Graphs & Transformations: Rational curve & oblique asymptote — NYJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
Curve \(C\) has equation \(y = \dfrac{px^2 + qx + r}{x-1}\), where \(p\), \(q\) and \(r\) are real numbers. Given that \(C\) has a turning point at \((3,\,7)\) and an oblique asymptote parallel to the line \(y = \dfrac{3}{2}x + \sqrt{5}\), find \(p\), \(q\) and \(r\).
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Since \(C\) has an oblique asymptote with gradient \(\dfrac{3}{2}\): \[ p = 1.5 \]
\(C\) contains the point \((3,\,7)\): \[\begin{aligned} 7 &= \frac{1.5(3)^2 + q(3) + r}{3 - 1}\\ 14 &= 13.5 + 3q + r\\ r &= 0.5 - 3q \end{aligned}\]
So \(y = \dfrac{1.5x^2 + qx + 0.5 - 3q}{x - 1}\).
Turning point at \((3,\,7)\): \[\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{(3x+q)(x-1) - (1.5x^2 + qx + 0.5 - 3q)(1)}{(x-1)^2}\\[4pt] 0 &= \frac{(9+q)(2) - (13.5 + 3q + 0.5 - 3q)}{(3-1)^2}\\ 0 &= (9+q)(2) - 14\\ q &= -2\\ r &= 0.5 - 3(-2) = 6.5 \end{aligned}\]