# Very Stuck. Help on Squares of functions!

• May 30th 2006, 05:16 AM
classicstrings
Very Stuck. Help on Squares of functions!
Hi I need some info on how to Sketch the square of functions.

Eg (x+6)(x+3)(x-4) Sketch the square and comment on the number of turning points.

Can someone find a website on this? I have searched google to no avail.

Thanks
• May 30th 2006, 08:47 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
Hi I need some info on how to Sketch the square of functions.

Eg (x+6)(x+3)(x-4) Sketch the square and comment on the number of turning points.

Can someone find a website on this? I have searched google to no avail.

Thanks

The function $\displaystyle f(x)=(x+6)^2(x+3)^2(x-4)^2$ is always greater than or
equal to zero, it is increasing as $\displaystyle x \to \pm \infty$, and touches the $\displaystyle x$ axis when
$\displaystyle x=4,\ -3$ and $\displaystyle -6$.

Each of the zeros is a turning point, so it must have an additional turning points
between them making at least 5 turning points. Also it can have at most
five turning points (as $\displaystyle f'(x)$ is a quintic and so has five zeros
which are potential turning points), hence it has exactly 5 turning points.

This should be sufficient to allow you to sketch the curve.

RonL
• May 30th 2006, 08:59 AM
Soroban
Hello, classicstrings!

An interesting problem . . . I've never been asked to do this.

Quote:

$\displaystyle f(x)\,=\,(x+6)(x+3)(x-4)$

Sketch the square and comment on the number of turning points.
$\displaystyle f(x)$ is a cubic with x-intercepts: -6, -3, 4.
There are two turning points.
Code:

                              |                               |                  *                               |               **            |                  *         -6 *      *          |                *       - - o - + - + - o - + - + - + - + - + - o - - -         *          -3*      |            * 4                         *    |          *         *                    *|      *                               |  **                               |
$\displaystyle g(x) \,= \,(x=6)^2(x+3)^2(x-4)^2$ has the same x-intercepts but of order two.

. . The graph is tangent to the x-axis there.

Also the entire graph is above (or on) the x-axis.
Code:

                              |     *                        |  **              *                 **            |*    *       *      *  *          *|        *          *       *    *    *      *  |          *      *       - - o*- + - + - o*- + - + - + - + - + -*o - - -         -6          -3      |              4
There are five turnning points.
• May 30th 2006, 09:02 AM
classicstrings
Thanks for your help guys! The diagrams + explanations rock! CHeers!!!

EDIT: If you read what i posted before forget it!

EDIT: Soroban do you post on other maths help boards? I think I have seen you elsewhere.
• May 30th 2006, 09:08 AM
classicstrings
Also where would the intercepts of the two lines be? Thanks