Question

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)?

Answer 1

`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`