## Task Condition

Derive the equation of the normal to the given curve at the point with abscissa \( x_{0} \).

\[
y=2 x^{2}-3 x+1, x_{\bar{u}}=1
\]
## Solution

Let's find $y^{\prime}: y^{\prime}=\left(2 x^{2}-3 x+1\right)^{\prime}=4 x-3$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=4 x_{\overline{0}}^{\prime}-3=4 \cdot 1-3=1$

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

$y-y_{\overline{\bar{\omega}}}=-\frac{1}{y_{\overline{\bar{U}}}^{\prime}} \cdot\left(x-x_{\overline{\bar{\omega}}}\right)_{\text {, where }}$

$y_{\bar{u}}^{\prime}=1$

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

We get:

$$
\begin{aligned}
& y-0=-\frac{1}{1} \cdot(x-1) \\
& y=-(x-1)+2 \\
& y=-x+1
\end{aligned}
$$

That is, the equation of the normal line is:

$y=-x+1$

## Problem Kuznetsov Differentiation 3-9