Question

Given the following polynomial;
f(x)=(x-2)^3
Find:
a. the zeros and the multiplicity of each
b. where the graph crosses or touches the x-axis
c. number of turning points
d. end behavior

I have no clue what I am doing. pleeeaaase help!!!

Answer 1

surely you can do at least part (b)! When x=2, x-2=0, so f(x) = 0 at x=2.

If I said f(x)=x^3 could you list the zeros, and their multiplicity? Sure you could - it'd be x=0, 3 times. Well, here we have x=2 is a zero, multiplicity 3, because
f(x) = (x-2)(x-2)(x-2)

Now, think of the graph of f(x)=x^3
You know it's kind of a snaky thing, which doesn't turn around anywhere - it just keeps rising as x gets bigger. This graph is exactly the same shape, but shifted 2 units to the right. No turning points.

as x gets huge, so does (x-2)^3 both positive and negatively.

To play around with graphs, you can visit wolframalpha.com and type in your function. Try various changes to see what happens.