1. ## Tangents

Hi, I am having trouble getting equation to tangent lines for functions, I compute the derivative first but then don't have a clue what to do after, or where to plug in/substitute the specified values. Here are the exact directions:

Find the equation of the line that is tangent to the graph of the given function at the specified point.

Y= -X^3 - 5X^2 + 3X - 1 ; (-1,-8)

The answer is Y= 10X + 2

Also another question, I don't know if I use the same technique as the above to answer it. Instructions: Find all points on the graph of the given function where the tangent line is horizontal.

F(X)= (X+1)(X^2-X-2)

2. The derivative outputs the slope of the original function for whatever x value you input, so the slope of y(x) at x=-1 is y'(-1). Once you know the slope you can use $\displaystyle y-y_1 = m(x-x_1)$ to get the whole equation of the line.
4. You know the tangent line goes through the point (-1,-8), so plug these values in for $\displaystyle x_1$ and $\displaystyle y_1$ in $\displaystyle y-y_1=m(x-x_1)$ and rearrange the equation into the right form.