Simultaneous linear equations: JPJC 2024 Promo Q3

$f(x)=\dfrac a{x^2}+bx+c$

At $(-1,4)$, $f(-1)=4$ $$\begin{aligned} \frac a{(-1)^2}+b(-1)+c&=4\\ a-b+c&=4 \quad\text{(1)} \end{aligned}$$ At $(2,-11)$, $f(2)=-11$ $$\begin{aligned} \frac a{(2)^2}+b(2)+c&=-11\\ \frac a4+2b+c&=-11 \quad\text{(2)} \end{aligned}$$ $y=\dfrac1{f(x)}$ has a vertical asymptote with equation $x=1$ implies $C$ has an $x$-intercept at $x=1$.

$f(1)=0$ $$\begin{aligned} \frac a{(1)^2}+b(1)+c&=0\\ a+b+c&=0 \quad\text{(3)} \end{aligned}$$

Or

$$f(x)=\frac{a+bx^3+cx^2}{x^2}, \qquad \frac1{f(x)}=\frac{x^2}{a+bx^3+cx^2}.$$ $y=\dfrac1{f(x)}$ has a vertical asymptote with equation $x=1$, $$\begin{aligned} a+b(1)^3+c(1)^2&=0\\ a+b+c&=0 \quad\text{(3)} \end{aligned}$$ Solving (1), (2) & (3) using GC: $a=12$, $b=-2$, $c=-10$.

Hence, $f(x)=\dfrac{12}{x^2}-2x-10$.