Thread: 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)

Answer is (1,-4) , (-1,0)

Thanks for the help

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 to get the whole equation of the line.

For the second one, the tangent line is horizontal if and only if the slope is zero, and since the derivative gives you the slope the zeros of the derivative will be the x coordinates for which the slope is zero. Once you have the x coordinates you can plug them into the original function to find their y coordinates.

How would I get the X and Y values? I understand how to find the slope, but what do I need to plug into to get X and Y values?

You know the tangent line goes through the point (-1,-8), so plug these values in for and in and rearrange the equation into the right form.