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k thx i think i figured it out.

ok, thx i figured out that i didn't plug the whole eq. into f(x+h), i just did (x+h)^3, when i needed (x+h)^3 -9. is there a way i can type like you? that looks a lot easier to read and make.

My quiz on Monday requires us to do the long way of finding derivatives, using: . . . . .f'(x) = lim_h->0 [f(x + h) - f(x)] / h I used the limit for the equation y = x³ - 9. I know the derivative is supposed to be y' = 3x², according to the Power Rule. But when...

yah, i just figured that out, thx.

i need to find the y coordinates too, so i just plug in those x values into y prime right?

oh you're right they do have 0 slope. i was thinking that 0 slope=undefined, but it is just horizontal, thanks.

Determine the points (if any) at which the function has a horizontal tangent line: . . .y = x^4 - 8x^2 + 2 I know that functions like "1/x^2" don't have any horizontal tangent lines, because the graph is never horizontal. Those are easy. For this one, I just tried plugging values in for x. I...