## Precalculus (10th Edition)

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. $f(x+h)=-2x^2-4xh-2h^2+x+h-1$
(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 $x$. $f(x+h)=-2(x+h)^2+(x+h)-1$ $f(x+h)=-2(x^2+2xh+h^2)+(x+h)-1$ $f(x+h)=-2x^2-4xh-2h^2+x+h-1$