Answer: The asymptotes are \(x=2\) and \(y=x-1\).
The curve has no points for \(-1\lt y\lt 3\).
The turning points are \((1,-1)\) and \((3,3)\).
Write the curve as \(y=\frac{x^2-3x+3}{x-2}\), with \(x\neq 2\).
Asymptotes
The denominator is zero at \(x=2\), so there is a vertical asymptote \(x=2\).
Now divide \(x^2-3x+3\) by \(x-2\):
\(x^2-3x+3=(x-2)(x-1)+1\), so
\(y=x-1+\frac{1}{x-2}\).
As \(x\to\pm\infty\), the term \(\frac{1}{x-2}\to 0\), so the oblique asymptote is \(y=x-1\).
No points for \(-1\lt y\lt 3\)
Rearrange the equation as a quadratic in \(x\):
\(y(x-2)=x^2-3x+3\)
\(xy-2y=x^2-3x+3\)
\(x^2-(y+3)x+(3+2y)=0\).
For real points on the curve, this quadratic must have real solutions in \(x\), so its discriminant must satisfy
\((y+3)^2-4(3+2y)\ge 0\).
Simplifying gives
\(y^2+6y+9-12-8y\ge 0\)
\(y^2-2y-3\ge 0\)
\((y-3)(y+1)\ge 0\).
Hence \(y\le -1\) or \(y\ge 3\). Therefore there are no points on \(C\) for \(-1\lt y\lt 3\).
Turning points
Differentiate using \(y=x-1+\frac{1}{x-2}\):
\(\frac{dy}{dx}=1-\frac{1}{(x-2)^2}\).
Set \(\frac{dy}{dx}=0\):
\(1-\frac{1}{(x-2)^2}=0\)
\((x-2)^2=1\)
so \(x=1\) or \(x=3\).
Find the corresponding \(y\)-values:
At \(x=1\), \(y=\frac{1-3+3}{1-2}=-1\).
At \(x=3\), \(y=\frac{9-9+3}{3-2}=3\).
So the turning points are \((1,-1)\) and \((3,3)\).
Sketch
The graph has two branches separated by the vertical asymptote \(x=2\). Since there are no points with \(-1\lt y\lt 3\), the left branch lies at or below \(y=-1\), with a turning point at \((1,-1)\), and tends to \(-\infty\) as \(x\to 2^-\). The right branch lies at or above \(y=3\), with a turning point at \((3,3)\), and tends to \(+\infty\) as \(x\to 2^+\). Both branches approach the oblique asymptote \(y=x-1\) for large \(|x|\).
