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Mathematics 2.2 Draw straightforward non-linear graphs

Quadratic graphs

The graph of a quadratic function is called a parabola. You need to interpret the following features in context.

• x intercepts
• y intercept
• axis of symmetry
• vertex or maximum or minimum value.

Drawing parabolas

To draw a parabola you need to identify the intercepts on the axes and the turning point.

• To find the y-intercept, put x = 0 into the equation and work out the y-coordinate.
• To find the x-coordinate, put y = 0 in the equation and solve the quadratic equation to get the x-coordinates.
• To find the vertex (turning point), add the x-intercepts together and divide by 2.

This gives the x-coordinate. Put this x-value into the equation to find the y-coordinate of the vertex.

Plot these points and join together with a smooth U-shaped curve.

• See Graphing quadratics.
• For transforming parabolas see Interpreting equations.
• Graphs of quadratic functions
• The meaning of the leading coefficient
• Quadratic functions (Java™ required)
• Slider graph (Java™ required)
• Graphing parabolas
• Practice with parabolas

Writing equations for parabolas

There are two forms of equation for the parabola:
form (a)y = a(x – b)(x – c)
form (b)y = a(x + b)² + c

When you first look at a graph, write down the coordinates you have (on the graph). These include intercepts and the vertex (turning point).

If you have the x-intercepts, then use form (a). Write it down:
y = a(x – b)(x – c)

If the graph intersects at 3 and –2 on the x-axis, replace ‘b’ and ‘c’ with these values.
Then the equation is y = a(x – 3)(x + 2)

One more point is needed to find ‘a’.
Let’s say the point (0, -12) is on this parabola. Substitute it into the equation:
–12 = a(0 – 3)(0 + 2)
–12 = a(–3)(2)
–12 = –6a
a = 2

So the equation is:
y = 2(x – 3)(x + 2)

Use form (b) when you have the turning point (or vertex) of the parabola.
Write down:
y = a(x + b)² + c

If the vertex is at point (1, 2), you know that the graph of y = ax² has been translated 1 unit horizontally and 2 units vertically.

Therefore the equation is:
(y – 2) = a(x – 1)²
or y = a(x – 1)² + 2

All you need to do now is work out the value of ‘a’.
Do that by choosing some other point on the parabola (not the vertex) and substitute its coordinates into the equation. You can then work out the value of ‘a’ and complete your equation.

For further information, see:

• Equations from graphs and related graphs

• Graphing quadratic equations.

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