## What you need to know

We’ve had a look at drawing quadratic graphs and some of their important features here (https://mathsmadeeasy.co.uk/gcse-maths-revision/plotting-quadratic-and-harder-graphs-gcse-revision-and-worksheets/) and here (https://mathsmadeeasy.co.uk/gcse-maths-revision/solving-quadratics-factoring/), and now we’re going to introduce a new feature: turning points.

The turning point of a graph (marked with a blue cross on the right) is the point at which the “turns around”, i.e. it goes from having a downward slope to having an upward slope. On a positive quadratic graph (one with a positive coefficient of $x^2$), the turning point is also the minimum point. In the case of a negative quadratic (one with a negative coefficient of $x^2$) where the graph is upside-down, it is the maximum point.

In this topic, we will look at two methods for finding the coordinates of the turning point:

– Using the symmetry of the graph;

– (HIGHER ONLY) Using completing the square.

Firstly, we will look at how to use the symmetry of a quadratic graph. To do this, we must understand that a quadratic graph has a single, vertical line of symmetry, and it passes right through the turning point. As a result, the two roots of the quadratic are equidistant to the turning point, or more usefully, the $x$ coordinate of the turning point sits right in the middle of the $x$ coordinates of the two roots.

Example: Use the symmetry of the graph to find the coordinates of the turning point of $y=x^2-3x+2$.

Firstly, we should factorise this quadratic and set it to zero to find the locations of the roots. Observing that $(-2)\times(-1)=2$ and $(-2)+(-1)=-3$, we get that

$x^2-3x+2=(x-1)(x-2)$

Setting this equal to zero gives

$(x-1)(x-2)=0$

This equation admits two roots: $x=1$ and $x=2$. Therefore, the $x$ coordinate must be 1.5, since it is halfway between 1 and 2. If it isn’t obvious what the halfway point between two values is, simply add them up and divide by 2.

Now that we’ve found the $x$ coordinate of the turning point, what remains is to the find the $y$. To do this, we need to substitute $x=1.5$ into the equation. Using a calculator, we get

$y=(1.5)^2-3(1.5)+2=-0.25$

Therefore, the coordinates of the turning point are $(1.5, -0.25)$. As you can see from the graph of the function, this answer looks right.

The remainder of this topic is higher only. Before reading on, make sure to click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/completing-square-gcse-maths-revision-worksheets/) to ensure you understand how to complete the square of a quadratic.

Before we get right into an example, we must make the following observation. If we’re looking for a minimum point on a graph (remember, the turning point of a positive quadratic is the minimum point), then we’re looking for the smallest value of $y$ that the graph reaches. Since the graph is determined by its equation, this is also the smallest value of $y$ that the equation can take. Let’s see a couple of examples.

Example: Use completing the square to find the coordinates of the turning point of $y=x^2+4x-12$.

Firstly, we must complete the square. Given that half of 4 is 2, our quadratic becomes

$y=(x+2)^2-4-12=(x+2)^2-16$.

Now, this is a positive quadratic, so we are looking for a minimum point. To find the coordinates of the minimum point, we will ask: what is the smallest value this quadratic can take?

The right-hand side of our equation comes in two parts. The first part, $(x+2)^2$, is always going to be positive no matter what $x$ we put in, because the result of squaring a real number is always positive. If $(x+2)^2$ is never negative, then the smallest number it can be is zero, meaning that the minimum value of the quadratic is $y=0-16=-16$.

Now we know the $y$ coordinate of our minimum point, what is the $x$ coordinate? The question is, what value of $x$ must we put into our equation to make $(x+2)^2=0$? The answer is $x=-2$. Therefore, the coordinates of our minimum point are $(-2,-16)$. Try plotting the graph to see for yourself.

Example: Complete the square to find the coordinates of the turning point of $y=2x^2-20x+14$.

