Answer:

1. (a) The asymptotes are \(x=1\) and \(y=x+2\).

2. (b) There is no point on \(C\) for which \(1

3. (c) The intersections with the axes are \((0,1)\), \(\left(\frac{-1+\sqrt5}{2},0\right)\) and \(\left(\frac{-1-\sqrt5}{2},0\right)\).

4. (d) The graph of \(y=\left|\frac{x^2+x-1}{x-1}\right|\) is obtained by reflecting the parts of \(C\) below the \(x\)-axis in the \(x\)-axis.

1. (a) Divide the numerator by \(x-1\):

   \(x^2+x-1=(x-1)(x+2)+1\), so \(y=x+2+\frac{1}{x-1}\).

   Hence the vertical asymptote is \(x=1\), and as \(|x|\) becomes large, \(\frac{1}{x-1}\to0\), so the oblique asymptote is \(y=x+2\).

2. (b) Rearrange the equation as a quadratic in \(x\):

   \(y(x-1)=x^2+x-1\), so \(x^2+(1-y)x+(y-1)=0\).

   For a point on the curve with a given value of \(y\), this quadratic must have a real solution for \(x\). Its discriminant is

   \((1-y)^2-4(y-1)=y^2-6y+5=(y-1)(y-5)\).

   If \(10\) and \(y-5<0\), so \((y-1)(y-5)<0\). The discriminant is therefore negative, so there is no real value of \(x\). Hence there is no point on \(C\) for which \(1

3. (c) For the \(y\)-axis intersection, put \(x=0\):

   \(y=\frac{-1}{-1}=1\), giving \((0,1)\).

   For the \(x\)-axis intersections, put \(y=0\). Then \(x^2+x-1=0\), so

   \(x=\frac{-1\pm\sqrt5}{2}\).

   Thus the intercepts are \((0,1)\), \(\left(\frac{-1+\sqrt5}{2},0\right)\) and \(\left(\frac{-1-\sqrt5}{2},0\right)\).

   For the sketch, use \(y=x+2+\frac{1}{x-1}\). The curve has asymptotes \(x=1\) and \(y=x+2\). For \(x<1\), the curve lies below the oblique asymptote, passes through both \(x\)-intercepts and \((0,1)\), and tends to \(-\infty\) as \(x\to1^-\). For \(x>1\), the curve lies above the oblique asymptote and tends to \(+\infty\) as \(x\to1^+\).

   Also, \(\frac{dy}{dx}=1-\frac{1}{(x-1)^2}\), so stationary points occur at \(x=0\) and \(x=2\). These are \((0,1)\) on the left branch and \((2,5)\) on the right branch, helping to fix the shape.

4. (d) The modulus graph is found by leaving the parts of \(C\) above the \(x\)-axis unchanged and reflecting the parts below the \(x\)-axis in the \(x\)-axis.

   The original curve is below the \(x\)-axis for \(x<\frac{-1-\sqrt5}{2}\) and for \(\frac{-1+\sqrt5}{2}

   As \(x\to\infty\), the right-hand branch still approaches \(y=x+2\). As \(x\to-\infty\), the reflected left-hand branch approaches \(y=-(x+2)=-x-2\).