Answer
$p_0(x)=2\\ p_1(x)=2-3(x-1)\\ p_2(x)=2-3(x-1)\\p_3(x)=2-3(x-1)+(x-1)^3$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_n(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
We are given that $f(1)=2$, $f'(1)=-3$, $f''(1)=0$, $f'''(1)=6$
Hence the nth Taylor polynomial for a function $p_n(x)$ for $x_0=1$ is:
$p_0(x)=2\\ p_1(x)=2-3(x-1)\\ p_2(x)=2-3(x-1)\\p_3(x)=2-3(x-1)+(x-1)^3$
