Hi I need some info on how to Sketch the square of functions.
Eg (x+6)(x+3)(x4) 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
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Hi I need some info on how to Sketch the square of functions.
Eg (x+6)(x+3)(x4) 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 $\displaystyle f(x)=(x+6)^2(x+3)^2(x4)^2$ is always greater than orQuote:
Originally Posted by classicstrings
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
Hello, classicstrings!
An interesting problem . . . I've never been asked to do this.
$\displaystyle f(x)$ is a cubic with xintercepts: 6, 3, 4.Quote:
$\displaystyle f(x)\,=\,(x+6)(x+3)(x4)$
Sketch the square and comment on the number of turning points.
There are two turning points.$\displaystyle g(x) \,= \,(x=6)^2(x+3)^2(x4)^2$ has the same xintercepts but of order two.Code:
 *

**  *
6 * *  *
  o  +  +  o  +  +  +  +  +  o   
* 3*  * 4
*  *
* * *
 **

. . The graph is tangent to the xaxis there.
Also the entire graph is above (or on) the xaxis.There are five turnning points.Code:
*  ** *
** * *
* * * * * *
* * * *  * *
  o* +  +  o* +  +  +  +  + *o   
6 3  4
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.
Also where would the intercepts of the two lines be? Thanks