Answer: (i) Comparing the given curve with the asymptote tells us that for large values of \(x\),

\(\displaystyle \frac{x^2+1}{ax+b} \sim 2x+1.\)

Now divide \(x^2+1\) by \(ax+b\):

\(\displaystyle \frac{x^2+1}{ax+b}=\frac{1}{a}x-\frac{b}{a^2}+\frac{\text{remainder}}{ax+b}.\)

So the slant asymptote is \(y=\frac{1}{a}x-\frac{b}{a^2}\). Matching this with \(y=2x+1\) gives

\(\frac{1}{a}=2 \Rightarrow a=\frac12,\)

and

\(-\frac{b}{a^2}=1 \Rightarrow -\frac{b}{(1/2)^2}=1 \Rightarrow -4b=1 \Rightarrow b=-\frac14.\)

So \(a=\frac12\) and \(b=-\frac14\).

(ii) The other asymptote is where the denominator is zero:

\(\frac12 x-\frac14=0 \Rightarrow x=\frac12.\)

So the other asymptote is \(x=\frac12\).

(iii) The curve is

\(\displaystyle y=\frac{x^2+1}{\frac12 x-\frac14}=\frac{4(x^2+1)}{2x-1}.\)

Useful features for the sketch:

• Vertical asymptote: \(x=\frac12\)
• Oblique asymptote: \(y=2x+1\)
• \(y\)-intercept: when \(x=0\), \(y=\frac{1}{-1/4}=-4\), so the curve crosses the \(y\)-axis at \((0,-4)\)

Also, since \(x^2+1>0\) for all real \(x\), the sign of \(y\) is determined by the denominator \(\frac12 x-\frac14\). Hence the curve is below the \(x\)-axis for \(x<\frac12\) and above it for \(x>\frac12\). As \(x\to \frac12^-\), \(y\to -\infty\), and as \(x\to \frac12^+\), \(y\to +\infty\). For large positive \(x\), the curve approaches \(y=2x+1\) from above; for large negative \(x\), it approaches the same line from below.

A correct sketch should show a vertical asymptote at \(x=\frac12\), the slant asymptote \(y=2x+1\), and the point \((0,-4)\).

We are given that the line \(y=2x+1\) is an asymptote of \(C\), where \(C\) has equation \(\displaystyle y=\frac{x^2+1}{ax+b}\).

Since the numerator has degree 2 and the denominator has degree 1, the curve has a slant asymptote found by division.

Divide \(x^2+1\) by \(ax+b\). The leading term of the quotient must be \(\frac{1}{a}x\), because

\(\displaystyle \frac{x^2}{ax}=\frac{1}{a}x.\)

So the asymptote has form

\(\displaystyle y=\frac{1}{a}x+c\)

for some constant \(c\). Since the asymptote is given as \(y=2x+1\), we match coefficients:

\(\displaystyle \frac{1}{a}=2 \Rightarrow a=\frac12.\)

Now find \(b\). With \(a=\frac12\), write the division explicitly:

\(\displaystyle \frac{x^2+1}{\frac12 x+b}.\)

Take the next term in the quotient as \(2x\), because

\(\displaystyle 2x\left(\frac12 x+b\right)=x^2+2bx.\)

Subtracting from \(x^2+1\) leaves a remainder of \(-2bx+1\). The next constant term in the quotient is chosen so that the linear term of the remainder matches the asymptote constant. More simply, using the general result for division of \(x^2+1\) by \(ax+b\), the asymptote is

\(\displaystyle y=\frac{1}{a}x-\frac{b}{a^2}.\)

Matching with \(y=2x+1\), and using \(a=\frac12\), gives

\(\displaystyle -\frac{b}{(1/2)^2}=1\)

so

\(\displaystyle -4b=1 \Rightarrow b=-\frac14.\)

Therefore

\(\displaystyle a=\frac12,\quad b=-\frac14.\)

(ii) The other asymptote is the vertical asymptote where the denominator is zero:

\(\displaystyle \frac12 x-\frac14=0 \Rightarrow x=\frac12.\)

So the other asymptote is \(\boxed{x=\frac12}\).

(iii) For the sketch, substitute the values of \(a\) and \(b\):

\(\displaystyle y=\frac{x^2+1}{\frac12 x-\frac14}=\frac{4(x^2+1)}{2x-1}.\)

Key features:

• Vertical asymptote: \(x=\frac12\)
• Oblique asymptote: \(y=2x+1\)
• \(y\)-intercept: at \(x=0\), \(y=\frac{1}{-1/4}=-4\), so the curve passes through \((0,-4)\)

Since \(x^2+1>0\) for all real \(x\), the sign of \(y\) depends on \(2x-1\). Hence:

• for \(x<\frac12\), \(y<0\),
• for \(x>\frac12\), \(y>0\).

As \(x\to \frac12^-\), \(y\to -\infty\), and as \(x\to \frac12^+\), \(y\to +\infty\).

Also, because the asymptote is \(y=2x+1\), the curve approaches this straight line for large \(|x|\). A correct sketch should therefore show two branches separated by the vertical asymptote, with the left-hand branch passing through \((0,-4)\) and tending to \(-\infty\) near \(x=\frac12\), and the right-hand branch tending to \(+\infty\) near \(x=\frac12\) and then approaching the line \(y=2x+1\).