Answer
$f(2)=12$
$f(-2)=16$
$f(a)=3a^{2}-a+2$
$f(-a)=3a^{2}+a+2$
$f(a+1)=3a^{2}+5a+4$
$2f(a)=6a^{2}-2a+4$
$f(2a)=12a^{2}-2a+2$
$f(a^{2})=3a^{4}-a^{2}+2$
$[f(a)]^{2}=9a^4-6a^3+13a^2-4a+4$
$f(a+h)=3a^2+6ah+3h^2-a-h+2$
Work Step by Step
$$\bf f(x)=3x^{2}-x+2$$
Replace $x$ with the given values to obtain the corresponding values of $f$.
$f(2)=3(2)^{2}-2+2=3(4)=\bf12$
$f(-2)=3(-2)^{2}+2+2=3(4) +4 =\bf16$
$f(a)=\bf3a^{2}-a+2$
$f(-a)=3(-a)^{2}-(-a)+2=\bf3a^{2}+a+2$
$f(a+1)=3(a+1)^{2}-(a+1)+2$
$=3(a^{2}+2a+1)-a-1+2$
$=\bf3a^{2}+5a+4$
$2f(a)=2(3a^{2}-a+2) = \bf6a^{2}-2a+4$
$f(2a)=3(2a)^{2}-(2a)+2$
$=3(4a^2)-2a+2$
$=\bf12a^{2}-2a+2$
$f(a^{2})=3(a^2)^{2}-(a^2)+2$
$=\bf3a^{4}-a^{2}+2$
$[f(a)]^{2}=(3a^{2}-a+2)^{2}$
$=9a^4-3a^3-3a^3+6a^2+a^2+6a^2-2a-2a+4$
$=\bf9a^4-6a^3+13a^2-4a+4$
$f(a+h)=3(a+h)^{2}-(a+h)+2$
$=3(a^2+2ah+h^2)-a-h+2$
$=\bf3a^2+6ah+3h^2-a-h+2$
