Question

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

Answer 1

The normal line is given by `y=-x-4`

Explanation:

Rewrite `f(x)=(2x^2+1)/x` to `2x+1/x` to make differentiation simpler.

Then, using the power rule, `f'(x)=2-1/x^2`.

When `x=-1`, the y-value is `f(-1)=2(-1)+1/-1=-3`. Thus, we know that the normal line passes through `(-1,-3)`, which we will use later.

Also, when `x=-1`, instantaneous slope is `f'(-1)=2-1/(-1)^2=1`. This is also the slope of the tangent line.

If we have the slope to the tangent `m`, we can find the slope to the normal via `-1/m`. Substitute `m=1` to get `-1`.

Therefore, we know that the normal line is of the form
`y=-x+b`

We know that the normal line passes through `(-1,-3)`. Substitute this in:
`-3=-(-1)+b`
`therefore b=-4`

Substitute `b` back in to get our final answer:
`y=-x-4`

You can verify this on a graph:
graph{(y-(2x^2+1)/x)(y+x+4)((y+3)^2+(x+1)^2-0.01)=0 [-10, 10, -5, 5]}