(a) The vertical asymptote occurs where the denominator is zero, so \(x = -1\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x+1}\) to get \(y = x - 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate \(y = \frac{x^2}{x+1}\) to get \(\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}\). Solve \(x^2 + 2x = 0\) to find \(x = 0\) and \(x = -2\). Substitute back to find \(y\) values: \((0, 0)\) and \((-2, -4)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve approaches the asymptotes and passes through the stationary points.

(d) For \(y = \frac{1}{f(x)}\), differentiate and set \(\frac{dy}{dx} = 0\) to find stationary points. The stationary point is \((-2, -\frac{1}{4})\).

(e) Sketch \(y = \frac{1}{f(x)}\) using the asymptotes and stationary points. Solve \(\frac{1}{f(x)} > f(x)\) by setting \(\frac{x^2}{x+1} = 1\) and \(\frac{x^2}{x+1} = -1\). Solve \(x^2 - x - 1 = 0\) to find critical points \(x = \frac{1}{2} \pm \frac{1}{2} \sqrt{5}\). The solution set is \(x < -1, -\frac{1}{2} \sqrt{5} < x < \frac{1}{2} \sqrt{5}, x \neq 0\).

(a) The vertical asymptote occurs where the denominator is zero: \(5x - 2 = 0 \Rightarrow x = \frac{2}{5}\). For the oblique asymptote, divide \(5x^2\) by \(5x - 2\) to get \(y = x + \frac{2}{5}\).

(b) Differentiate \(y = \frac{5x^2}{5x-2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(5x-2)(10x) - (5x^2)(5)}{(5x-2)^2}\). Set \(\frac{dy}{dx} = 0\) to find stationary points: \(5x^2 - 4x = 0 \Rightarrow x(5x - 4) = 0\). Thus, \(x = 0\) or \(x = \frac{4}{5}\). The coordinates are \((0,0)\) and \(\left( \frac{4}{5}, \frac{8}{5} \right)\).

(c) Sketch the curve showing the asymptotes and stationary points.

(d) Solve \(\left| \frac{5x^2}{5x-2} \right| < 2\). This gives two cases: \(\frac{5x^2}{5x-2} < 2\) and \(\frac{5x^2}{5x-2} > -2\). Solving these inequalities gives the intervals \(-1 - \frac{1}{3}\sqrt{5} < x < -1 + \frac{1}{3}\sqrt{5}\) and \(1 - \frac{1}{3}\sqrt{5} < x < 1 + \frac{1}{3}\sqrt{5}\).

Answer:

1. The curve has vertical asymptote \(x=-7\), horizontal asymptote \(y=a\), and passes through \((0,0)\). For \(x<-7\) the branch lies above \(y=a\); for \(x>-7\) the branch comes from \(-\infty\), passes through \((0,0)\), and approaches \(y=a\) from below.

2. The modulus graph is found by reflecting the part of the graph below the \(x\)-axis in the \(x\)-axis.

   The required set of values is \(x<-7\), \(-77\).

1. Rewrite the function as \(y=\frac{ax}{x+7}=a-\frac{7a}{x+7}\).

   Hence the vertical asymptote is \(x=-7\), and the horizontal asymptote is \(y=a\). Since \(a>0\), the curve crosses the axes at \((0,0)\).

   For \(x<-7\), \(\frac{x}{x+7}>1\), so the branch lies above \(y=a\) and tends to \(+\infty\) as \(x\to -7^-\). For \(-70\), the curve is positive and approaches \(y=a\) from below.

2. For \(y=\left|\frac{ax}{x+7}\right|\), reflect the part of the graph from part (a) that lies below the \(x\)-axis in the \(x\)-axis. The vertical asymptote remains \(x=-7\), the horizontal asymptote remains \(y=a\), and the graph meets the \(x\)-axis at \((0,0)\).

   To solve \(\left|\frac{ax}{x+7}\right|>\frac{a}{2}\), use \(a>0\) and divide through by \(a\): \(\left|\frac{x}{x+7}\right|>\frac{1}{2}\).

   The boundary points occur when \(\frac{x}{x+7}=\frac{1}{2}\) or \(\frac{x}{x+7}=-\frac{1}{2}\).

   Solving \(\frac{x}{x+7}=\frac{1}{2}\) gives \(2x=x+7\), so \(x=7\).

   Solving \(\frac{x}{x+7}=-\frac{1}{2}\) gives \(2x=-(x+7)\), so \(3x=-7\), and therefore \(x=-\frac{7}{3}\).

   Using the intervals separated by \(x=-7\), \(x=-\frac{7}{3}\), and \(x=7\), the inequality is satisfied for \(x<-7\), \(-77\).

Answer: (a) \(x=-\frac{3}{2}-\frac{\sqrt{5}}{2}\), \(x=-\frac{3}{2}+\frac{\sqrt{5}}{2}\), and \(y=0\).

(b) No stationary points.

(c) Intersections: \((0,2)\) and \((-2,0)\). Asymptotes: \(x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}\), \(y=0\).

(d) Reflect all parts of \(C\) below the \(x\)-axis above the \(x\)-axis.

(e) \(-\frac{7}{4}-\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}-\frac{\sqrt{5}}{2}\), \(-\frac{3}{2}-\frac{\sqrt{5}}{2}\lt x\lt-\frac{5}{2}\), \(-\frac{7}{4}+\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}+\frac{\sqrt{5}}{2}\), or \(-\frac{3}{2}+\frac{\sqrt{5}}{2}\lt x\lt0\).

