Answer: approximately 63.43 degrees
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Work Shown:
Locate the intersection points
f(x) = g(x)
x^2 + 2x + 1 = 1
x^2 + 2x = 1-1
x^2 + 2x = 0
x(x+2) = 0
x = 0 or x+2 = 0
x = 0 or x = -2
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Compute the derivative
f(x) = x^2 + 2x + 1
f ' (x) = 2x + 2
Then plug in each solution x value we found earlier
f ' (0) = 2(0) + 2 = 2
The slope of the tangent line at the intersection point (0,1) is m = 2
The tangent line is y = 2x+1
The angle between the lines y = 1 and y = 2x+1 is arctan(2) = 63.43 degrees approximately
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Plug x = -2 into the derivative function
f ' (x) = 2x+2
f ' (-2) = 2(-2)+2
f ' (-2) = -2
The slope of the tangent line at (-2,1) is m = -2
The tangent line here is y = -2x-3
The angle between the lines y = 1 and y = -2x-3 is also 63.43 through similar reasoning as before.
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See the diagram below.