# What is the equation of the normal line of f(x)=x^2-4x at x=1?

Nov 15, 2015

$y = \frac{1}{2} x - \frac{7}{2}$

#### Explanation:

A normal line is a perpendicular line. In order to find it at a certain point, we need to find the tangent line at that point, and then find the line passing through that point whose slope, multiplied by the slope of the tangent line, is equal to $- 1$.

The slope of the tangent line at a point is given by the derivative of the function, evaluated at that point.

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

Thus the slope of the tangent line of $f \left(x\right)$ at $x = 1$ is $- 2$ and the slope of the normal line ${m}_{n}$ has the property
$- 2 {m}_{n} = - 1$
$\implies {m}_{n} = \frac{1}{2}$

The equation of a line with slope $m$ passing through the point $\left({x}_{1} , {y}_{1}\right)$ is given by $y - {y}_{1} = m \left(x - {x}_{1}\right)$.
As we are looking for the normal line at $x = 1$, we need it to pass through the point $\left(1 , f \left(1\right)\right) = \left(1 , - 3\right)$.

Putting this together with the slope gives us
$y + 3 = \frac{1}{2} \left(x - 1\right)$

Thus the equation of the normal line is
$y = \frac{1}{2} x - \frac{7}{2}$