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

May 20, 2016

$y = \frac{1}{4} x$

#### 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:
$f \left(x\right) = {x}^{3} - 4 x$
$f ' \left(x\right) = 3 {x}^{2} - 4$
$f ' \left(0\right) = 3 {\left(0\right)}^{2} - 4$
$f ' \left(0\right) = - 4$

That means the slope of the tangent line of ${x}^{3} - 4 x$ 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 $- 4$ = $\frac{1}{4}$

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 $y = m x + b$, where $x$ and $y$ are points, $m$ is the slope, and $b$ is the $y$-intercept. We know the slope - it's $\frac{1}{4}$. All we need is the $y$-intercept.

Note that if we find a point $\left(x , y\right)$, 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 $f \left(x\right)$ at $x = 0$
$f \left(0\right) = {\left(0\right)}^{3} - 4 \left(0\right)$
$f \left(0\right) = 0$

Our point is $\left(0 , 0\right)$.

Making substitutions in $y = m x + b$, we have:
$0 = \left(\frac{1}{4}\right) \left(0\right) + b$
$0 = b$

The $y$-intercept is $0$, so the equation of the normal line is $y = \frac{1}{4} x$.