# What is the equation of the normal line of f(x)=(x-3)/(x+4) at x=3?

Mar 30, 2017

$y = - 7 x + 21$

#### Explanation:

As long as $f ' \left(x\right) \ne 0$, the normal line of $f \left(x\right)$ at $x$ can be written on the form $y = a x + m$, where $a$ is the slope and $m$ is the intercept with the $y$-axis (see further down what happens for the case $f ' \left(x\right) = 0$). If we first find $a$, then we can find $m$ through elimination.

We know that the normal line of $f \left(x\right)$ is perpendicular to the tangent line of $f \left(x\right)$ at $x = 3$. Therefore, if we can find the slope of $f \left(x\right)$ at $x = 3$, then we can also find the slope of the normal line at $x = 3$.

Denote the slope of the tangent line $b$. If $a , b \ne 0$, then $a \cdot b = - 1$ (see further down for explanation).

The slope of the tangent line at a value $x$ is by definition $f ' \left(x\right)$, which we can compute using the product rule (or quotient rule) .

$f ' \left(x\right) = \frac{1}{x + 4} - \frac{x - 3}{x + 4} ^ 2 = \frac{x + 4}{x + 4} ^ 2 - \frac{x - 3}{x + 4} ^ 2 = \frac{\left(x + 4\right) - \left(x - 3\right)}{x + 4} ^ 2 = \frac{7}{x + 4} ^ 2.$

Therefore $b = f ' \left(3\right) = \frac{7}{7} ^ 2 = \frac{1}{7}$, and we find the slope of the normal line by solving $a \cdot b = - 1$ for $a$, which gives that
$a = - 7$.

Since the normal line passes through $x = 3$, we know that the point $\left(3 , f \left(3\right)\right)$ must lie on the line. Evaluating $f \left(3\right) = 0$, and inserting in the equation for the normal line we get that
$0 = a \cdot 3 + m$.
Inserting $a = - 7$ and solving for $m$ gives that
$m = 21$,
and we have found that the normal line is given by
$y = - 7 x + 21$.

Comment 1: In case $f \left(x\right) = 0$, the tangent to $f \left(x\right)$ is flat, which means that the normal line is vertical. Then we must use the general equation for the line
$p y + q x = r$,
which, since the normal line is vertical must have $p = 0$. To find $q$ and $r$, insert a pair of $x$ and $y$-values that you know lie on the line, like we did earlier.

Comment 2: In case $f \left(x\right)$ is not differentiable at $x$, then the function does not have a normal line there.

Comment 3: As for showing the formula $a \cdot b = - 1$, you need either geometry, trigonometry or linear algebra. Here are some proofs.