Question 215945
Find the equation of the line that is tangent to the graph of y = x3 at the point (-2 , -8). 
Answer in the form: Ax + By + C = 0, where A, B and C are relatively prime integers and A > 0. 
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Slope at x = -2:
y' = 3x^2 = 12
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Equation of line m = 12 thru point (-2,-8)
y = mx + b
-8 = 12*-2 + b = -24 + b
b = 16
y = 12x + 16
12x - y + 16 = 0