Answer
a. $ f(0) = -1$
b. $ f(1) = -2$
c. $ f(-1) = -4$
d. $ f(-x)=-2x^2-x-1$
e. $ -f(x)=2x^2-x+1$
f. $ f(x+1)=-2x^2-3x-2$
g. $ f(2x)=-8x^2+2x-1$
h. $ f(x+h)=-2x^2-4xh-2h^2+x+h-1$
Work Step by Step
(a).
To Evaluate $f(0)$ in the given function, substitute $0$ in the place of $x$.
$f(0)=-2(0)^2+(0)-1$
$f(0)=0+0-1$
$f(0)=-1$
(b).
To Evaluate $f(1)$ in the given function, substitute $1$ in the place of $x$.
$f(1)=-2(1)^2+(1)-1$
$f(1)=-2(1)+1-1$
$f(1)=-2$
(c).
To Evaluate $f(-1)$ in the given function, substitute $-1$ in the place of $x$.
$f(-1)=-2(-1)^2+(-1)-1$
$f(-1)=-2(1)-1-1$
$f(-1)=-2-1-1$
$f(-1)=-4$
(d)
To Evaluate $f(-x)$ in the given function, substitute $-x$ in the place of $x$.
$f(-x)=-2(-x)^2+(-x)-1$
$f(-x)=-2(x)-x-1$
$f(-x)=-2x^2-x-1$
(e)
To Evaluate $-f(-x)$ in the given function,
$-f(x)=-(-2x^2+x-1)$
$-f(x)=2x^2-x+1$
(f)
To Evaluate $f(x+1)$ in the given function, substitute $x+1$ in the place of $x$.
$f(x+1)=-2(x+1)^2+(x+1)-1$
$f(x+1)=-2(x^2+2x+1)+(x+1)-1$
$f(x+1)=-2x^2-4x-2+x+1-1$
$f(x+1)=-2x^2-3x-2$
(g).
To Evaluate $f(2x)$ in the given function, substitute $2x$ in the place of $x$.
$f(2x)=-2(2x)^2+(2x)-1$
$f(2x)=-2(4x^2)+2x-1$
$f(2x)=-8x^2+2x-1$
(h)
To Evaluate $f(x+h)$ in the given function, substitute $x+h$ in the place of $x$.
$f(x+h)=-2(x+h)^2+(x+h)-1$
$f(x+h)=-2(x^2+2xh+h^2)+(x+h)-1$
$f(x+h)=-2x^2-4xh-2h^2+x+h-1$
