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
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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
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