Answer: (i) The denominator is zero when \(x+1=0\), so the vertical asymptote is \(x=-1\).

To find the oblique asymptote, divide \(px^2+4x+1\) by \(x+1\): \(px^2+4x+1=(x+1)(px+4-p)+(p-3)\). Hence \(y=px+4-p+\frac{p-3}{x+1}\), so the oblique asymptote is \(y=px+4-p\).

(ii) For the \(x\)-axis to be a tangent, the equation \(y=0\) must have a repeated root. So solve \(px^2+4x+1=0\) and require its discriminant to be zero: \(4^2-4p(1)=16-4p=0\). Thus \(p=4\).

(iii) For \(p=1\), \(y=\frac{x^2+4x+1}{x+1}=x+3-\frac{2}{x+1}\). Differentiating gives \(\frac{dy}{dx}=1+\frac{2}{(x+1)^2}\), which is always positive for \(x\neq -1\). Therefore \(C\) has no turning points.

The \(x\)-intercepts satisfy \(x^2+4x+1=0\), so \(x=\frac{-4\pm\sqrt{16-4}}{2}=-2\pm\sqrt{3}\). Hence the points of intersection with the \(x\)-axis are \((-2-\sqrt{3},0)\) and \((-2+\sqrt{3},0)\).

(i) Write the numerator in terms of \(x+1\):

\(px^2+4x+1=(x+1)(px+4-p)+(p-3)\).

So

\(y=\frac{px^2+4x+1}{x+1}=px+4-p+\frac{p-3}{x+1}\).

As \(x\to -1\), the denominator of the last term tends to zero, so there is a vertical asymptote \(x=-1\). Also, as \(x\to\pm\infty\), the term \(\frac{p-3}{x+1}\to 0\), so the oblique asymptote is \(y=px+4-p\).

(ii) For the \(x\)-axis to be tangent to \(C\), the equation \(y=0\) must have a repeated root:

\(\frac{px^2+4x+1}{x+1}=0 \iff px^2+4x+1=0\), with \(x\neq -1\).

A repeated root occurs when the discriminant is zero:

\(4^2-4(p)(1)=0\)

\(16-4p=0\)

\(p=4\).

So the required value is \(p=4\).

For sketching in this case,

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

The asymptotes are \(x=-1\) and \(y=4x\). The curve meets the \(x\)-axis where \(4x^2+4x+1=(2x+1)^2=0\), so it touches the axis at \(x=-\frac12\). Thus the tangent point is \(\left(-\frac12,0\right)\). The graph has two branches, one either side of \(x=-1\), each approaching the asymptotes appropriately.

(iii) Now take \(p=1\):

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

Differentiate:

\(\frac{dy}{dx}=1+\frac{2}{(x+1)^2}.\)

Since \((x+1)^2\gt 0\) for all \(x\neq -1\), we have \(\frac{dy}{dx}\gt 0\) everywhere on the curve. Therefore \(C\) has no turning points.

To find the \(x\)-intercepts, set \(y=0\):

\(x^2+4x+1=0\).

Using the quadratic formula,

\(x=\frac{-4\pm\sqrt{16-4}}{2}=\frac{-4\pm\sqrt{12}}{2}=-2\pm\sqrt3\).

So the exact points of intersection with the \(x\)-axis are \((-2-\sqrt3,0)\) and \((-2+\sqrt3,0)\). The sketch should show the vertical asymptote \(x=-1\), oblique asymptote \(y=x+2\), and two increasing branches passing through these intercepts.