What is the equation of the line tangent to #f(x)=(x-1)^3-3x^2 # at #x=1#? Calculus Derivatives Tangent Line to a Curve 1 Answer Bdub May 6, 2016 #f(x)=-6x+3# Explanation: #f'(x)=3(x-1)^2-6x# #f'(1)=3(1-1)^2-6(1)=-6# #f(1)=(1-1)^3-3(1)^2=-3# #f(x)=f(a)+f'(a)(x-a)# #f(x)=-3+(-6)(x-1)# #f(x)=-3-6x+6# #f(x)=-6x+3# Answer link Related questions How do you find the equation of a tangent line to a curve? How do you find the slope of the tangent line to a curve at a point? How do you find the tangent line to the curve #y=x^3-9x# at the point where #x=1#? How do you know if a line is tangent to a curve? How do you show a line is a tangent to a curve? How do you find the Tangent line to a curve by implicit differentiation? What is the slope of a line tangent to the curve #3y^2-2x^2=1#? How does tangent slope relate to the slope of a line? What is the slope of a horizontal tangent line? How do you find the slope of a tangent line using secant lines? See all questions in Tangent Line to a Curve Impact of this question 1036 views around the world You can reuse this answer Creative Commons License