# What is the equation of the normal line of f(x)=(x-2)/(x-3)^2 at x=5?

Nov 27, 2016

$y = 2 x - \frac{37}{4}$

#### Explanation:

We have $f \left(x\right) = \frac{x - 2}{x - 3} ^ 2$

Using the quotient rule we have:

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

When $x = 5 \implies f ' \left(5\right) = \frac{1 - 5}{5 - 3} ^ 3 = - \frac{4}{2} ^ 3 = - \frac{1}{2}$

So the gradient of the tangent when $x = 5$ is ${m}_{T} = \frac{1}{2}$

The tangent and normal are perpendicular so the product of their gradients is $- 1.$

So the gradient of the normal when $x = 5$ is ${m}_{N} = 2$

Also $f \left(5\right) = \frac{5 - 2}{5 - 3} ^ 2 = \frac{3}{2} ^ 2 = \frac{3}{4}$

So the normal passes through (5,3/4) and has gradient $2$, so using $y = {y}_{1} = m \left(x - {x}_{1}\right)$ the required equation is;

$y - \frac{3}{4} = 2 \left(x - 5\right)$
$y - \frac{3}{4} = 2 x - 10$
$y = 2 x - \frac{37}{4}$