CraigsMaths

Parabolas

CraigsMaths
Mathematics Teaching for Learning
Engineering Maths B, Quadratic Equations
August 5th, 2008 by Craig Rose
Permalink: http://www.craigsmaths.com/ea003-engineering-mathematics-b/parabolas/

A parabola is a curve that is described by a quadratic equation. You may be presented with a quadratic equation in one of several forms and then asked to:

Sketch a graph of the equation
Find the local maxima or local minima
Find the vertex coordinates
Find the x-intercept(s) if they exist
Find the the y-intercept
Find the axis of symmetry

This article will describe the simplest approach for the different forms of the quadratic equation.

First step

Before you proceed your first step should be to rearrange the terms of the equation so that they match one of the forms.

For example

$3x^{2}+5x=7-5x$

should be rearranged to match the general form thus:

$y=8x^{2}+5x-7$

Since the standard form provides so much information directly you may want to optionally complete the square to get a general form equation into the standard form.

Hint: Learn the methods for the general form first. Although the arithmetic may be a bit more involved than other approaches the complete solutions are contained in the easily remembered quadratic formula.

General form

$y=ax^{2}+bx+c$

The following equation holds all the clues:

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

This is called the quadratic formula and it gives the values of $x$ when $y=0$ .

How many x-intercepts are there?

First calculate the discriminant $\Delta=b^{2}-4ac$

If $\Delta=0$ then there is only one real solution for $x$ . Hence there is only one x-intercept.

If $\Delta<0$ then there are no real solutions for $x$ . Hence there are no x-intercepts.

if $\Delta>0$ then there are two real solutions for $x$ . Hence there are two x-intercepts.

Hint: Note that the discriminant is the part of the quadratic formula that we take the square root of. The three cases above (one, two and no solutions) come from the facts that:

The square root of a negative number has no real solutions.

The square root of a positive number has two real solutions.

The square root of zero has only one solution.

What are the x-intercepts?

Since the x-intercepts (where $y=0$ ) are given by the quadratic formula:

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Simply substitute the values of $a$ , $b$ and $c$ into this.

What is the y-intercept?

There will only be one y-intercept. This is where $x=0$ .

Substituting $x=0$ into the original equation we get a y-intercept of:

$y=0x^{2}+0x+c=c$

Where is the vertex (or local maxima or minima)?

The vertex can be found where:

$x=\frac{-b}{2a}$

Then calculate $y$ by simply substituting this value into the original equation.

Hint: If you look at the quadratic formula you can see the terms that give the value for $x$ at the vertex.

Is the vertex a minima or maxima?

If $a<0$ then it is a local maxima. The opening of the curve points down.

If $a>0$ then it is a local minima. The opening of the curve points up.

What is the axis of symmetry?

This is the vertical line which passes through the vertex.

Hence the axis of symmetry is $x=\frac{-b}{2a}$ .

Factored form.

$y=a\left(bx+m\right)\left(cx+n\right)$

As with all quadratics you could immediately expand this out to the general form and apply the formulae that we discussed above. But there are a couple of shortcuts you can take.

What are the x-intercepts?

The x-intercepts are immediately obvious when we put $y=0$ :

$0=a\left(bx+m\right)\left(cx+n\right)$

The possible values that will make this true are $\frac{-m}{b}$ and $\frac{-n}{c}$ .

Hint: In the factored form there will always be at least one x-intercept. This arises because the case with no intercepts (in the general form) has no factors.

What is the y-intercept?

This occurs when $x=0$ .

$y=a\left(0+m\right)\left(0+n\right)=amn$

Where is the vertex (or local maxima or minima)?

Expand the equation out into the general form and find the vertex as for the general form.

$x=\frac{-b}{2a}$

There are other methods but this is two fairly simple steps.

Standard form

$y=a(x-h)^{2}+k$ where $a\neq0$

Once again you could immediately expand this out to the general form and apply the formulae that we discussed above. But this form can give us all we need to know fairly directly and so we should take advantage of this.

Vertex

The vertex for the parabola is at $(h,k)$ .

Maxima or minima?

The sign of $a$ will tell us this.

If $a$ is negative then the vertex is a maxima.

Conversely, if $a$ is positive then the vertex is a minima.

Axis of symmetry

The axis of symmetry will be the line $x=h$ .

$x$ intercepts

To find the values for $x$ where the parabola touchs or crosses the X-axis we need to put $y=0$ .

$0=a\left(x-h \right)^2+k$

Solving this for $x$ we have:

$x=\pm\sqrt{\frac{-k}{a}}+h$

If $\frac{k}{a}>0$ then there are no real solutions and there are no $x$ intercepts.

If $\frac{k}{a}=0$ then there is only one real solution and the parabola touches the X-axis at $x=h$ . Note that the definition of a quadratic includes $a\neq0$ .