Find the angle measurements of the intersections for the two equations f(x) = 4x - 5 and g(x) = 2x^2 - 5. 63 7 20 76 90

 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.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15