# What is the equation of the line normal to f(x)=3x^2 - 12x-5  at x=1?

So the equation normal to $f \left(x\right)$ at $x = 1$ is

$y = \frac{1}{6} x - \frac{85}{6}$

#### Explanation:

The normal is a line at right angles to the tangent.

First we need to calculate the $\mathrm{df} \frac{1}{\mathrm{dx}}$ hence we have that

$f ' \left(x\right) = 6 x - 12 \implies f ' \left(1\right) = = - 6$

So the normal will have a gradient of $m \cdot \left(- 6\right) = - 1 \implies m = \frac{1}{6}$

hence at point $\left(1 , f \left(1\right)\right)$ or $\left(1 , - 14\right)$ the equation is

$y = m \cdot x + k \implies y = \frac{1}{6} \cdot x + k$

replacing the value of point $\left(1 , f \left(1\right)\right)$ we get

$- 14 = \frac{1}{6} + k \implies k = - 14 - \frac{1}{6} \implies k = - \frac{85}{6}$