Condition of the problem

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

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

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

Then:

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

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

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right), \text { where } \\
& y_{0}^{\prime}=-4 \\
& y_{0}=y\left(x_{0}\right)=2 x_{0}^{2}+3=2 \cdot(-1)^{2}+3=2+3=5
\end{aligned}
$$

We get:

$y-5=-4 \cdot(x-(-1))$

$y=-4 \cdot(x+1)+5$

$y=-4 x+1$

Thus, the equation of the tangent line is:

$y=-4 x+1$

## Problem Kuznetsov Differentiation 3-13