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 \(y^2=x^2(x+2)\).
Find exactly the possible values of the gradient at the point where \(x=0\).[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 \(x=-2\).