Determine whether f(x) = x sqrt(x + 9) is even, odd, or neither - algebraically, graphically, and numerically by comparing f(x) with f(-x).

Step-by-step solution

1. 1

   Recall the two tests. $f$ is even if $f(-x) = f(x)$ for all $x$ in the domain (symmetry about the $y$-axis), and odd if $f(-x) = -f(x)$ (symmetry about the origin). If neither identity holds everywhere, the answer is "neither" — and most functions are neither.

2. 2

   Check the domain first. The radicand must satisfy $x + 9 \ge 0$, so the domain is $[-9, \infty)$. This interval is not symmetric about $0$: $x = 20$ is in the domain but $x = -20$ is not. For an even or odd function the domain has to be symmetric, so this observation alone already forces the answer, before any algebra.

3. 3

   Compute $f(-x)$ algebraically. Replace $x$ by $-x$ everywhere:

   $f(-x) = (-x)\sqrt{-x+9} = -x\sqrt{9-x}$

   Compare with $f(x) = x\sqrt{x+9}$ and $-f(x) = -x\sqrt{x+9}$. The radicals differ — $\sqrt{9-x}$ versus $\sqrt{x+9}$ — so $f(-x)$ equals neither one.

4. 4

   Confirm numerically. Evaluate at two symmetric pairs:

   $x$	$f(x)$
   $1$	$1\sqrt{10} \approx 3.16$
   $-1$	$-1\sqrt{8} \approx -2.83$
   $4$	$4\sqrt{13} \approx 14.42$
   $-4$	$-4\sqrt{5} \approx -8.94$

   If $f$ were even we would need $f(-1) = f(1)$, but $-2.83 \neq 3.16$. If $f$ were odd we would need $f(-1) = -f(1) = -3.16$, but $f(-1) = -2.83$. Both fail, and the $\pm 4$ pair fails the same way.

5. 5

   Read the graph the same way. The curve starts at $(-9, 0)$, dips to a minimum near $x = -6$, and rises without bound to the right; there is nothing to the left of $x = -9$. So it is symmetric neither about the $y$-axis nor about the origin.

6. 6

   Conclude. All three tests agree: $f(x) = x\sqrt{x+9}$ is neither even nor odd. The near-miss is instructive — $x\sqrt{x^2+9}$ would be odd, because squaring $x$ makes the radicand immune to the sign change.