Question

LET f(x) =-2x^2+x-5
a. find f(-x). is f even or odd neither
b. find f(x-h)-f(x) all over h , h cant equal 0

Answer 1

f(-x)=-2(-x)^2+(-x)-5
f(-x)=-2x^2-x-5
f(x-h)-f(x) all over h= -2(x-h)^2+x-h-5+2x^2-x+5 all over h
-2(x-h)^2-h+2x^2 all over h

Answer 2

\[=\frac{f(x-h)-f(x)}{h}=\frac{-2(x-h)^2+x-h-5-(-2x^2+x-5)}{h}\]
\[=\frac{-2(x^2-2xh+h^2)+x-h-5+2x^2-x+5}{h}\]
\[=\frac{-2x^2+4xh-h^2+x-h-5+2x^2-x+5}{h}\]
\[=\frac{4xh-h-h^2}{h}\]
\[=4x-1-h\]

Answer 3

f is neither even nor odd, because it has an even and an odd exponent. you can check that
\[f(-x)=-2(-x)^2+(-x)-5=-2x^2-x-5\] and since this is neither f nor -f it is not even or odd