## Where to?

– Worksheets and Exam Questions

– Revise Turning Points of Quadratic Graphs

## Turning Points of Quadratic Graphs

Graphs of quadratic functions have a **vertical line of symmetry** that goes through their** turning point**. This means that the turning point is located exactly half way between the x-axis intercepts (if there are any!).

There are two methods to find the turning point, Through **factorising** and **completing the square**.

Make sure you are happy with the following topics:

**What is the turning point?**

The turning point of a graph (marked with a **blue cross** on the right) is the point at which the graph “turns around”.

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**.

**Method 1:** Through Factorising

Use the symmetry of the graph to find the coordinates of the turning point of the following quadratic:

y=x^2-3x+2

**[3 marks]**

**Step 1: ** Factorise the quadratic, setting it to zero to find the locations of the roots.

Doing this we get:

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

Setting this equal to zero gives

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

Solving we get the following roots.

x=\textcolor{red}{1} and x=\textcolor{blue}{2}

**Step 2:** Find the x coordinate.

We find halfway between the to x coordinates

\dfrac{\textcolor{red}{1}+\textcolor{blue}{2}}{2} = \textcolor{green}{1.5}

Therefore, the x coordinate must be \textcolor{green}{1.5}

**Step 3:** Finding the y coordinate.

We substitute x=1.5 into the equation.

y=(\textcolor{green}{1.5})^2-3(\textcolor{green}{1.5})+2=\textcolor{orange}{-0.25}

Therefore, the coordinates of the turning point are:

(\textcolor{green}{1.5}, \textcolor{orange}{-0.25})

As you can see from the graph of the function, this answer looks correct.

**Method 2: ****Completing the Square **

Use completing the square to find the coordinates of the turning point of the following quadratic:

y=x^2+4x-12

**[3 marks]**

**Step 1:** Complete the square, this gives us the following:

y=(x+2)^2-4-12

(x\textcolor{red}{+2})^2\textcolor{blue}{-16}

**This is a positive quadratic, so we are looking for a minimum point.**

**Step 2:** Find the y coordinate of the minimum point.

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\textcolor{blue}{-16}=\textcolor{blue}{-16}.

The y coordinate will always be equal to the \text{\textcolor{blue}{correction number}}

y = \textcolor{blue}{-16}

**Step 3:** Find the x coordinate of the minimum point.

The question is, what value of x must we put into our equation to make (x+2)^2=0? The answer is x=-2.

The x coordinate will equal -1 \times \text{\textcolor{red}{Number inside the bracket}}

x = -1 \times \textcolor{red}{2} = \textcolor{orange}{-2}

Therefore, the coordinates of our minimum point are:

(\textcolor{orange}{-2},\textcolor{blue}{-16})

## Maths Exam Worksheets

£3.99### Example Questions

**Question 1:** Sketch the graph of y=x^2+5x+6, clearly marking on the coordinates of the roots and of the turning point.

**[2 marks]**

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 point, 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.

**Question 2:** (HIGHER ONLY) Use completing the square to find the coordinates of the turning point of

y=x^2-3x+11

**[2 marks]**

The coefficient of x is -3, and half of -3 is -\dfrac{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(\dfrac{3}{2}, \dfrac{35}{4}\right).

**Question 3:** (HIGHER ONLY) Use completing the square to find the coordinates of the turning point of

y=-x^2+10x-3

**[2 marks]**

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).

**Question 4:** Complete the square to find the coordinates of the turning point of y=2x^2+20x+14.

**[2 marks]**

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+7-25\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).

### Worksheets and Exam Questions

#### (NEW) Circle Graphs and Tangents Exam Style Questions - MME

Level 8-9 New Official MME### Drill Questions

#### Turning Points of Quadratic Graphs - Drill Questions

#### Completing the Square and turning points - Drill Questions

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.