# 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)
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.