N2010 P2 Q3

Junior College 2

11 marks

Given that $y=x\sqrt{x+2}$, find $\frac{\mathrm dy}{\mathrm dx}$, expressing your answer as a single algebraic fraction. Hence show that there is only one value of $x$ for which the curve $y=x\sqrt{x+2}$ has a turning point, and state this value.[5]
A curve has equation $y^2=x^2(x+2)$.
1. Find exactly the possible values of the gradient at the point where $x=0$.[2]
2. Sketch the curve $y^2=x^2(x+2)$.[2]
On a separate diagram sketch the graph of $y=\mathrm f'(x)$, where $\mathrm f(x)=x\sqrt{x+2}$. State the equations of any asymptotes.[2]

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Answer:(i) $y'=(3x+4)/(2\sqrt{x+2})$, turning point at $x=-4/3$. (ii)(a) $\pm\sqrt2$. (iii) Asymptote $x=-2$.

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