# What is the equation of the line that is normal to f(x)=(2x-4)^2/e^x at  x=2 ?

May 13, 2017

The vertical line $x = 2$

#### Explanation:

The normal line passes through the point $\left(2 , f \left(2\right)\right)$. Note that $f \left(2\right) = {\left(4 - 4\right)}^{2} / {e}^{2} = 0$, to the point is $\left(2 , 0\right)$.

The normal line is perpendicular to the tangent line. So, using the derivative, we can find the slope of the tangent line at $x = 2$, then take its opposite reciprocal to find the slope of the normal line.

To find this derivative, let's use the quotient rule:

$f ' \left(x\right) = \frac{\left(\frac{d}{\mathrm{dx}} {\left(2 x - 4\right)}^{2}\right) {e}^{x} - {\left(2 x - 4\right)}^{2} \left(\frac{d}{\mathrm{dx}} {e}^{x}\right)}{{e}^{x}} ^ 2$

We'll need the chain rule for $\frac{d}{\mathrm{dx}} {\left(2 x - 4\right)}^{2}$:

$f ' \left(x\right) = \frac{2 {\left(2 x - 4\right)}^{1} \left(\frac{d}{\mathrm{dx}} \left(2 x - 4\right)\right) {e}^{x} - {\left(2 x - 4\right)}^{2} {e}^{x}}{e} ^ \left(2 x\right)$

Note that $\frac{d}{\mathrm{dx}} \left(2 x - 4\right) = 2$:

$f ' \left(x\right) = \frac{4 \left(2 x - 4\right) {e}^{x} - {\left(2 x - 4\right)}^{2} {e}^{x}}{e} ^ \left(2 x\right)$

We can simplify, if we want:

$f ' \left(x\right) = \frac{\left(2 x - 4\right) {e}^{x} \left(4 - \left(2 x - 4\right)\right)}{e} ^ \left(2 x\right)$

$f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(8 - 2 x\right)}{e} ^ x$

$f ' \left(x\right) = \frac{4 \left(x - 2\right) \left(3 - x\right)}{e} ^ x$

So we see that $f ' \left(2\right) = 0$. This presents a tiny problem: we can't take the opposite reciprocal of $0$. But, a slope of $0$ means that the tangent line is horizontal at $x = 2$. Thus, the perpendicular normal line will be a vertical line. The vertical line that passes through the point $\left(2 , 0\right)$ is the line $x = 2$.