For f(x)=(2x-3)^3, find the equation of the tangent line at x=0 and x=2

Answer 1

f(x) = (2x-3)^2 at x=0 and x=2: (0,9), (2,1)

slope = the derivative =

f'(x) = 6(2x-3)^2

f'(0) = 6(0-3)^2 = 6(9) = 54

f'(2) =6(4-3)^2 = 6(1) = 6

equation of the tangent line at x=0 is y-9 = 54(x-0)

or y=54x +9

equation of the tangent line at x=2 is y-1 = 6(x-2)

or y = 6x-11