Answer: (a) The other asymptote is \(x=1\).

(b) \(a=2\) and \(b=-7\).

(c) \(c=7\).

(d) \(k\) must satisfy \(-7\lt k\lt 1\).

Write the function as \(y=\frac{ax^2+bx+c}{x-1}\). Since the denominator is zero at \(x=1\), the curve has a vertical asymptote \(x=1\).

For the oblique asymptote, divide \(ax^2+bx+c\) by \(x-1\):

\(ax^2+bx+c=(x-1)(ax+a+b)+(a+b+c)\), so

\(y=ax+a+b+\frac{a+b+c}{x-1}\).

As \(x\to\infty\), the fractional term tends to \(0\), so the oblique asymptote is \(y=ax+a+b\). This is given as \(y=2x-5\), so comparing coefficients gives \(a=2\) and \(a+b=-5\). Hence \(2+b=-5\), so \(b=-7\).

Using these values,

\(y=\frac{2x^2-7x+c}{x-1}\).

Now divide again by \(x-1\):

\(2x^2-7x+c=(x-1)(2x-5)+(c-5)\), so

\(y=2x-5+\frac{c-5}{x-1}\).

Differentiate:

\(y'=2-\frac{c-5}{(x-1)^2}\).

At a turning point when \(x=2\), we have \(y'=0\), so

\(2-\frac{c-5}{(2-1)^2}=0\).

Thus \(2-(c-5)=0\), so \(c=7\).

Finally, with \(c=7\),

\(y=\frac{2x^2-7x+7}{x-1}\).

If \(y=k\), then

\(\frac{2x^2-7x+7}{x-1}=k\),

so

\(2x^2-7x+7=k(x-1)\),

\(2x^2-(7+k)x+(7+k)=0\).

For the line \(y=k\) not to intersect the curve, this quadratic must have no real roots, so its discriminant is negative:

\((7+k)^2-4\cdot 2\cdot(7+k)\lt 0\).

That is

\((7+k)^2-8(7+k)\lt 0\),

\(k^2+6k-7\lt 0\),

\((k+7)(k-1)\lt 0\).

Therefore \(-7\lt k\lt 1\).