National 5MathsSQARevision NotesAlgebraic SkillsQuadratic FunctionsFeatures of Quadratic Graphs

Features of Quadratic Graphs (SQA National 5 Maths): Revision Note

Exam code: X847 75

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 1 December 2025

Turning point & axis of symmetry of a quadratic graph

What does a quadratic graph look like?

A quadratic graph is a smooth curve with a vertical axis of symmetry
If the the equation of the graph is given in the form $y = a x^{2} + b x + c$ , then
- A positive number in front of $x^{2}$ gives a $\cup$ -shaped curve
- A negative number in front of $x^{2}$ gives a $\cap$ -shaped curve
The shape made by a quadratic graph is known as a parabola
A quadratic graph will always cross the $y$ -axis
A quadratic graph intersects the $x$ -axis twice, once, or not at all
- The points where the graph intersects the $x$ -axis are called the roots
If the graph is a $\cup$ -shape, then it has a minimum point
If the graph is a $\cap$ -shape, then it has a maximum point
Minimum and maximum points are both examples of turning points
- The turning point of a parabola is sometimes also referred to as its vertex

Diagram showing a positive quadratic curve with a minimum point and a negative quadratic curve with a maximum point.

How can I find the turning point and axis of symmetry of a quadratic graph?

The easiest way to find the turning point and axis of symmetry of a quadratic graph is by completing the square on its equation
- On such questions in the exam the equation will be given to you in the form $y = x^{2} + p x + q$
Complete the square to rewrite the quadratic in the form $y = {(x + a)}^{2} + b$
- If $y = {(x + a)}^{2} + b$ then the turning point is at $(- a, b)$
  - Notice the negative sign with the x-coordinate
- This can also help you create the equation of a quadratic when given the turning point
The axis of symmetry of a quadratic graph always goes through the turning point
- So if the turning point is $(- a, b)$ , then the axis of symmetry is $x = - a$

Completing the square Notes Diagram 3, A Level & AS Level Pure Maths Revision Notes

Worked Example

(a) Express $x^{2} + 10 x + 7$ in the form ${(x + a)}^{2} + b$ .

(b) Hence, or otherwise, state the coordinates of the turning point of the graph of $y = x^{2} + 10 x + 7$ .

The diagram shows the graph of $y = x^{2} + 10 x + 7$ .

A line PQ has been drawn parallel to the $x$ -axis, where:

P lies on the $y$ -axis
P and Q lie on the graph of $y = x^{2} + 10 x + 7$ .

Graph of a parabola opening upwards, intersecting the y-axis at point P. A horizontal line intersects the parabola at points P and Q. Axes are labelled x and y with origin at O.

Answer:

Part (a)

Use $x^{2} + p x = {(x + \frac{p}{2})}^{2} - {(\frac{p}{2})}^{2}$ to rewrite $x^{2} + 10 x$

Here $p = 10$ , so $\frac{p}{2} = 5$

$\begin{array}{rcl} x^{2} + 10 x & = & {(x + 5)}^{2} - 5^{2} \\ = & {(x + 5)}^{2} - 25 \end{array}$

Substitute that back into the original expression

$\begin{array}{rcl} x^{2} + 10 x - 17 & = & {(x + 5)}^{2} - 25 + 7 \\ = & {(x + 5)}^{2} - 18 \end{array}$

$\begin{array}{rcl} {(x + 5)}^{2} - 18 \end{array}$

Part (b)

Use the fact that if $y = {(x + a)}^{2} + b$ then the turning point is at $(- a, b)$

Don't forget the negative sign with the x-coordinate!

$(- 5, - 18)$

Part (c)
Answering this part of the question relies on the symmetry of the parabola

From part (b), you know that the turning point is $(- 5, - 18)$

That means the axis of symmetry is at $x = - 5$

Point P is the $y$ -axis intercept of the parabola, with $x$ -coordinate 0

You can find its $y$ -coordinate by substituting $x = 0$ into the equation of the curve

$y = {(0)}^{2} + 10 (0) + 7 = 7$

So the coordinates of point P are $(0, 7)$

Graph of a parabola opening upwards, intersecting the y-axis at point P. A horizontal line intersects the parabola at points P and Q. The turning point of the parabola has been labelled (-5, -18), and the vertical line x=-5 has been drawn through it. Point P has been labelled (0-, 7). Axes are labelled x and y with origin at O.

Line PQ is parallel to the $x$ -axis

So P and Q have the same $y$ coordinate

And point P is 5 units to the right of the axis of symmetry

So by the symmetry point Q will be five units to the left of the axis of symmetry

Point Q has coordinates (-10, 7)

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