To complete the square on this, we first take a factor of 2 out of the whole quadratic:

$y=2(x^2+10x+7)$

Now, we complete the square on the inside section. Half of 10 is 5, so we get

$y=2\left[(x+5)^2-25+7\right]=2\left[(x+5)^2-18\right]=2(x+5)^2-36$

Having the extra number in front of the bracket doesn’t actually change anything. We’re still looking for a minimum point, and the minimum value that $2(x+5)^2$ can take is zero, so the minimum value of the quadratic is $y=0-36=-36$.

The $x$ coordinate that is required to get this minimum value has to make the expression $2(x+5)^2$ equal to zero, so it must be $x=-5$. Therefore, the coordinates of the turning point (minimum point) are $(-5, -36)$.

If you aren’t sure about this, have another read of these two examples to make sense of it. Once you’ve got used to it, it just becomes a process of reading off the coordinates of the turning point once you have completed the square.

So, the main thing to remember is, when the result of completing the square is

$a(x+b)^2+c,$

the turning point of $y=a(x+b)^2+c$ has coordinates $(-b, c)$.

## Example Questions

#### 1) Sketch the graph of $y=x^2+5x+6$, clearly marking on the coordinates of the roots and of the turning point.

Firstly, we must find the roots of this quadratic by factorising it and setting it equal to zero. Observing that $2\times3=6$ and $2+3=5$, we get that

$x^2+5x+6=(x+2)(x+3)$

Therefore, to find the roots we set

$(x+2)(x+3)=0$

This clearly admits two roots: $x=-2$ and $x=-3$. Then, to find the coordinates of the turning, we need the halfway point between the roots, which is

$\dfrac{-2+(-3)}{2}=-2.5$

This is the $x$ coordinate of the turning point. To find the $y$ coordinate, we put this value back into the equation to get

$y=(-2.5)^2+5(-2.5)+6=-0.25$

Then, the resulting sketch of the graph should look like

Note: because the question asked you for a sketch, it doesn’t have to be a perfect drawing, it just has to have the correct shape and the correctly identified & labelled roots and turning points.

#### 2) (HIGHER ONLY) Use completing the square to find the coordinates of the turning point of$y=x^2-3x+11$

The coefficient of $x$ is -3, and half of -3 is $-\frac{3}{2}$. Therefore, completing the square, we get

$x^2 - 3x + 11 = \left(x - \dfrac{3}{2}\right)^2 + 11 - \left(-\dfrac{3}{2}\right)^2 = \left(x - \dfrac{3}{2}\right)^2 + \dfrac{35}{4}$

Thus, the coordinates of the minimum point are $\left(\frac{3}{2}, \frac{35}{4}\right)$.

#### 3) (HIGHER ONLY) Use completing the square to find the coordinates of the turning point of$y=-x^2+10x-3$.

The difference here is that this is a negative quadratic, so the graph is going to have a maximum value rather than a minimum. In terms of completing the square, all we have to do is take a factor of -1 out of the quadratic

$-x^2 + 10x - 3 = -(x^2 - 10x + 3)$,

and then complete the square on the inside of the bracket as normal. So, half of -10 is -5, therefore

$-(x^2 - 10x + 3) = -\left[(x - 5)^2 + 3 - (-5)^2\right].$

This simplifies to

$-\left[(x - 5)^2 - 22\right] = -(x - 5)^2 + 22$.

Here we can see why it’s a maximum point: the first part of the expression has a minus sign in front of it, meaning it is the negative of a square so must always be negative. So, the maximum exists where $-(x-5)^2$ is zero, which means that coordinates of the maximum point (and thus, the turning point) are $(5, 22)$.

## Turning Points of Quadratic Graphs Revision and Worksheets

Completing the Square and turning points
Level 8-9

## Turning Points of Quadratic Graphs Teaching Resources

GCSE Maths tutors who are looking for exceptional resources for the specific topic of turning points for quadratic graphs can view and access many worksheets and online tests. These resources work really well as part of a tuition course where material is covered in lessons and worksheets for specific topics like quadratic graphs can be set for homework.