Answer: (i) The curve has a vertical asymptote at \(x=-1\). Dividing the numerator by \(x+1\),

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

So the oblique asymptote is \(y=px+4-p\).

(ii) For the \(x\)-axis to be tangent, the equation \(y=0\) must have a repeated root. This occurs when \(p=4\). Hence the \(x\)-axis is tangent to \(C\) when \(p=4\).

(iii) When \(p=1\),

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

Then

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

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

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

\(\displaystyle \frac{x^2+4x+1}{x+1}=0 \iff x^2+4x+1=0.\)

So

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

Hence the points of intersection with the \(x\)-axis are \(\left(-2-\sqrt3,0\right)\) and \(\left(-2+\sqrt3,0\right)\).

(i) To find the asymptotes, rewrite the function by algebraic division:

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

Now

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

so

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

As \(x\to -1\), the term \(\frac{p-3}{x+1}\) becomes unbounded, so the vertical asymptote is

\(\displaystyle x=-1.\)

Also, as \(x\to\pm\infty\), the fraction \(\frac{p-3}{x+1}\to 0\), so the oblique asymptote is

\(\displaystyle y=px+4-p.\)

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

\(\displaystyle y=\frac{px^2+4x+1}{x+1},\)

we need the quadratic \(px^2+4x+1=0\) to have a repeated root.

Its discriminant is

\(\displaystyle 4^2-4(p)(1)=16-4p.\)

Set this equal to zero:

\(\displaystyle 16-4p=0 \implies p=4.\)

So the \(x\)-axis is a tangent when \(\boxed{p=4}.\)

For this case,

\(\displaystyle y=\frac{4x^2+4x+1}{x+1}=4x\!\,+\!0+\frac{1}{x+1},\)

or more usefully, the numerator factors as

\(\displaystyle 4x^2+4x+1=(2x+1)^2,\)

so \(y=0\) at \(x=-\tfrac12\) with a double root. The graph therefore touches the \(x\)-axis at \(\left(-\tfrac12,0\right)\), has vertical asymptote \(x=-1\), and oblique asymptote \(y=4x\).

(iii) When \(p=1\),

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

Differentiating gives

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

Since \(\frac{2}{(x+1)^2} \gt 0\) for all \(x\neq -1\), it follows that

\(\displaystyle \frac{dy}{dx} \gt 1\)

for all points on the curve. Therefore \(\frac{dy}{dx}\) is never zero, so the curve has no turning points.

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

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

Using the quadratic formula,

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

So the exact intersections with the \(x\)-axis are

\(\displaystyle \left(-2-\sqrt3,0\right)\) and \(\displaystyle \left(-2+\sqrt3,0\right).\)

The sketch should show two branches separated by the vertical asymptote \(x=-1\), approaching the oblique asymptote \(y=x+2\), and crossing the \(x\)-axis at these two points.