# Math Help - Graphing

1. ## Curve Sketching Challenge

What do the graphs
a) P(x) = x(3x+2)(x-3)(x+2)

and

b) P(x) = ((1-x)^3)(x-3)

look like and there keys points

2. Originally Posted by nath_quam
What do the graphs
a) P(x) = x(3x+2)(x-3)(x+2)

look like and there keys points
It goes to infinity as x goes to +/- infinity.

It has four distince roots at 0, -2/3, 3 and -2.

It has a local max/min between each pair of roots.

RonL

3. Originally Posted by nath_quam
b) P(x) = ((1-x)^3)(x-3)
It goes to -infinity as x goes to +/- infinity

It has a root of multiplicity 3 at x=1 (it is negative as x approches 1 from
below, is tangent to the axis at x=1, and is positive
as x moves away from 1 above), and a simple root at x=3

It has a maximum between the roots.

RonL

4. Thanks Captain is any one able to graph this and send it to me

5. Hello, nath_quam!

What do the graphs looks like and what are their key points?

$a)\;P(x) \:= \:x(3x+2)(x-3)(x+2)$

$b)\; P(x) \:= \1-x)^3(x-3)" alt="b)\; P(x) \:= \1-x)^3(x-3)" />
$a)\;P(x) \:=\:x(3x+2)(x-3)(x+2)$ has x-intercepts: $-2,\;-\frac{2}{3},\;0,\;3$

As $x\to\infty,\;\;P(x) \to \infty$ . . . The graph rises to the right.

As $x\to-\infty,\;\;P(x) \to \infty$ . . . The graph rises to the left.

You should be able to sketch the graph now . . .
Code:
                      |                 *
*               |
|                *
*           ** |
----o-------o-----o---------------o--
*     *      | *            *
**        |    *       *
|        * *
|
You can use the derivative to locate the exact positions
. . of the maximum and the two minimums.

$b)\;\;P(x)\:=\1-x)^3(x-3)" alt="b)\;\;P(x)\:=\1-x)^3(x-3)" /> has x-intercepts: $1,\;3$

The graph rises to the right and to the left.

We find that $(1,0)$ is an inflection point
. . and there is a minimum at $x = \frac{5}{2}$
Code:
        *     |
*    |                     *
*  |
*                    *
--------+---o--------------o---
|     *           *
|      *       *
|         * *
|

6. ## Maximum

Thanks but with the second graph wouldn't the point 5/2 be a maximum

7. Originally Posted by Soroban

$b)\;\;P(x)\:=\1-x)^3(x-3)" alt="b)\;\;P(x)\:=\1-x)^3(x-3)" /> has x-intercepts: $1,\;3$

The graph rises to the right and to the left.

We find that $(1,0)$ is an inflection point
. . and there is a minimum at $x = \frac{5}{2}$
Code:
        *     |
*    |                     *
*  |
*                    *
--------+---o--------------o---
|     *           *
|      *       *
|         * *
|
The co-efficient of $x^4$ is negative, so the graph should be
the other way up.

Nice ASCII art work as usual

RonL