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
