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FM November 2022 p11 q07
4252
The curve \(C\) has equation \(y = \frac{5x^2}{5x-2}\).
Find the equations of the asymptotes of \(C\).
Find the coordinates of the stationary points on \(C\).
Sketch \(C\).
Sketch the curve with equation \(y = \left| \frac{5x^2}{5x-2} \right|\) and find in exact form the set of values of \(x\) for which \(\left| \frac{5x^2}{5x-2} \right| < 2\).
Solution
(a) The vertical asymptote occurs where the denominator is zero: \(5x - 2 = 0 \Rightarrow x = \frac{2}{5}\). For the oblique asymptote, divide \(5x^2\) by \(5x - 2\) to get \(y = x + \frac{2}{5}\).
(b) Differentiate \(y = \frac{5x^2}{5x-2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(5x-2)(10x) - (5x^2)(5)}{(5x-2)^2}\). Set \(\frac{dy}{dx} = 0\) to find stationary points: \(5x^2 - 4x = 0 \Rightarrow x(5x - 4) = 0\). Thus, \(x = 0\) or \(x = \frac{4}{5}\). The coordinates are \((0,0)\) and \(\left( \frac{4}{5}, \frac{8}{5} \right)\).
(c) Sketch the curve showing the asymptotes and stationary points.
(d) Solve \(\left| \frac{5x^2}{5x-2} \right| < 2\). This gives two cases: \(\frac{5x^2}{5x-2} < 2\) and \(\frac{5x^2}{5x-2} > -2\). Solving these inequalities gives the intervals \(-1 - \frac{1}{3}\sqrt{5} < x < -1 + \frac{1}{3}\sqrt{5}\) and \(1 - \frac{1}{3}\sqrt{5} < x < 1 + \frac{1}{3}\sqrt{5}\).
