Building a rational function from a zero, asymptote and hole

Factor every numerator and denominator first. Each feature comes from a factor: - a zero at $x=3$ needs $(x-3)$ in the numerator and not in the denominator, - a vertical asymptote at $x=2$ needs $(x-2)$ in the denominator and not cancelled, - a hole at $x=1$ needs $(x-1)$ in both, so it cancels.

The denominator $x^2-3x+2$ factors as $(x-1)(x-2)$. For choice A the numerator $x^2-4x+3$ factors as $(x-1)(x-3)$. So $$h(x)=\frac{(x-1)(x-3)}{(x-1)(x-2)}$$ The $(x-1)$ cancels, which leaves a hole at $x=1$. What remains is $\frac{x-3}{x-2}$ for $x\ne 1$, with a zero at $x=3$ and a vertical asymptote at $x=2$. Every requirement is met.

The location of the hole is $(1,\frac{1-3}{1-2})=(1,2)$, although the question only needs its $x$-value.

Each choice

• A (correct). Every feature lines up: the shared factor $(x-1)$ makes the hole, the leftover $(x-2)$ below makes the asymptote, and the $(x-3)$ on top makes the zero.
• B. The numerator here is $(x-2)(x-3)$, so the factor that cancels is $(x-2)$. That puts the hole at $x=2$ and the asymptote at $x=1$, the reverse of what was asked. The right numbers appear, but it matters which factor cancels.
• C. It has the zero at 3 and the asymptote at 2, but nothing cancels. Without a shared factor there is no hole, and $x=1$ becomes a second vertical asymptote instead.
• D. This denominator is $(x-1)(x-3)$, so $(x-3)$ cancels with the numerator. That leaves a hole at $x=3$ rather than a zero there, an asymptote at $x=1$, and nothing at $x=2$.