Some graphs are their own mirror image. You met symmetry in the Parent Functions: $x^2$ is symmetric about the $y$-axis, and $x^3$ looks the same after a half turn about the origin. Algebra can detect this symmetry without drawing anything.
Key idea
$f$ is even when $f(-x) = f(x)$ for every $x$: the graph is symmetric about the $y$-axis. $f$ is odd when $f(-x) = -f(x)$ for every $x$: the graph is symmetric about the origin. Most functions are neither.
Why it works
Replacing $x$ with $-x$ reflects a graph across the $y$-axis. If that gives back the same formula, the graph is its own mirror image: even. If it gives back the negative of the formula, then reflecting across the $y$-axis and then across the $x$-axis (a half turn) returns the graph to itself: odd. The names come from powers: $(-x)^n = x^n$ when $n$ is even and $(-x)^n = -x^n$ when $n$ is odd.
Worked examples
Worked example: An even polynomial
Is $f(x) = x^4 - 3x^2 + 2$ even, odd or neither?
- $f(-x) = (-x)^4 - 3(-x)^2 + 2$ $= x^4 - 3x^2 + 2$.
- That is exactly $f(x)$. (The constant 2 is $2x^0$, an even power.)
Answer: Even
Worked example: An odd polynomial
Is $g(x) = x^3 - 5x$ even, odd or neither?
- $g(-x) = (-x)^3 - 5(-x)$ $= -x^3 + 5x$.
- Compare with $-g(x) = -x^3 + 5x$. They are equal.
Answer: Odd
Worked example: Neither
Is $h(x) = x^2 + x$ even, odd or neither?
- $h(-x) = x^2 - x$.
- That is not $h(x) = x^2 + x$, and not $-h(x) = -x^2 - x$.
- Numbers confirm it: $h(1) = 2$ but $h(-1) = 0$.
Answer: Neither
Common mistake
Deciding from the exponents you happen to see. $x^2 + x$ has an even power, but it also has an odd one, so it is neither. A hidden constant counts too: $x^3 + 1$ is neither. A quick test can rule things out: if $f(-1) \ne f(1)$ it is not even, and if $f(-1) \ne -f(1)$ it is not odd. One number can disprove symmetry but never prove it; for that, compute $f(-x)$.