Answer: (i) The points of intersection with the axes are \((-1,0)\), \((-6,0)\) and \((0,-3)\).

(ii) The asymptotes are \(x=2\) and \(y=x+9\).

(iii) A sketch of \(C\) should show two branches, with vertical asymptote \(x=2\), oblique asymptote \(y=x+9\), crossing the axes at \((-6,0)\), \((-1,0)\) and \((0,-3)\).

(i) To find where the curve meets the axes, we use:

• the x-axis: set \(y=0\)
• the y-axis: set \(x=0\)

For the \(x\)-axis,

\(0=\frac{x^{2}+7x+6}{x-2}\).

A fraction is zero when its numerator is zero, provided the denominator is not zero. So solve

\(x^{2}+7x+6=0\).

Factorising gives

\((x+1)(x+6)=0\).

Hence \(x=-1\) or \(x=-6\).

So the curve meets the \(x\)-axis at \((-1,0)\) and \((-6,0)\).

For the \(y\)-axis, set \(x=0\):

\(y=\frac{0^2+7(0)+6}{0-2}=\frac{6}{-2}=-3\).

So the curve meets the \(y\)-axis at \((0,-3)\).

(ii) The denominator is zero when \(x=2\), and there is no factor of \(x-2\) in the numerator to cancel it, so there is a vertical asymptote

\(x=2\).

To find the oblique asymptote, divide \(x^2+7x+6\) by \(x-2\):

\(x^2+7x+6=(x-2)(x+9)+24\).

Therefore

\(y= x+9+\frac{24}{x-2}\).

As \(x\to\pm\infty\), the fraction tends to 0, so the oblique asymptote is

\(y=x+9\).

(iii) For the sketch, draw the asymptotes \(x=2\) and \(y=x+9\) first.

Then plot the intercepts \((-6,0)\), \((-1,0)\) and \((0,-3)\).

The curve has two separate branches, one on each side of the vertical asymptote \(x=2\). From

\(y=x+9+\frac{24}{x-2}\),

when \(x>2\), the term \(\frac{24}{x-2}\) is positive, so the right-hand branch lies above the slant asymptote and rises to \(+\infty\) as \(x\to2^+\).

When \(x<2\), the term \(\frac{24}{x-2}\) is negative, so the left-hand branch lies below the slant asymptote and falls to \(-\infty\) as \(x\to2^-\).

A correct sketch should show these features and pass through the intercepts found in part (i).