How to find the equation of the tangent line to the curve #f(x) = (x^3-3x +1)(x+2)# at the point (1, -3)?

1 Answer
Aug 5, 2018

#y=x-4#

Explanation:

First, expand the given function so that we can derive it with ease:

#f(x)=(x^3-3x+1)(x+2)#
#f(x)=x^4+2x^3-3x^2-5x+2#

Then, we differentiate the function to find its gradient/derivative(s):

#f'(x)=4x^3+6x^2-6x-5#

Since we have been given the coordinates of a point #(1,-3)# in the curve / tangent line, we know that we need to find the gradient of the tangent line at #x=1#:

#f'(1)=4*1^3+6*1^2-6*1-5#
#f'(1)=4+6-6-5#
#f'(1)=1#

Finally, we use the gradient-intercept formula #y_2-y_1=m(x_2-x_1)# to find the equation of the tangent line:

#y--3=1(x-1)#
#y+3=x-1#
#y=x-4#