## Task Condition

Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.

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

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

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

Then:

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

Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:

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

$y_{0}^{\prime}=1$

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

We get:

$$
\begin{aligned}
& y-1=1 \cdot(x-3) \\
& y=x-3+1
\end{aligned}
$$

$y=x-2$

That is, the equation of the tangent line is:

$y=x-2$

Problem Kuznetsov Differentiation 3-22