Taylor Polynomials My question is this, 
Compute $T_2(x)\ $ at $x=0.8$ for $y=e^x$
I have figured out at that $T_2(x)$ equals:
$$e^{0.8}+e^{0.8}(x-0.8)+\frac{e^{0.8}((x-0.8)^2)}{2}$$
The second part of the question asks to compute the error,
$\big|\ e^x\ -T_2(x)\big|$ at $ x=-0.3 $
This is where I have trouble... Can anyone offer some assistance with computing the error? Thank you!

By the Taylor remainder formula, the Lagrange form of the remainder gives:
$$ R_2(x) = \frac{\exp(0.3)}{6} (x-0.8)^{3}\approx 0.225(x^3-2.4x^2+1.92x-0.512)\\
\implies \boxed{R_2(x)=0.225x^3-0.54x^2+0.432x-0.1152}$$
