# What is the equation of the normal line of f(x)=(x+2)/(x-4) at x=3?

Apr 5, 2016

Equation of the normal line is $x - 2 y - 13 = 0$

#### Explanation:

At $x = 3$, $f \left(3\right) = \frac{3 + 2}{3 - 4} = \frac{5}{-} 1 = - 5$

Hence normal is desired at point $\left(3 , - 5\right)$ on the curve.

Slope of tangent is given by $f ' \left(x\right)$

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

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

Hence slope of tangent at $\left(3 , - 5\right)$ is $- \frac{2}{3 - 4} ^ 2 = - 2$.

Hence slope of normal would be $- \frac{1}{-} 2 = \frac{1}{2}$

And equation of the normal line using point slope form is

$\left(y - \left(- 5\right)\right) = \frac{1}{2} \left(x - 3\right)$ or

$2 \cdot \left(y + 5\right) = \left(x - 3\right)$ or $x - 2 y - 13 = 0$