Answer: The asymptotes are \(x=1\) and \(y=2x+3\).

There is no point on the curve for which \(1\lt y\lt 9\).

Rewrite the function by dividing \(2x^2+x-1\) by \(x-1\):

\(y=\frac{2x^2+x-1}{x-1}=2x+3+\frac{2}{x-1}.\)

So the vertical asymptote is \(x=1\), and since \(\dfrac{2}{x-1}\to 0\) as \(x\to \pm\infty\), the oblique asymptote is \(y=2x+3\).

Now show there are no points with \(1\lt y\lt 9\). Starting from

\(y=\frac{2x^2+x-1}{x-1},\)

multiply through by \(x-1\):

\(y(x-1)=2x^2+x-1.\)

Rearranging gives a quadratic in \(x\):

\(2x^2+(1-y)x+(y-1)=0.\)

For a point on the curve to exist for a given value of \(y\), this quadratic must have real roots. So its discriminant must satisfy

\((1-y)^2-4(2)(y-1)\ge 0.\)

Hence

\((1-y)^2-8(y-1)\ge 0.\)

Simplifying,

\(y^2-10y+9\ge 0,\)

so

\((y-1)(y-9)\ge 0.\)

Therefore, if \(1\lt y\lt 9\), then \((y-1)(y-9)\lt 0\), so the discriminant is negative and there are no real values of \(x\).

Hence there is no point on \(C\) for which \(1\lt y\lt 9\).