Question 222·Medium·Advanced Math

The denominator $x-2$ is zero at $x=2$, so use $1$ instead. Type $x_1=1$, then type the original fraction with $x_1$ in place of $x$. Desmos prints $7$, so any equivalent expression must also give $7$ at $1$. Don't test at $2$: the original fraction is undefined there.

Type $x_1^2-4$, $x_1^2+4$, and $x_1^2+2$. Desmos prints $-3$, $5$, and $3$, respectively. None equals the original's $7$, so one failed input rules out each of them.

Type $y=\frac{x^3-8}{x-2}$ and $y=x^2+2x+4$. The curves overlap, rather than crossing at only one input. So $x^2+2x+4$ matches the original for its allowed inputs. The original still excludes $x=2$, even though that missing point may not be visible. Choice A.

Since $8=2^3$, the numerator is a difference of cubes. Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x$ and $b=2$:

The $2x$ term matters: it comes from $ab=x\cdot2$ in the rule. Leaving it out would change the numerator.

Because the given condition says $x\ne2$, the factor $x-2$ isn't zero. Divide the numerator and denominator by that whole factor:

So the equivalent expression is $x^2+2x+4$ for the inputs the original allows. Choice A.