# Slope of tangent line

#### ascendancy523

I was wondering if someone could check my work. Here's my problem:

Find the slopes of the tangent line to the curve $$\displaystyle y=x^3-3x$$ at the points where x=-2,-1,0,1,2.

So $$\displaystyle m_{tan}=\frac{[(c+h)^3-3(c+h)]-(c^3-3c)}{h}$$

$$\displaystyle =\frac{(c^3+3c^2h+3ch^2+h^3)-3c-3h-c^3+3c)}{h}$$

$$\displaystyle =\frac{3c^2h+3ch^2-3h+h^3}{h}$$

$$\displaystyle =\frac{h(3c^2+3ch-3+h^2)}{h}$$

lim h->0 $$\displaystyle 3c^2+3ch-3+h^2=3c^2-3$$

so at x=-2, $$\displaystyle m_{tan}=9$$
x=-1, $$\displaystyle m_{tan}=0$$
x=0, $$\displaystyle m_{tan}=-3$$
x=1, $$\displaystyle m_{tan}=0$$
x=2, $$\displaystyle m_{tan}=9$$

Thank you for any help!

#### Ackbeet

MHF Hall of Honor
Looks good to me.