2020 CJC Promo Q8
2020 CJC Promo Q8
9 marks
Promo
The curve \(C\) has equation
\(y=\frac{3{{x}^{2}}-2x-1}{x+1}\) for \(x\in \mathbb{R}\), \(x\ne -1\).
- Find, using an algebraic method, the range of values that \(y\) cannot take.[3]
- Sketch \(C\), stating clearly the equations of asymptotes, the coordinates of the axial intercepts and the coordinates of the turning points.[4]
- Verify that \((-1,-8)\) lies on the graph \(y=m(x+1)-8\), where \(m\in \mathbb{R}\). Hence find the range of values of \(m\) for which the equation \(\frac{3{{x}^{2}}-2x-1}{x+1}=m(x+1)-8\) has \(2\) distinct real roots.[2]
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Answer:(i) \(-8-4\sqrt3<y<-8+4\sqrt3\) (ii) Asymptotes \(x=-1\) and \(y=3x-5\), intercepts \(\left(-\dfrac13,0\right)\), \((1,0)\) and \((0,-1)\), turning points \(\left(-1-\dfrac{2\sqrt3}{3},-8-4\sqrt3\right)\) and \(\left(-1+\dfrac{2\sqrt3}{3},-8+4\sqrt3\right)\) (iii) \((-1,-8)\) lies on the line for every \(m\), and \(m>3\)
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