Answer: The asymptotes are the vertical asymptote \(x=2\) and the oblique asymptote \(y=x+2\).

There are no points on \(C\) for which \(0\lt y\lt 8\).

The turning points are \((0,0)\) and \((4,8)\).

We begin by rewriting the curve:

\(y=\dfrac{x^2}{x-2}=x+2+\dfrac{4}{x-2}\).

So the denominator shows a vertical asymptote at \(x=2\), and the rewritten form shows that as \(|x|\to\infty\), the curve approaches the line \(y=x+2\). Hence the asymptotes are \(x=2\) and \(y=x+2\).

To show there are no points on \(C\) with \(0\lt y\lt 8\), treat the defining equation as a quadratic in \(x\):

\(y=\dfrac{x^2}{x-2}\Rightarrow y(x-2)=x^2\Rightarrow x^2-yx+2y=0\).

For a given value of \(y\), this quadratic must have a real solution in \(x\) for there to be a point on the curve. Its discriminant is

\(\Delta = (-y)^2-4(1)(2y)=y^2-8y=y(y-8)\).

If \(0\lt y\lt 8\), then \(y\gt 0\) and \(y-8\lt 0\), so \(y(y-8)\lt 0\). Thus \(\Delta\lt 0\), so there are no real solutions for \(x\). Therefore there are no points on \(C\) with \(0\lt y\lt 8\).

For the turning points, differentiate:

\(y=x+2+\dfrac{4}{x-2}\), so \(\dfrac{dy}{dx}=1-\dfrac{4}{(x-2)^2}\).

Set \(\dfrac{dy}{dx}=0\):

\(1-\dfrac{4}{(x-2)^2}=0 \Rightarrow (x-2)^2=4\Rightarrow x=0\text{ or }x=4\).

Now find the corresponding \(y\)-values:

• if \(x=0\), then \(y=\dfrac{0^2}{0-2}=0\), giving \((0,0)\);
• if \(x=4\), then \(y=\dfrac{4^2}{4-2}=8\), giving \((4,8)\).

So the turning points are \((0,0)\) and \((4,8)\). The sketch should show two branches separated by the vertical asymptote \(x=2\), with the oblique asymptote \(y=x+2\), and passing through \((0,0)\) and \((4,8)\).