics Higher Tier Pack 3 • Paper 2 • Question 6

(a) Complete the table of values for the function 𝑦 = 3 − 𝑥 − 𝑥². (b) Draw the graph of 𝑦 = 3 − 𝑥 − 𝑥² for 𝑥-values between −2 and 2. (c) What is the 𝑥-coordinate of the turning point of the graph?

04:16

### Video Transcript

Part a) Complete the table of values for the function 𝑦 equals three minus 𝑥 minus 𝑥 squared. Part b) Draw the graph of 𝑦 equals three minus 𝑥 minus 𝑥 squared for 𝑥 values between negative two and two. And part c) What is the 𝑥-coordinate of the turning point of the graph?

So for part a, what we need to do is substitute the values of 𝑥 from the table into our function. So first of all, we’re gonna start with 𝑥 equals negative two. So we can have 𝑦 is equal to three minus then we’ve got negative two then minus then we’ve got negative two all squared.

So then we’ve got 𝑦 equals three add two. That’s because we got three minus a negative. And if you minus a negative, it turns into a plus. Then minus four, and that’s because negative two squared is four, because a negative multiplied by a negative is a positive.

So therefore, we’re gonna get 𝑦 equals one, because three add two is five, minus four is one. So that’s our first value filled in. So we’re gonna move on to the second value of 𝑥, which is 𝑥 equals zero.

So if we substitute in 𝑥 equals zero, we’re gonna get 𝑦 equals three minus zero minus zero squared. So therefore, 𝑦 is gonna be equal to three. And then we have 𝑥 is equal to one. So if we substitute in 𝑥 is equal to one, we’re gonna get 𝑦 is equal to three minus one minus one squared. So we’re just gonna get 𝑦 is equal to one. That’s because you have three minus one and then minus another one, which gives us one.

And then finally we’re gonna substitute in 𝑥 equals two. And we’re gonna get 𝑦 is equal to three minus two minus two squared, which gives us 𝑦 is equal to three minus two, which gonna be one, then minus four, which gives us negative three. So therefore, we finished part a and completed the table of values. Now let’s move on to part b.

In part b, we’ve got to draw the graph. So to draw the graph, what we’re gonna do is mark on the points. So the first one is negative two, one. So remembering that we go along the 𝑥-axis then up the 𝑦-axis, then we’ve got negative one, three. Then we have zero, three. Then we have one, one then finally two, negative three.

So the next thing we need to do is draw a smooth curve or parabola. One thing we can notice is that our parabola or curve is gonna be an n shape or inverted U. And this is correct because we’ve got a negative 𝑥 squared term in our quadratic. And if it’s negative 𝑥 squared, then it’s an inverted U or an n shape. If it’s not, if it’s positive 𝑥 squared, then it’s a U-shaped parabola.

So we’ve done that and drawn the graph between 𝑥 values negative two and two. So that’s part b completed. Now let’s move on to part c.

For part c, what we need to do is identify the 𝑥-coordinate of the turning point of our graph. Well, first of all, we have to have a look at what the turning point is. And I’ve marked it on our parabola or our graph. And the turning point is a point at which the gradient will change from positive to negative or negative to positive. And it’s also a maximum or minimum point on the graph. But it is in this type of graph, and we can see that it’s the maximum point of our graph.

So we can see this is the turning point of the graph because, first of all, it’s the maximum point of the graph and we can see. But also our graph is a symmetrical parabola. And we can see that the distance between the two points and the turning point that I’ve identified is 0.5. So therefore, it’s going to be the point in the middle of these two.

So therefore, if we read down and onto the 𝑥-axis on our graph, we can see that this turning point is at the 𝑥-coordinate negative a half. So therefore, we can say that the 𝑥-coordinate of the turning point of the graph is negative a half.