# Graphing

• Jun 13th 2006, 10:59 PM
nath_quam
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
• Jun 14th 2006, 10:52 PM
CaptainBlack
Quote:

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
• Jun 14th 2006, 10:59 PM
CaptainBlack
Quote:

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
• Jun 14th 2006, 11:19 PM
nath_quam
Thanks Captain is any one able to graph this and send it to me
• Jun 15th 2006, 04:50 AM
Soroban
Hello, nath_quam!

Quote:

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

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

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

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

As $\displaystyle x\to-\infty,\;\;P(x) \to \infty$ . . . 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.

$\displaystyle b)\;\;P(x)\:=\:(1-x)^3(x-3)$ has x-intercepts: $\displaystyle 1,\;3$

The graph rises to the right and to the left.

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

        *    |         *    |                    *           *  |               *                    *       --------+---o--------------o---               |    *          *               |      *      *               |        * *               |
• Jun 15th 2006, 04:59 AM
nath_quam
Maximum
Thanks but with the second graph wouldn't the point 5/2 be a maximum
• Jun 15th 2006, 05:01 AM
CaptainBlack
Quote:

Originally Posted by Soroban

$\displaystyle b)\;\;P(x)\:=\:(1-x)^3(x-3)$ has x-intercepts: $\displaystyle 1,\;3$

The graph rises to the right and to the left.

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

        *    |         *    |                    *           *  |               *                    *       --------+---o--------------o---               |    *          *               |      *      *               |        * *               |

The co-efficient of $\displaystyle x^4$ is negative, so the graph should be
the other way up.

Nice ASCII art work as usual:)

RonL