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2004 Paper 3 Q2
D: 1700.0
B: 1516.0
curve sketching
asymptotes
oblique asymptote
rational function
double root
tangent
vertical asymptote
The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]
- Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.
- Show that the equation \(\f ( x ) =0\) has a double root.
- Sketch the curve.
Solution:
- Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
- \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
