Find the angle(s) of intersection between the equations f (x) = x^2 + 2x + 1 and g(x) = 1.

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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.