Answer
(a) $f(2)=-3$
(b) $f(x)+f(2)=x^2-4x-2$
(c) $f(-x)=x^2+4x+1$
(d) $-f(x)=-x^2+4x-1$
(e) $f(x+2)=x^2-3$
(f) $\dfrac{f(x+h)-f(x)}{h}= 2x+h-4,\space x\ne0$
Work Step by Step
(a) $f(2)=2^2-4(2)+1=4-8+1=-3$
(b) $f(x)+f(2)=x^2-4x+1+(-3)=x^2-4x-2$
(c) $f(-x)=(-x)^2-4(-x)+1=x^2+4x+1$
(d) $-f(x)=-(x^2-4x+1)=-x^2+4x-1$
(e) $f(x+2)=(x+2)^2-4(x+2)+1=x^2+2x+2x+4-4x-8+1=x^2-3$
(f) $\dfrac{f(x+h)-f(x)}{h}=$
$\dfrac{(x+h)^2-4(x+h)+1-(x^2-4x+1)}{h}=$
$\dfrac{x^2+2xh+h^2-4x-4h+1-x^2+4x-1}{h}=$
$\dfrac{2xh+h^2-4h}{h}=$
$\dfrac{h(2x+h-4)}{h}=$
$2x+h-4,\space x\ne0$
