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

Mar 11, 2018

$y = \frac{1}{19} x + \frac{592}{19}$

#### Explanation:

The function passes through the point

Denoting the point

$\left(x = - 3 , y = f \left(- 3\right)\right)$

by

$\left({x}_{1} , {y}_{1}\right)$

The tangent to that point has instantaneous slope
$f ' \left(- 3\right)$

A line perpendicular to the tangent will have slope
$- \frac{1}{f ' \left(- 3\right)}$

The line with this slope passing through $\left({x}_{1} , {y}_{1}\right)$ will be the normal to the function at that point.

Denoting its slope by $m$, t will have equation

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

For the normal, to the function $f \left(x\right)$ at $x = - 3$,

${x}_{1} = - 3$
${y}_{1} = f \left(- 3\right)$
$m = - \frac{1}{f ' \left(- 3\right)}$

Noting

$f \left(- 3\right) = 3 {\left(- 3\right)}^{2} - \left(- 3\right) + 1$
$= 27 + 3 + 1$
$= 31$

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

so that
$f ' \left(- 3\right) = 6 \left(- 3\right) - 1$
$= - 19$

and
$- \frac{1}{f ' \left(- 3\right)} = \frac{1}{19}$

The equation of the normal to $f \left(x\right)$ at $x = - 3$ is:

$\left(y - 31\right) = \frac{1}{19} \left(x - \left(- 3\right)\right)$

that is

$y = \frac{1}{19} x + \frac{3}{19} + 31$

that is

$y = \frac{1}{19} x + \frac{592}{19}$