The curve $C$ has equation $y=\frac{{{x}^{2}}+2x}{x+a}$, where $a$ is a constant such tha

Question

The curve $C$ has equation $y=\frac{{{x}^{2}}+2x}{x+a}$, where $a$ is a constant such that $a\ne 0$ and $a\ne 2$. Find the equations of the asymptotes of $C$. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$. Hence find the set of values of $a$ for which $C$ has no stationary points. Sketch $C$ for $a=-1$, showing clearly the axial intercepts, asymptotes and stationary points (if any). Hence, or otherwise, find the range of values of $k$ such that the equation ${{x}^{3}}+2{{x}^{2}}=kx\left( x-1 \right)$ has no real root.

Accepted Answer

(i) $x=-a$ and $y=x+2-a$ (ii) $\dfrac{dy}{dx}=\dfrac{x^2+2ax+2a}{(x+a)^2}$, with no stationary points for $0<a<2$ (iii) For $a=-1$, asymptotes $x=1$ and $y=x+3$, intercepts $(-2,0)$ and $(0,0)$, turning points $(1-\sqrt3,4-2\sqrt3)$ and $(1+\sqrt3,4+2\sqrt3)$. As written, no such $k$ exists because $x=0$ is always a root. If no nonzero real root was intended, $4-2\sqrt3<k<4+2\sqrt3$

2008 AJC Promo Q9

2008 AJC Promo Q9

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