(x) = x ^3 - 12 x + 11, write the equation of the line tangent to the curve at x = 1
f'(x) = 3x^2 - 12
f'(1) = -9
f(1) = 0
so you want the line through (1,0) with slope -9:
y = -9(x-1)
f'(1) = -9
f(1) = 0
so you want the line through (1,0) with slope -9:
y = -9(x-1)