CraigsMaths

General Quadratic Form

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/general-quadratic-form/

The general quadratic form is:

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

This article will describe how to find the vertex, x-intercepts and y-intercepts for this form of a quadratic equation.

Vertex

The vertex can be found from:

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

You can then find $y$ by substituting $x$ into the original equation or by:

$y=\frac{-\left(b^{2}-4ac \right)}{4a}$

Maxima or minima?

The sign of $a$ will tell us this.

If $a$ negative then the vertex is a maxima. (The “opening” of the parabola points down).

Conversely, if $a$ positive then the vertex is a minima. (The “opening” of the parabola points up).

Axis of symmetry

The axis of symmetry will be the line $x=\frac{-b}{2a}$ .

$x$ intercepts

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

$0=ax^{2}+bx+c$

$b^{2}-4ac$ is the discriminant.

If the discriminant is negative then there are no real solutions for $x$ and hence no $x$ intercepts.

If the discriminant is $0$ the there is one real solution for $x$ . This is the point at which the parabola just touchs the x-axis.

If the descriminant is positive then there are two real solutions for $x$ . Hence the parabola crosses the x-axis at two points.

Then we can find $x$ by:

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

$y$ can be found by substituting ALL found values of $x$ into the original equation.

$y$ intercept

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

$y=c$ is therefore the $y$ intercept.

Tags: general form, maths tutor hobart, parabola, quadratic

Tuesday, August 5th, 2008 at 9:16 pm
Category: Engineering Maths B, Quadratic Equations.
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General Quadratic Form

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