Answer
$f(x)=21-36(x+2)+25(x+2)^2-8(x+2)^3+(x+2)^4$
Work Step by Step
We have $f(x)=x^4+x^2+1 \implies f(-2)=21$; $f'(x)=4x^3+2x \implies f'(-2)=-36$;
$f''(x)=12x^2+2 \implies f''(-2)=50\\ f'''(x)=24x \implies f'''(2)=24$
Therefore, the Taylor's series at $x=-2$ is as follows:
$f(x)=f(-2)+f'(-2) (x+2)+\dfrac{f''(-2)(x+2)^2 }{2!}+\dfrac{f''(-2)(x+2)^3 }{3!}+\dfrac{f''(-2)(x+2)^4 }{4!}$
or, $f(x)=21-36(x+2)+25(x+2)^2-8(x+2)^3+(x+2)^4$
