Find the slope (calculus)

how do i find the slope for the following equation y=x^3-14x p (1,-13)
it got this far
h^3+3h^2+3h+1-14-14h-1+14
h^3+3h^2+3h-14h
h(h^2+3h-11)/h
h^2+3h-11
h(h+3)-11

Ah, thats the limit definition of the derivative, I see :D but where's your x?

Here's the formula:
lim[ f(x + h) - f(x) ] / h
(h-->0)

Note: the limit here is always h--> 0, but because thats tedious, im just gonna let you know that when i write lim, it means lim(h-->0) okay? I'll let you know when you evaluate 0 for the limit okay?

= lim { [(x + h)^3 - 14(x + h)] - [x^3 - 14x] } / h
= lim { [(x^3 + 3x²h + 3xh² + 3h^3 - 14x - 14h] - x^3 + 14x } / h
= lim { [x^3 - x^3 + 3x²h + 3xh² + 3h^3 - 14x + 14x - 14h] } / h
= lim [3x²h + 3xh² + 3h^3 - 14h] / h
= lim [h(3x² + 3xh + 3h² - 14)] / h
= lim (3x² + 3xh + 3h² - 14)

Now plug in 0 for h.

= 3x² + 0 + 0 - 14
= 3x² - 14

Now that we got the slope equation, we will plug in x = 1, (don't worry about the y value, that'll kick in later)

= 3(1)² - 14
= 3 - 14
= -11

Yay!

Therefore, your slope of y = x^3 - 14x at x = 1 is -11
Hope this helps :D

And yes,
when y = x^3 - 14x
dy/dx = 3x² - 14
It's called derivatives, it's much funner (lol) than doing limits xD

y' = 3x^2 - 14
3(1)^2 - 14 = 3 - 14 = -11

you ask for the slope of the tangent at p
y´=3x^2-14 =-11