Find the point on the graph of the given function at which the slope of the tangent line is the given slope.

f(x)=x^3+6x^2+20x+13; slope of the tangent line=8

A ( -2 , 9 )

Step-by-step explanation:

Idea: You find the first derivative of f(x), and then set it equal to the desired slope. You'll find some x. That we will use to find the point.

f'(x) = 3x^2 + 12x + 20

f'(x) = 8

3x^2 + 12x + 20 = 8

3x^2 + 12x + 12 = 0

3 ( x^2 + 4x + 4 ) = 0

3 ( x + 2 )^2 = 0

x + 2 = 0

x = -2

So, the desired point is:

A ( -2, f(-2) ) --> A ( -2 , 9 )

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation: