SOLUTION: Graph (x+3) (x-2)^2=0 (use maxs, mins, x and y intercepts)

Question 134056: Graph (x+3) (x-2)^2=0 (use maxs, mins, x and y intercepts)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
doesn't actually have a graph other than the three values (-3, 2, and 2) on the number line that satisfy the equation. You can, however, graph , and I suspect that is what you really meant. (I sincerely hope your instructor didn't pose the question that way. If s/he did, go get your money back)

Zeros, or x-intercepts at -3, 2, and 2 (we should have an extrema at 2)

Expand:

Extreme points where f'(x) = 0, so:

f'(x) = . Set the first derivative equal to zero:

, hence critical points are at (,) and (,).

We know that , so we have a critical point at (2,0).

And we have an extreme point at ,)

f"(x)=
f"(2)= => (2,f(2)) is a local minimum
f"(-4/3)= => (,) is a local maximum

The y-intercept is found by evaluating f(0):
hence the y-intercept is (0,12)

Using this information, we can make a rough sketch of the graph: