Homework Help: Polynomial Graph Behavior

1. Oct 26, 2008

jacksonpeeble

1. The problem statement, all variables and given/known data
Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper behavior. No calculator allowed.

P(x)=x(x-3)(x+2)

2. Relevant equations
P(x)=x(x-3)(x+2)

3. The attempt at a solution
P(x)=x(x-3)(x+2)
0(x-3)(x+2)=0
x(3-3)(x+2)=0
x(x-3)(-2+2)=0
Zeros: 0, 3, -2
Degree: 1+1+1=3 (Odd)
Leading Coefficient: 1 (Positive)
Multiplicities: 1, 1, 1

Therefore, I have X-Intercepts at all of the zeros. However, I am confused as to how I determine what direction the graph goes at different intervals and whether it bounces or goes through the zeros. Any help is greatly appreciated.

2. Oct 26, 2008

rock.freak667

Find the stationary points by putting P'(x)=0

3. Oct 27, 2008

Enjoicube

There's a reason this is the pre-cal thread, no matter how much some like to ignore the title. Although it is a lot easier with calculus,I suggest you find out what lies in between the zeroes (positive values or negative values). In fact,as a good way to guess, pick x values halfway between the zeroes for substitution.

4. Oct 27, 2008

HallsofIvy

If x> 3, then all of x, x-3, and x+2 are positive so x(x-3)(x+ 2) is positive.
If 0< x< 3, then x and x+2 are positive but x-3 is negative so x(x-3)(x+2) is negative.
If -2< x< 0, then x+ 2 is positive but x and x-3 are negative so x(x-3)(x+2) is positive.
If x< -2 then all of x, x-3, and x+2 are negative so x(x-3)(x+2) is negative.

That should be exactly what you need.

5. Oct 27, 2008

jacksonpeeble

What's that supposed to mean? This work is from my class, called "Honors Trigonometry and Pre-calculus."

Thank you to HallsOfIvy and everyone else that contributed, that was exactly what I needed!

6. Oct 27, 2008

Enjoicube

P'(x) means the derivative with respect to x, which I am guessing you wouldn't have learned by now because that is in calculus.