Locate the minimum after a rational transformation

The curve C1C_1 with equation y=(x+2)2x+1y=\frac{(x+2)^2}{x+1} is transformed onto the curve C2C_2 with equation y=f(x)y=f(x). The curve C1C_1 has a minimum turning point (0,4)(0,4) which corresponds to the point with coordinates (a,b)(a,b) on the curve C2C_2, where a,b>0a,b>0. Given that f(x)f(x) has the form p2x2px−1+q\frac{p^2x^2}{px-1}+q, where p,qp,q are positive constants, express p,qp,q in terms of a,ba,b. marks

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

Solution

Q

Finding a transformation

The curve C1C_1 with equation y=(x+2)2x+1y=\dfrac{(x+2)^2}{x+1} is transformed onto the curve C2C_2 with equation y=f(x)y=f(x). The curve C1C_1 has a minimum turning point (0,4)(0,4) which corresponds to the point (a,b)(a,b) on C2C_2, where a,b>0a,b>0. Given that f(x)=p2x2px−1+qf(x)=\dfrac{p^2x^2}{px-1}+q with p,qp,q positive constants, express pp and qq in terms of aa and bb.

Explore this idea

Hint

  1. Simplify: replacing xx by x−2x-2 in C1C_1 gives a much simpler curve. What is it?

  2. Then compare with p2x2px−1\frac{p^2x^2}{px-1}: it is that curve with xx replaced by pxpx.

  3. Follow the minimum point (0,4)(0,4) through each step, then match the result with (a,b)(a,b).

Answer · Q

p=2/a,q=b−4p=2/a,\quad q=b-4.

The stated p,q>0 additionally requires a>0,b>4.

Explore the idea

Three steps from C₁ to C₂ (drawn with a = 1, b = 7)

With a = 1 and b = 7 we get p = 2 and q = 3. Watch the minimum point travel: (0,4) → (2,4) → (1,4) → (1,7).

Teal: C₁: y = (x+2)²/(x+1). No comparison overlay.

Start: C₁: y = (x+2)²/(x+1)

The original curve with its minimum (0,4)(0,4) and vertical asymptote x=−1x=-1.

  • Curve after the selected step
  • Image of the minimum point
Ideas to try
  1. Which step changes only the xx-coordinate of the minimum? Which changes only yy?
  2. Change the question: if a=4a=4, what must pp be?