How do i tell if y=x^3+x^2 is even or odd?

If a function is even, then f(-x) = x.

If a function is odd, then f(-x) = -x.

y = x³ + x² → f(x) = x³ + x² → -f(x) = -(x³ + x²) = -x³ - x²

f(-x) = (-x)³ + (-x)² = [(-1)(x)]³ + [(-1)(x)]² = (-1)³x³ + (-1)²x²

= -1x³ + 1x² =-x³ + x²

f(-x) ≠ f(x) and f(-x) ≠ -f(x)

y = x³ + x² is not odd and not even

neither

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

10 - 75

step-by-step explanation:

there's nothing attached so it's most likely anything in that range.