Question

The slope of the the tangent line to a curve at a point. Calc derivative help

Answer 1

Hello from MrBillDoesMath!

Discussion:

The slope of the tangent to the curve y = 4x^3   is given by the first derivative.

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y' = 4 (3x^2) =  12x^2

At (-3,-108), the slope is   12x^2 = 12(-3)^2 = 12*9 = 108

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y = mx + b

From the first part, m = 108 so y = 108x + b.

Substituting  y = -108 when x = -3 gives

-108 =  108(-3) + b =>   add 3* 108 to both sides

-108 + 108(3) = b   =>   as 3 -1  = 2

108*2 = b = 216

y = 108x + 216

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Thank you,

MrB