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
Ahnaf R
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

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