Question

What is the equation of the line normal to `f(x)=(2x^2 + 1) / x` at `x=-2`?

Answer 1

`y = -4/7x -79/14`

Explanation:

slope function `= f'(x)`

Find the derivative:
Use the Quotient Rule `(u/v)' = (vu' - uv')/v^2`

Let `u = 2x^2 + 1`, `u' = 4x`
Let `v = x`, `v' = 1`

`f'(x) = (x*4x - (2x^2+1)(1))/x^2 = (4x^2 - 2x^2 - 1)/x^2 = (2x^2 - 1)/x^2`

Find the tangent and normal slopes:
Tangent Slope, `m = f'(-2) = (2(-2)^2-1)/(-2)^2 = 7/4`

Normal slope, `-1/m = -4/7`

Find normal equation:
1. Find the point: `f(-2) = (2(-2)^2 + 1)/-2 = -9/2`
2. Use `y - y_1 = m(x-x_1)`: `y - (-9/2) = -4/7(x - (-2))`
3. Simplify: `y +9/2 = -4/7x -8/7`;
4. write in `y`-intercept form: `y = -4/7x -79/14`