## Condition of the problem

To derive the equation of the normal to the given curve at the point with abscissa $x_{0}$.

$y=2 x^{2}+3 x-1, x_{0}=-2$
## Solution

Let's find $y^{\prime}:$

$y^{\prime}=\left(2 x^{2}+3 x-1\right)^{\prime}=4 x+3$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=4 x_{0}+3=4 \cdot(-2)+3=-8+3=-5$

Since $y^{\prime}\left(x_{0}\right) \neq 0$, the equation of the normal line is:

$y-y_{0}=-\frac{1}{y_{0}^{\prime}} \cdot\left(x-x_{0}\right)$, where

$y_{0}^{\prime}=-5$

$y_{0}=y\left(x_{0}\right)=2 x_{0}^{2}+3 x_{0}-1=2 \cdot(-2)^{2}+3 \cdot(-2)-1=8-6-1=1$

We get:

$y-1=-\frac{1}{-5} \cdot(x-(-2))$

$y-1=\frac{1}{5} \cdot(x+2)$

$y-1=\frac{x}{5}+\frac{2}{5}$

$y=\frac{x}{5}+1 \frac{2}{5}$

Thus, the equation of the normal line is:

$y=\frac{x}{5}+1 \frac{2}{5}$

## Problem Kuznetsov Differentiation 3-2