Question

What is the equation of the line tangent to `f(x)=(2x^3 - 4) / x` at `x=-4`?

Answer 1

`y=-63/4(x+4)-33`

Explanation:

1. Find the derivative of f(x):
   Using the quotient rule...
   `f'(x)=((x)(6x^2)-(2x^3-4)(1))/x^2`
   Simplifying that...
   `f'(x)=(4x^3+4)/x^2`

2. Plug `x=-4` into your `f'(x)` equation:
   `f'(-4)=(4(-4)^3+4)/(-4)^2=-63/4`
   This is the slope of the tangent line.

3. Find your `y` coordinate by plugging `x=-4` into your original equation:
   `f(-4)=(2(-4)^3-4)/(-4)=33`

4. Using `x=-4`, `y=33`, and `f'(-4)=-63/4`, form your tangent line equation:
   `y=-63/4(x+4)-33`