# What is the equation of the line tangent to  f(x)=(-x^2-1)/(x+4)  at  x=5 ?

Jan 9, 2018

$y + \frac{26}{9} = - \frac{64}{81} \left(x - 5\right)$

#### Explanation:

the slope of the tangent line at $x = 5$ is the derivative of the function evaluated at $x = 5$
use quotient rule or $\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$ to find the derivative:

$\frac{d}{\mathrm{dx}} \left(\frac{- {x}^{2} - 1}{x + 4}\right)$
$= \frac{- 2 x \cdot \left(x + 4\right) - 1 \cdot \left(- {x}^{2} - 1\right)}{{\left(x + 4\right)}^{2}}$

when $x = 5$:

$= \frac{- 2 \left(5\right) \cdot \left(5 + 4\right) - 1 \cdot \left(- {5}^{2} - 1\right)}{{\left(5 + 4\right)}^{2}}$

$= \frac{- 90 + 26}{81}$
$= - \frac{64}{81}$

find the point $\left(5 , f \left(5\right)\right)$ to complete the equation of the tangent line:
$f \left(5\right) = \frac{- {5}^{2} - 1}{5 + 4}$
$f \left(5\right) = \frac{- 26}{9}$

use point slope form to find the equation of the line:
$y - \left(- \frac{26}{9}\right) = - \frac{64}{81} \left(x - 5\right)$
$y + \frac{26}{9} = - \frac{64}{81} \left(x - 5\right)$