where does the function cut the yintercept f(x) =
+2?
or when I get the f '(x) -9?
I've never heard that terminology. I will guess that "cut the y-intercept" means something like "cuts" (or crosses) "the y-axis". If this is correct, then you need to plug zero in for x to find the y-value of the y-intercept.
I don't know what you're trying to do with the derivative, so I'm afraid I cannot comment on that. Sorry!