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FM Nov 2024 p13 q06
4155
The curve C has equation \(y = \frac{4x^2 + x + 1}{2x^2 - 7x + 3}\).
Find the equations of the asymptotes of C.
Find the coordinates of any stationary points on C.
Sketch C, stating the coordinates of any intersections with the axes.
Sketch the curve with equation \(y = \left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) and state the set of values of \(k\) for which it has 4 distinct real solutions.
Solution
(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 2\).
(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\). Simplifying the numerator gives \(-3x^2 + 2x + 1 = 0\). Solving this quadratic gives the stationary points \(\left( -\frac{1}{3}, \frac{1}{3} \right)\) and \((1, -3)\).
(c) The sketch should show the asymptotes at \(x = \frac{1}{2}, x = 3\), and \(y = 2\). The curve intersects the y-axis at \((0, \frac{1}{3})\).
(d) For \(y = \left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at four points. This occurs when \(k > 3\).
