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

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

A sketch should show two branches separated by the line \(x=2\), approaching the line \(y=x+2\) for large \(|x|\), and passing through \(0,0\) and \(4,8\).

First rewrite the function by dividing \(x^2\) by \(x-2\):

\(x^2 = (x-2)(x+2)+4\), so

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

So the vertical asymptote is where the denominator is zero:

\(x-2=0 \Rightarrow x=2.\)

And since \(\frac{4}{x-2} \to 0\) as \(x \to \pm\infty\), the oblique asymptote is

\(y=x+2.\)

To find turning points, differentiate:

\(y' = 1 - \frac{4}{(x-2)^2}.\)

Set \(y'=0\):

\(1 - \frac{4}{(x-2)^2}=0 \Rightarrow (x-2)^2=4.\)

Hence

\(x-2=\pm 2\), so \(x=0\) or \(x=4\).

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

For \(x=0\), \(y=\frac{0^2}{0-2}=0\), so one turning point is \((0,0)\).

For \(x=4\), \(y=\frac{4^2}{4-2}=\frac{16}{2}=8\), so the other turning point is \((4,8)\).

For the sketch, draw the axes, the asymptotes \(x=2\) and \(y=x+2\), and plot \((0,0)\) and \((4,8)\). The branch for \(x\lt 2\) passes through \((0,0)\) and tends to \(x=2\) and \(y=x+2\). The branch for \(x\gt 2\) has a turning point at \((4,8)\) and also approaches the same asymptotes.