On-line Math 21

4.5 Curve Sketching

Example 2 f(x) = 2x³-3x²-12x .

Solution

Intercepts

Here we get two for one. The y -intercept is at f(0) = 0 , so (0,0) is both an x - and y -intercept. To find the other x -intercepts, we solve 2x³-3x²-12x = 0 .

0

=

2x³-3x²-12x

=

x(2x²-3x-12)

=

x(2x-?)(x+??).

Again I can't factor this. But, plugging in to the quadratic formula, 2x²-3x-12 = 0 when

x

=

3ą
Ö

9-4·2·(-12)

4

=

3ą ___
Ö105

4

@

3.3 or -1.8,

where the last are just rough estimates, based on
___
Ö105

@ 10.1

. At any rate, then (3.3,0) and (-1.8,0) are the approximate coordinates of the other x -intercepts.

Increasing/decreasing, and critical points

Computing the derivative, and factoring it,

f˘(x)

=

6x²-6x-12

=

6(x²-x-2)

=

6(x-2)(x+1),

so the derivative is 0 at x = 2 and x = -1 . Now, as before I'm going to draw the number line, and see where f˘ is positive, and where it is negative.

Don't forget to find the critical values.

f(-1)

2(-1)³-3(-1)²-12(-1)

-2-3+12

and

f(2)

2(2³)-3(2²)-12(2)

16-12-24

-20,

so the critical points and values are (-1,7) and (2,-20) .

Draw the graph

Then use this information to draw a fair representation of the graph.

The first information you plot are the intercepts; [Link to ex2-1.gif]
Then plot the critical points (and values). I always mark it with a horizontal line to remind myself that it is a critical point. [Link to ex2-2.gif]
Then fill in the graph, connecting the dots and making sure that the graph you draw is increasing/decreasing where the number line indicates it should be. [Link to ex2-3.gif]

[Each of these items should trigger the appearance of a new drawing with that information added.

File translated from T_EX by T_TH, version 2.61.
On 21 Dec 2000, 00:26.