Locate the minimum after a rational transformation · source reference

Printed question

The curve $C_1$ with equation $y=\frac{(x+2)^2}{x+1}$ is transformed onto the curve $C_2$ with equation $y=f(x)$. The curve $C_1$ has a minimum turning point $(0,4)$ which corresponds to the point with coordinates $(a,b)$ on the curve $C_2$, where $a,b>0$. Given that $f(x)$ has the form $\frac{p^2x^2}{px-1}+q$, where $p,q$ are positive constants, express $p,q$ in terms of $a,b$. [4]

Question booklet attribution: [2021/ASRJC/Prelim/I/Q7(i)]. Solution booklet attribution: [ASRJC/2021/Prelim/I/Q7(i)].

Printed working

$y=\frac{(x+2)^2}{x+1}\xrightarrow{x\mapsto x-2}y=\frac{x^2}{x-1}$ (Translation of 2 units in the positive $x$-direction).

$\xrightarrow{x\mapsto px}y=\frac{(px)^2}{px-1}=\frac{p^2x^2}{px-1}$ (Scaling parallel to $x$-axis by factor $\frac1p$).

$\xrightarrow{y\mapsto y-q}y=\frac{p^2x^2}{px-1}+q$ (Translation of $q$ units in the positive $y$-direction).

The minimum turning point on $C_1$, $(0,4)$ corresponds to $(a,b)$ on $C_2$.

Hence $a=\frac1p(0+2)\Rightarrow p=\frac2a$; $b=4+q\Rightarrow q=b-4$.

Printed answer key

$p=\frac2a$; $q=b-4$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.