## Solution

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

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

Then:

$$
y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=\frac{4-1}{2}=\frac{3}{2}
$$

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

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

$y_{0}^{\prime}=\frac{3}{2}$

$y_{0}=y\left(x_{0}\right)=\frac{4^{2}-2 \cdot 4-3}{4}=\frac{16-8-3}{4}=\frac{5}{4}$

We get:

$$
\begin{aligned}
& y-\frac{5}{4}=\frac{3}{2} \cdot(x-4) \\
& y=\frac{3}{2} \cdot x-6+\frac{5}{4} \\
& y=\frac{3}{2} \cdot x-\frac{19}{4}
\end{aligned}
$$

That is, the equation of the tangent line is:

$y=\frac{3}{2} \cdot x-\frac{19}{4}$

## Problem Kuznetsov Differentiation 3-30
## Task Condition

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

$y=\frac{x^{2}-2 x-3}{4}, x_{0}=4$