Sketching graphs

Plotting a graph takes time. Often mathematicians just want to know the key features.

These are: shape, location and some key points (such as where the graph crosses the axes or turning points).

Aim to recognise the equations and graphs of quadratics, cubics, reciprocals, exponentials and circles.

Sketching a quadratic graph using factors

If a quadratic equation can be factorised, the factors can be used to find where the graph crosses the x-axis.

Example

Sketch y = x^2 + x − 6

The quadratic factorises to give (x + 3)(x − 2) so the solutions of the equation x^2 + x − 6 = 0 are x = -3 and x = 2.

The graph of y = x^2 + x − 6 crosses the x-axis at x = -3 and x = 2.

The coefficient of x^2 is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

To find where the graph crosses the y-axis, work out y when x = 0. So:

0^2 + 0 – 6 = -6 so the graph crosses the y-axis at y = -6.

Sketching equation y = x to the power of 2 + x − 6

Example

Sketch y = x^2 − 6x + 9

The quadratic factorises to give (x − 3)(x − 3) so the only solution of the equation x^2 − 6x + 9 = 0 is x = 3.

The graph of y = x^2 − 6x + 9 touches the x-axis at x = 3.

The coefficient of x^2 is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

To find where the graph crosses the y-axis, work out y when x = 0:

0^2 – 6 \times 0 + 9 = 9 so the graph crosses the y-axis at y = 9.

A sketch of y = x to the power of 2 − 6x + 9

Sketching a quadratic using the turning point and the line of symmetry - Higher

The coordinates of the turning point and the equation of the line of symmetry can be found by writing the quadratic in completed square form.

Example

Sketch y = x^2 – 6x + 4.

The coefficient of x^2 is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

Writing y = x^2 – 6x + 4 in completed square form gives y = (x – 3)^2 – 5.

Squaring positive or negative numbers always gives a positive value. The lowest value given by a squared term is 0, which means that the minimum value of the term (x – 3)^2 – 5 is given when x = 3. This also gives the equation of the line of symmetry for the quadratic graph.

The value of y when x = 3 is -5. This value is always the same as the constant term in the completed square form of the equation.

So the graph of y = x^2 – 6x + 4 has a line of symmetry with equation x = 3 and a minimum turning point at (3, -5).

When x = 0, y = 4. So the graph crosses the y-axis at (0, 4).

A sketch of y = x2 – 6x + 4 plotted as 0, 4 on the y axis and 3, -5 on the x axis.
Question

Sketch the graph of y = x^2 – 2x – 3, labelling the points of intersection and the turning point.

The coefficient of x^2 is positive, so the graph will be a positive U-shaped curve with a minimum turning point.

Factorising y = x^2 – 2x – 3 gives y = (x + 1)(x – 3) and so the graph will cross the x-axis at x = -1 and x = 3.

The graph will cross the y-axis at (0, -3).

Writing y = x^2 – 2x – 3 in completed square form gives y = (x – 1)^2 – 4, so the coordinates of the turning point are (1, -4).

The turning point could also be found by using symmetry as it will have an x value halfway between x = -1 and x = 3, so when x = 1.

Sketching the turning point on a graph showing y = x to the power of 2 – 2x – 3
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