Question 562·Hard·Math

The x-intercepts $(a,0)$ and $(b,0)$ have output $0$, so $g(a)=0$ and $g(b)=0$. You need the two input values $a$ and $b$; multiplying their $y$-coordinates would give $0$ instead.

Type $f(x)=x^3-9x$, then $g(x)=f(x+2)-f(x-2)$. Desmos draws $g$ crossing the $x$-axis twice. Keep the whole second function output after the minus sign; changing that subtraction changes the graph.

Add $y=0$ to mark the $x$-axis, then click both places where it meets the graph of $g$. Desmos labels them about $(-1.291,0)$ and $(1.291,0)$. So the two inputs are approximately $a=-1.291$ and $b=1.291$, in either order.

Type $-1.291(1.291)$; Desmos shows $-1.666681$. Then type $-\frac{5}{3}$; it shows about $-1.666667$. The small gap comes from the rounded intercept labels. The matching exact product is $ab=-\frac{5}{3}$. Choice A.

In $f(x+2)$, the entire input is $x+2$; in $f(x-2)$, it is $x-2$. Substitute both into the given rule:

Group the cubes and factor $-9$ from the other terms: $g(x)=\big[(x+2)^3-(x-2)^3\big]-9\big[(x+2)-(x-2)\big]$. This keeps the subtraction of the second output attached to every term in it.

Expand both cubes, keeping parentheses around the subtracted cube:

Cancel the $x^3$ and $12x$ terms: $g(x)=(12x^2+16)-36$. Combine the constants: $g(x)=12x^2-20$. The cubic terms cancel, so its intercepts are roots of a quadratic.

For a quadratic $Ax^2+Bx+C=0$, its two roots multiply to $C/A$: expanding $A(x-a)(x-b)$ makes its constant $Aab=C$. Here $g(x)=0$ has leading coefficient $12$ and constant $-20$, so: $ab=\frac{-20}{12}$. Reduce by $4$: $ab=-\frac{5}{3}$. The negative constant keeps the product negative. Choice A.