Parabolas have a vertex, which is either the maximum or the minimum point depending on the concavity of the parabola. The x-value of the vertex is given by the axis of symmetry.

The vertex of the parabola defined by y=ax^2 is (0,0).

If we translate the entire parabola upwards by k units, and right by h units, then the equation becomes y=a\left(x-h\right)^2+k and the vertex is (h,k). We call this the vertex form of the parabola.

Examples

Example 1

Consider the graph of the parabola below.

-4

-3

-2

-1

-8

-6

-4

-2

How many x-intercepts does this parabola have?

Worked Solution

Create a strategy

Count the number of times the parabola crosses the x-axis.

Apply the idea

The parabola does not cross the x-axis. So the parabola has 0 x-intercepts.

What are the coordinates of the parabola's vertex?

Worked Solution

Create a strategy

Find the coordinates of the turning point of the parabola.

Apply the idea

The coordinates of the parabola's vertex is (1,-5).

Is the vertex the maximum or minimum of this parabola?

Worked Solution

Create a strategy

Look at the concavity of the parabola.

Apply the idea

The parabola is concave down, so the vertex is the maximum of the parabola.

Example 2

Consider the equation y=4x^2+24x+42.

Write the equation in the form y=a(x-h)^2+k.

Worked Solution

Create a strategy

Use completing the square

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle 4x^2+24x+42	Write the equation
\displaystyle y	\displaystyle =	\displaystyle 4\left(x^2+6x\right)+42	Factorise 4 out of the first two terms
	\displaystyle =	\displaystyle 4\left(x^2+6x+3^2-3^2\right)+42	Add and subtract half of 6 squared
	\displaystyle =	\displaystyle 4\left(x^2+6x+3^2\right)+4\times (-3^2)+42	Take -3^2 out of the brackets
	\displaystyle =	\displaystyle 4\left(x^2+6x+9\right)+6	Evaluate the constant terms
	\displaystyle =	\displaystyle 4\left(x+3\right)^2+6	Factorise the perfect square

Identify the coordinates of the vertex.

Worked Solution

Create a strategy

Use the vertex form of the equation.

Apply the idea

From part (a), the vertex form of the equation is y=4\left(x+3\right)^2+6.

So the coordinates of the vertex are (-3,6).

Idea summary

The equation of a parabola can be written in the form y=a\left(x-h\right)^2+k. This is called the vertex form of the parabola.

If the equation is in this form then the vertex will be at (h,k).

We can convert an equation of a parabola from general form to vertex form by completing the square.

6.06 Plotting parabolas using vertex form

Ideas

Parabolas in vertex form

Examples

Example 1

Example 2

Outcomes

MA5.3-9NA

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