Question

What is the equation of the line that is normal to `f(x)= x^3-4x` at `x=0`?

Answer 1

`y=1/4x`

Explanation:

A normal line is a straight line perpendicular to a tangent line. Since perpendicular lines have opposite reciprocal slopes, the way we find a normal line is to find the slope of a tangent line and then do the opposite reciprocal stuff.

But what is the slope of the tangent line at `x=0`? It's the derivative, of course! Let's start there:
`f(x)=x^3-4x`
`f'(x)=3x^2-4`
`f'(0)=3(0)^2-4`
`f'(0)=-4`

That means the slope of the tangent line of `x^3-4x` at `x=0` is `-4`.

The slope of the normal line is perpendicular to this, so we take the opposite reciprocal of `-4` to find that slope:
Opposite reciprocal of `-4` = `1/4`

However, we're being asked for the equation of the normal line, not just the slope. As a result, we still have some more work to do. Recall that straight lines (the normal line is straight) take the form `y=mx+b`, where `x` and `y` are points, `m` is the slope, and `b` is the `y`-intercept. We know the slope - it's `1/4`. All we need is the `y`-intercept.

Note that if we find a point `(x,y)`, we can solve for `b`, because we will have an equation with 3 knowns and 1 unknown. And furthermore, since a normal line is perpendicular to the tangent line, it must intersect the curve. So we can find our point by evaluating `f(x)` at `x=0`
`f(0)=(0)^3-4(0)`
`f(0)=0`

Our point is `(0,0)`.

Making substitutions in `y=mx+b`, we have:
`0=(1/4)(0)+b`
`0=b`

The `y`-intercept is `0`, so the equation of the normal line is `y=1/4x`.