## Task Condition

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

$y=\frac{x^{2}-3 x+6}{x^{2}}, x_{0}=3$
## Solution

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

$$
y^{\prime}=\left(\frac{x^{2}-3 x+6}{x^{2}}\right)^{\prime}=\left(1-\frac{3}{x}+\frac{6}{x^{2}}\right)^{\prime}=\frac{3}{x^{2}}-2 \cdot \frac{6}{x^{3}}=\frac{3}{x^{2}}-\frac{12}{x^{3}}
$$

Then:

$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{3}{x_{0}^{2}}-\frac{12}{x_{0}^{3}}=\frac{3}{3^{2}}-\frac{12}{3^{3}}=\frac{1}{3}-\frac{1}{9}=-\frac{1}{9}$

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

$$
\begin{aligned}
& y-y_{\overline{0}}=-\frac{1}{y_{\overline{0}}^{\prime}} \cdot\left(x-x_{\overline{0}}\right), \text { where } \\
& y_{\overline{0}}^{\prime}=-\frac{1}{9} \\
& y_{\overline{0}}=y\left(x_{\overline{0}}\right)=\frac{x_{\overline{0}}^{2}-3 x_{0}+6}{x_{\overline{0}}^{2}}=\frac{3^{2}-3 \cdot 3+6}{3^{2}}=\frac{6}{9}=\frac{2}{3}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& y-\frac{2}{3}=-\frac{1}{\left(-\frac{1}{9}\right)} \cdot(x-3) \\
& y=9 \cdot(x-3)+\frac{2}{3} \\
& y=9 x-27+\frac{2}{3} \\
& y=9 x-26 \frac{1}{3}
\end{aligned}
$$

Thus, the equation of the normal line is:

$y=9 x-26 \frac{1}{3}$

## Problem Kuznetsov Differentiation 3-10