Asked by Rory

find the equation to the tangent to the curve y = (4x + 3)^5 at the point (-0.5, 1), giving your answer in the form y = mx + c
Answered by Reiny
dy/dx = 5(4x+3)^4 (4) = 20(4x+3)^4
when x = -1/2
dy/dx = 20(1)^4 = 20

1 = 20((-1/2) + b
1 = -10+b
b = 11

y = 20x + 11
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