Answer
$T_{2}=1+x+\frac{1}{2} x^{2}$
$0.0185$
Work Step by Step
Since
\begin{array}{ll}
{f(x)=e^{x},} & {f(0)=1} \\
{f^{\prime}(x)=e^{x},} & {f^{\prime}(0)=1} \\
{f^{\prime \prime}(x)=e^{x},} & {f^{\prime \prime}(0)=1}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=1+(x-0)+\frac{1}{2}(x-0)^{2} \\
&=1+x+\frac{1}{2} x^{2}
\end{aligned}
and
$$\left|f(-0.5)-T_{2}(-0.5)\right|=|0.606531-0.625|=0.0185$$
