Answer
$n=0$: ${\ \ \ }$ ${p_0} = 1$
$n=1$: ${\ \ \ }$ ${p_1} = 1 - 2x$
$n=2$: ${\ \ \ }$ ${p_2} = 1 - 2x + 2{x^2}$
$n=3$: ${\ \ \ }$ ${p_3} = 1 - 2x + 2{x^2} - \frac{4}{3}{x^3}$
Work Step by Step
We compute the derivatives of $f\left( x \right) = {{\rm{e}}^{ - 2x}}$ for $n = 0,1,2,3$:
$\begin{array}{*{20}{c}}
n&{}&{{f^{\left( n \right)}}\left( x \right)}&{}&{{f^{\left( n \right)}}\left( 0 \right)}\\
0&{}&{{{\rm{e}}^{ - 2x}}}&{}&1\\
1&{}&{ - 2{{\rm{e}}^{ - 2x}}}&{}&{ - 2}\\
2&{}&{4{{\rm{e}}^{ - 2x}}}&{}&4\\
3&{}&{ - 8{{\rm{e}}^{ - 2x}}}&{}&{ - 8}
\end{array}$
By Definition 9.7.3, the $n$th Taylor polynomial about $x = {x_0}$ for $f$ is
${p_n}\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \dfrac{{f{\rm{''}}\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2} + \dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{\left( {x - {x_0}} \right)^n}$
The first four distinct Taylor polynomials about $x = {x_0} = 0$:
$n=0$: ${\ \ \ }$ ${p_0} = f\left( 0 \right) = 1$
$n=1$: ${\ \ \ }$ ${p_1} = f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right) = 1 - 2x$
$n=2$: ${\ \ \ }$ ${p_2} = f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right) + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{\left( {x - 0} \right)^2} = 1 - 2x + 2{x^2}$
$n=3$: ${\ \ \ }$ ${p_3} = f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right) + \dfrac{{f{\rm{''}}\left( 0 \right)}}{{2!}}{\left( {x - 0} \right)^2} + \dfrac{{f{\rm{'''}}\left( 0 \right)}}{{3!}}{\left( {x - 0} \right)^3} = 1 - 2x + 2{x^2} - \dfrac{4}{3}{x^3}$
Please see the figure attached for the graphs of $f\left( x \right) = {{\rm{e}}^{ - 2x}}$, ${p_0}$, ${p_1}$, ${p_2}$, ${p_3}$.
