# What is the equation of the normal line of f(x)=(x-3)/(x+4) at x=-1?

May 27, 2018

$21 y + 27 x + 55 = 0$

#### Explanation:

$f \left(x\right) = \frac{x - 3}{x + 4}$

Recall: the quotient rule is given by $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v u ' - u v '}{v} ^ 2$ if $y = \frac{u}{v}$

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

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

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

When $x = - 1$

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

$f ' \left(- 1\right) = \frac{7}{9}$

You just found the gradient of the tangent but we want the normal. Recall that there is a rule ${m}_{1} {m}_{2} = - 1$ where ${m}_{1}$ is a gradient and ${m}_{2}$ is another gradient and when you multiply it together, it equals to $- 1$

ie. $\frac{7}{9} {m}_{2} = - 1$

${m}_{2} = - \frac{9}{7}$

When $x = - 1$, y is equal to $f \left(- 1\right) = \frac{- 1 - 3}{- 1 + 4} = - \frac{4}{3}$
$\left(- 1 , - \frac{4}{3}\right)$

$\left(y + \frac{4}{3}\right) = - \frac{9}{7} \left(x + 1\right)$
$7 y + \frac{28}{3} = - 9 x - 9$
$7 y + 9 x + \frac{55}{3} = 0$
$21 y + 27 x + 55 = 0$