How do I find the slope of the tangent to the graph of the function below? y=x^4 +8x^3 -2x - 2; x= -2?

Differentiate
dy/dx = 4x^3 + 24x^2 - 2

When x = -2
dy/dx = 4(-2)^3 + 24(-2)^2 - 2
dy/dx = -32 + 96 - 2
dy/dx = 62 The slope!!!!

How do I find the slope of the tangent to the graph of the function below? y=x^4 +8x^3 -2x - 2; x= -2?

By taking the derivative of that function first (the derivative is the slope), and then substituting -2 for x in the derivative and find the slope of the tangent at that point...


Step 1: Taking the derivative of the function.

I don't know if you have to do this through the definition of the derivative or using rules...I don't want to expand out all of that so I will use the power rule for derivative.

Take the derivative of each term using the power rule: ax^b = (a*b)x^(b-1)....

y = x^4 + 8x^3 - 2x - 2

dy/dx = 4x^(4-1) + (8*3)x^(3-1) - (2*1)x^(1-1) - 0

dy/dx = 4x^3 + 24x^2 - 2


Step 2: Substitute -2 for x in the derivative, evaluate and find the slope of the tangent at that point...

dy/dx = 4(-2)^3 + 24(-2)^2 - 2

dy/dx = -32 + 96 - 2

dy/dx = 62 <-- answer...

So the slope of the tangent at x = -2 is 62.