CraigsMaths

Standard Quadratic Form

CraigsMaths
Mathematics Teaching for Learning
Engineering Maths B, Quadratic Equations
August 1st, 2008 by Craig Rose
Permalink: http://www.craigsmaths.com/ea003-engineering-mathematics-b/standard-quadratic-form/

This is the “standard” form of the quadratic equation:

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

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

Vertex

$h$ and $k$ represent translations of the parabola $y=x^{2}$ . This parabola has it’s vertex at $(0,0)$ . $k$ translates the parabola in the y-direction (ie up or down). The sign of $k$ determines the whether it is up or down. A positive $k$ will translate it up whilsts a negative $k$ will translate it down. Similarily $h$ is a translation in the x-direction (ie left or right). A positive $h$ will translate it right whilsts a negative $h$ will translate it left.

The vertex $(0,0)$ of the parabola $y=x^{2}$ will translate by the respective amounts of $h$ and $k$ . Hence 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. (The “opening” of the parabola points down).

Conversely, if $a$ is 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 $x=h$ .

$x$ 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 $a\neq0$ .