(a)

Vertical asymptotes occur where the denominator is zero:

\(x^2+3x+1=0.\)

Thus

\[ x=\frac{-3\pm\sqrt{9-4}}{2} =-\frac{3}{2}\pm\frac{\sqrt{5}}{2}. \]

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is

\(y=0.\)

(b)

Differentiate using the quotient rule:

\[ \frac{dy}{dx} = \frac{(x^2+3x+1)(1)-(x+2)(2x+3)}{(x^2+3x+1)^2}. \]

Stationary points would require the numerator to be zero:

\[ x^2+3x+1-(x+2)(2x+3)=0. \]

This simplifies to

\[ -x^2-4x-5=0, \]

or

\[ x^2+4x+5=0. \]

The discriminant is

\[ 4^2-4(1)(5)=16-20=-4\lt0, \]

so there are no real roots and hence no stationary points.

(c)

The \(y\)-intercept is found by setting \(x=0\):

\[ y=\frac{2}{1}=2, \]

so the curve passes through \((0,2)\).

The \(x\)-intercept is found by setting the numerator to zero:

\[ x+2=0\Rightarrow x=-2, \]

so the curve passes through \((-2,0)\).

The sketch should show vertical asymptotes at \(x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}\), horizontal asymptote \(y=0\), and the correct branch positions.

(d)

The graph of \(y=\left|f(x)\right|\) is obtained from \(y=f(x)\) by leaving all parts above the \(x\)-axis unchanged and reflecting all parts below the \(x\)-axis in the \(x\)-axis.

For this curve, the asymptotes stay the same, and the point \((-2,0)\) remains on the graph, forming the correct sharp point after reflection.

(e)

Solve the boundary equation:

\[ \left|\frac{x+2}{x^2+3x+1}\right|=2. \]

This gives

\[ \frac{x+2}{x^2+3x+1}=2 \quad\text{or}\quad \frac{x+2}{x^2+3x+1}=-2. \]

For \(\frac{x+2}{x^2+3x+1}=2\):

\[ x+2=2x^2+6x+2 \Rightarrow -2x^2-5x=0. \]

So

\[ x=0 \quad\text{or}\quad x=-\frac{5}{2}. \]

For \(\frac{x+2}{x^2+3x+1}=-2\):

\[ x+2=-2x^2-6x-2 \Rightarrow 2x^2+7x+4=0. \]

So

\[ x=-\frac{7}{4}\pm\frac{\sqrt{17}}{4}. \]

The expression is undefined at

\[ x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}. \]

Using the sketch or sign intervals gives

\[ -\frac{7}{4}-\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}-\frac{\sqrt{5}}{2}, \]

\[ -\frac{3}{2}-\frac{\sqrt{5}}{2}\lt x\lt-\frac{5}{2}, \]

\[ -\frac{7}{4}+\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}+\frac{\sqrt{5}}{2}, \]

\[ -\frac{3}{2}+\frac{\sqrt{5}}{2}\lt x\lt0. \]

Mark scheme

Answer: (a) \(x=-1\), \(y=x\)

(b) \((0,1)\) and \((-2,-3)\)

(c) Sketch with vertical asymptote \(x=-1\), oblique asymptote \(y=x\), stationary points \((0,1)\) and \((-2,-3)\), and correct branch behaviour.

(d) Take the right branch of part (c) and reflect it in the \(y\)-axis; the graph is symmetric about \(x=0\) and smooth at \((0,1)\).

(e) \(-1-\sqrt3

(a)

The denominator is zero when \(x=-1\), so \(x=-1\) is the vertical asymptote.

Divide:

\(\frac{x^2+x+1}{x+1}=x+\frac1{x+1}.\)

As \(|x|\to\infty\), \(\frac1{x+1}\to0\), so the oblique asymptote is

\(y=x.\)

(b)

Write

\(y=x+\frac1{x+1}.\)

Then

\(\frac{dy}{dx}=1-\frac1{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}.\)

Set \(\frac{dy}{dx}=0\):

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

So \(x=0\) or \(x=-2\). The corresponding values are

\(y(0)=1,\qquad y(-2)=-3.\)

Therefore the stationary points are \((0,1)\) and \((-2,-3)\).

(c)

The curve can be written as

\(y=x+\frac1{x+1}.\)

So the asymptotes are \(x=-1\) and \(y=x\).

The right branch, for \(x>-1\), approaches \(+\infty\) as \(x\to-1^+\), has a smooth minimum at \((0,1)\), then approaches \(y=x\) from above.

The left branch, for \(x<-1\), approaches \(y=x\) from below as \(x\to-\infty\), has a smooth maximum at \((-2,-3)\), then tends to \(-\infty\) as \(x\to-1^-\).

(d)

Let \(t=|x|\). Then

\(y=\frac{t^2+t+1}{t+1},\qquad t\ge0.\)

This is the same as the right-hand branch of \(C\) from part (c), reflected in the \(y\)-axis. The graph is symmetric about \(x=0\), has a smooth minimum at \((0,1)\), and approaches \(y=|x|\) for large \(|x|\).

(e)

Let \(t=|x|\), where \(t\ge0\). The inequality becomes

\(\frac{t^2+t+1}{t+1}<3.\)

Since \(t+1>0\), multiply through:

\(t^2+t+1<3t+3.\)

So

\(t^2-2t-2<0.\)

The roots are

\(t=1\pm\sqrt3.\)

Since \(t\ge0\), the relevant critical value is \(t=1+\sqrt3\). Therefore

\(|x|<1+\sqrt3,\)

so

\(-1-\sqrt3

Mark scheme