A Level Maths, KS5

Quadratic Graphs – A Level Maths Revision

The good news is that, at A Level, there is no new content on Quadratic Graphs. However, you will need to have a thorough understanding of the GCSE content.

To further revise GCSE content on linear and quadratic graphs, try this preparation for A Level resource or these multiple-choice prior knowledge questions. If you think you’ve grasped it, try this A Level task to test your understanding, or these exam-style questions on the whole of AS level coordinate geometry.

All quadratic graphs are parabolas: a symmetrical curve. The orientation of this parabola depends on whether the coefficient of x2 is positive or negative.

Quadratic graphs - coefficient of x²

You can use your skills working with quadratic equations to find other key features of the graph:

  1. The y-intercept. This is found by setting x = 0 and solving the given equation for y.
  2. The x-intercept(s). This is found by setting y = 0 and solving the given equation for x. There may be 0, 1, or 2 x-intercepts.
  3. The turning point. For a quadratic function, this will be the minimum (lowest point – if the coefficient of x2 is positive) or the maximum (highest point – if the coefficient of x2 is negative). This can either be found by working out the average of the x-intercept values (where they exist) or by completing the square. A graph with an equation rearranged into the form y = a(x + b)2 + c has a turning point at (-b, c).

Example Question

Sketch the graph of y = x2 – 7x + 10, clearly indicating any points of intersection with the axes and the location of the turning point of the curve.

The coefficient of x2 is 1. This is positive, so our graph will be u-shaped.

The y-intercept is found by substituting x = 0 into the equation:

y = x^2 - 7x + 10

y = 0^2 - 7 \times 0 + 10

y = 10

The x-intercepts are found by substituting y = 0 then solving the resulting equation:

y = x^2 - 7x + 10

0 = x^2 - 7x + 10

0 = (x - 2)(x - 5)

x = 2, x = 5

To find the turning point, either find the average of the two x-intercepts or complete the square.

Finding the average gives us an x-coordinate of:

\frac{2\,+\,5}{2} = \frac{7}{2}

And a y-coordinate of:

y = x^2 - 7x + 10

y = (\frac{7}{2})^2 - 7 \times \frac{7}{2} + 10

y = \frac{49}{4} - \frac{49}{2} + 10 = \frac{49}{4} - \frac{98}{4} + \frac{40}{4} = -\frac{9}{4}

Giving a turning point of (\frac{7}{2}, -\frac{9}{4}) .

Finding the turning point using completing the square will give the same value:

\begin{aligned} x^2 - 7x + 10 &= (x - \frac{7}{2})^2 - (\frac{7}{2})^2 + 10 \\ &= (x - \frac{7}{2})^2 - \frac{49}{4} + \frac{40}{4} &= (x - \frac{7}{2})^2 - \frac{9}{4} \end{aligned}

As a graph with an equation in the form y = a(x + b)2 + c has a turning point at (-b, c), the turning point in this example is (\frac{7}{2}, -\frac{9}{4}) .

The question also asks you to sketch the graph. When sketching a graph, it does not need to be to scale but should be the right shape and roughly in proportion, as shown:

Sketching a graph

Practice Questions

For each graph, find the x-intercept, y-intercept and turning point:

1. y = x2 + 4x – 5.

2. y = x2 + 8x – 9

3. y = –x2 + 7x

4. y = 2x2 + 17x + 8

5. y = 5x2 – 20x + 15

1. x-intercepts: (-5, 0), (1, 0); y-intercept: (0, -5); turning point: (-2, -9);

2. x-intercepts: (-9, 0), (1, 0); y-intercept: (0, -9); turning point: (-4, -25);

3. x-intercepts: (0, 0), (7, 0); y-intercept: (0, 0); turning point: (3.5, 12.25);

4. x-intercepts: (-0.5, 0), (-8, 0); y-intercept: (0, 8); turning point: (-4.25, -28.125);

5. x-intercepts: (1, 0), (3, 0); y-intercept: (0, 15); turning point: (2, -5);

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