Match $f(x)=\sqrt{x}$ at $x_{0}=1$ by a quadratic polynomial, i.e., find constants $a_{0}, a_{1}, a_{2}$ such that
$$
p_{2}(x)=a_{0}+a_{1} x+a_{2} x^{2} \approx f(x)
$$
for values of $x$ near $x_{0}=1$
We will determine the coefficients $a_{0}, a_{1}, a_{2}$ by matching derivatives of $f$ at $x_{0}=1$, i.e., we will enforce ( 3 conditions for 3 coefficients)
$$
\begin{aligned}
&p_{2}(1)=f(1)=1 \
&p_{2}^{\prime}(1)=f^{\prime}(1)=\frac{1}{2} \
&p_{2}^{\prime \prime}(1)=f^{\prime \prime}(1)=-\frac{1}{4}
\end{aligned}
$$
since we know $f^{\prime}(x)=\frac{1}{2 \sqrt{x}}, f^{\prime \prime}(x)=-\frac{1}{4 x^{3 / 2}}$.
MATH4076 COURSE NOTES ：
Since our assumption implies
$$
\begin{aligned}
&p_{2}^{\prime}(x)=a_{1}+2 a_{2} x, \
&p_{2}^{\prime \prime}(x)=2 a_{2}
\end{aligned}
$$
we obtain a system of three linear equations in the three unknowns $a_{0}, a_{1}$ and $a_{2}$ :
$p_{2}(1)=a_{0}+a_{1}+a_{2}=1$
$p_{2}^{\prime}(1)=a_{1}+2 a_{2}=\frac{1}{2}$
$p_{2}^{\prime \prime}(1)=\quad 2 a_{2}=-\frac{1}{4} .$
Solving this triangular system we get $a_{2}=-\frac{1}{8}, a_{1}=\frac{3}{4}$, and $a_{0}=\frac{3}{8}$ so that
$$
p_{2}(x)=\frac{3}{8}+\frac{3}{4} x-\frac{1}{8} x^{2}
$$