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

Nov 11, 2015

I found: $y = \frac{1}{5} x + \frac{56}{5}$

#### Explanation:

First we evaluate the derivative of the function at $x = 4$:
$f ' \left(x\right) = \frac{\left(2 x - 1\right) \left(x - 3\right) - \left({x}^{2} - x\right)}{x - 3} ^ 2$ using the Quotient Rule.
Evaluate it at $x = 4$:
$f ' \left(4\right) = \frac{\left(\left(7 \cdot 1\right) - 12\right)}{1} = - 5 = m$

This is the slope of the TANGENT to our curve at $x = 4$; to get the NORMAL we use $m ' = - \frac{1}{m} = \frac{1}{5}$ as slope.

The point where our normal passes on the curve represented by our function, has coordinates:
${x}_{0} = 4$
$f \left(4\right) = {y}_{0} = \frac{16 - 4}{1} = 12$
and the equation of the line through this point with slope $m '$ is given as:
$\left(y - {y}_{0}\right) = m ' \left(x - {x}_{0}\right)$
$y - 12 = \frac{1}{5} \left(x - 4\right)$
$y = \frac{1}{5} x + \frac{56}{5}$