Question: 1. (1 point) Let h(x) =2-4x3, h' (2) = _ Use this to find the equation of the tangent line to the curve y =

1. (1 point) Let h(x) =2-4x3, h' (2) = _ Use

1. (1 point) Let h(x) =2-4x3, h' (2) = _ Use this to find the equation of the tangent line to the curve y = 2 -4x' at the point (2, -30) and write your answer in the form: y = mx + b, where m is the slope and b is the y-intercept. Correct Answers: -48 . y = -48*x | 66

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