Asked by Scott
determine the equation of the tangent to the graph of y =-2x^3 -4x + 5 when x=2
Answers
Answered by
mathhelper
y = -2x^3 - 4x + 5
dy/dx = -6x^2 - 4
when x = 2, dy/dx = -6(4) - 4 = -28
also when x = 2, y = -2(8) -4(2) + 5 = -19, so our point of contact is (2,-19)
equation of tangent is
y + 19 = -28(x-2)
doctor this up any way you need.
dy/dx = -6x^2 - 4
when x = 2, dy/dx = -6(4) - 4 = -28
also when x = 2, y = -2(8) -4(2) + 5 = -19, so our point of contact is (2,-19)
equation of tangent is
y + 19 = -28(x-2)
doctor this up any way you need.
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