Answered

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

Answer :

AlexVavvas

Answer:

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 )

The point on the graph should be A ( -2 , 9 )

Calculation of the point:

The equation is [tex]f'(x) = 3x^2 + 12x + 20[/tex]

here,

f'(x) = 8

[tex]3x^2 + 12x + 20 = 8\\\\3x^2 + 12x + 12 = 0\\\\3 ( x^2 + 4x + 4 ) = 0\\\\3 ( x + 2 )^2 = 0[/tex]

x + 2 = 0

x = -2

Now,  the desired point is:

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

Learn more about the slope here: https://brainly.com/question/11586768

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