# For the function f(x)=(x+8)^2, find the equation of the tangent line to the graph of f at x=-2. if the equation of the tangent line is written as y=mx+b what are m and b?

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We have f(x) = (x + 8)^2. We have to find the equation of the tangent to the graph at the point where x = -2.

The slope of the tangent drawn to a graph f(x) at the point (x, f(x)) is the value of the first derivative of the slope at that point.

f(x) = (x + 8)^2 = x^2 + 64 + 16x

f'(x) = 2x + 16

At x = -2

f'(x) = -2*2 + 16 = 12

The slope of the required line is 12. Also it passes through the point (-2 , 36)

The equation of the tangent is (y - 36) / ( x + 2) = 12

=> y - 36 = 12x + 24

=> y = 12x + 60

**Therefore we get m = 12 and b = 60**

We'll write the equation of the tangent line as:

y - f(-2) = m[x - (-2)] (1)

m is the slope of the tangent line and it represents the tangent line to the graph of the function f(x) = (x+8)^2, for x = -2.

m = f'(-2)

For the beginning, we'll calculate f'(x):

f'(x) = 2(x+8)

Now, we'll determine f'(-2):

f'(-2) = 2*(-2+8)

f'(-2) = 12

We'll calculate the value of the function for x = -2:

f(-2) = (-2+8)^2

f(-2) = 36

Now, we'll substitute the found values into the equation (1):

y - 36 = 12(x+2)

We'll add 36 both sides:

y = 12(x+2) + 36

We'll remove the brackets:

y = 12x + 24 + 36

y = 12x + 60

**The equation of the tangent line is: y = 12x + 60**