Answer: (i) The asymptote is \(y=5\).

(ii) The range is \(-\frac{1}{3} \leqslant y \lt 5\).

(iii) The stationary point is \(\left(-\frac{1}{2},-\frac{1}{3}\right)\).

(iv) The curve crosses the \(y\)-axis at \((0,1)\). A sketch should show a horizontal asymptote \(y=5\), the stationary point \(\left(-\frac{1}{2},-\frac{1}{3}\right)\), and the intercept \((0,1)\).

(i) Rewrite the function by dividing through by \(x^{2}+x+1\):

\(y=\frac{5x^{2}+5x+1}{x^{2}+x+1}=5-\frac{4}{x^{2}+x+1}.\)

As \(x\rightarrow \pm\infty\), the fraction \(\frac{4}{x^{2}+x+1}\rightarrow 0\), so \(y\rightarrow 5\). Hence the asymptote is \(y=5\).

(ii) Start from

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

Multiply through by \(x^{2}+x+1\):

\(yx^{2}+yx+y=5x^{2}+5x+1.\)

Rearrange as a quadratic in \(x\):

\((y-5)x^{2}+(y-5)x+(y-1)=0.\)

For there to be real values of \(x\), the discriminant must be non-negative:

\((y-5)^{2}-4(y-5)(y-1)\geqslant 0.\)

Factorising gives

\((y-5)(3y+1)\leqslant 0.\)

This holds when \(-\frac{1}{3}\leqslant y\leqslant 5\). However, \(y=5\) is not actually attained, because that would require \(\frac{4}{x^{2}+x+1}=0\), which is impossible for any real \(x\). Therefore

\(-\frac{1}{3}\leqslant y\lt 5.\)

(iii) Differentiate using the quotient rule:

\(y' = \frac{(x^{2}+x+1)(10x+5)-(5x^{2}+5x+1)(2x+1)}{(x^{2}+x+1)^{2}}.\)

For stationary points, set the numerator equal to zero:

\((x^{2}+x+1)(10x+5)-(5x^{2}+5x+1)(2x+1)=0.\)

Expanding and simplifying gives

\(4(2x+1)=0,\)

so \(x=-\frac{1}{2}.\)

Substitute into the curve:

\(y=\frac{5(\frac{1}{4})+5(-\frac{1}{2})+1}{\frac{1}{4}-\frac{1}{2}+1}=\frac{\frac{5}{4}-\frac{5}{2}+1}{\frac{3}{4}}=\frac{-\frac{1}{4}}{\frac{3}{4}}=-\frac{1}{3}.\)

So the stationary point is \(\left(-\frac{1}{2},-\frac{1}{3}\right)\).

(iv) When \(x=0\),

\(y=\frac{1}{1}=1,\)

so the curve intersects the \(y\)-axis at \((0,1)\).

For the sketch, draw a horizontal asymptote at \(y=5\). The curve lies below this line for all real \(x\), has minimum value \(-\frac{1}{3}\) at \(\left(-\frac{1}{2},-\frac{1}{3}\right)\), and passes through \((0,1)\).