(a) The vertical asymptote is at \(x = -2\). The oblique asymptote is found by dividing \(x^2 + ax + 1\) by \(x + 2\), giving \(y = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2}\). Simplifying gives \(\frac{dy}{dx} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\). Setting the numerator to zero gives \(x^2 + 4x + 2a - 1 = 0\). The discriminant is \(16 - 4(2a-1) = 20 - 8a\), which is less than zero for \(a > \frac{5}{2}\), so no stationary points exist.

(c) The curve intersects the y-axis at \((0, \frac{1}{2})\). The asymptotes are \(x = -2\) and \(y = x + a - 2\).

(d)(i) Sketch the curve \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes and a reflection in the x-axis where the original curve is negative.

(d)(ii) Draw the line \(y = a\) on the sketch.

(d)(iii) Solve \(\frac{x^2 + ax + 1}{x + 2} = a\) and \(\frac{x^2 + ax + 1}{x + 2} = -a\). This gives the quadratic \(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\). The roots are \(-a - \sqrt{a^2 - 2a - 1} < x < -a + \sqrt{a^2 - 2a - 1}\). Solving for \(a\) gives \(a = 5\).

(a) The vertical asymptote occurs where the denominator is zero: \(x + 2 = 0 \Rightarrow x = -2\).

The horizontal asymptote is found by dividing the leading coefficients: \(y = \frac{x^2 + ax + 1}{x + 2} = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\).

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

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

The discriminant \(16 - 4(2a-1) = 20 - 8a < 0\) implies no real roots, hence no stationary points.

(c) The curve intersects the \(y\)-axis at \(x = 0\):

\(y = \frac{0^2 + a(0) + 1}{0 + 2} = \frac{1}{2}\).

Asymptotes: vertical at \(x = -2\), oblique at \(y = x + a - 2\).

(d)(i) Sketch \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes.

(ii) Draw the line \(y = a\) on the sketch.

(iii) Solve \(\left| \frac{x^2 + ax + 1}{x + 2} \right| < a\) for given intervals:

\(\frac{x^2 + ax + 1}{x + 2} = a\) or \(\frac{x^2 + ax + 1}{x + 2} = -a\).

\(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\).

\(-a - \sqrt{a^2 - 2a - 1} < x < -\sqrt{2a - 1}\) and \(-a + \sqrt{a^2 - 2a - 1} < x < \sqrt{2a - 1}\).

Given intervals imply \(a = 5\).

(a) To find the vertical asymptote, set the denominator equal to zero: \(x + 1 = 0\), giving \(x = -1\).

For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + x + 9}{x + 1}\):

\(y = x + \frac{9}{x + 1}\), so the oblique asymptote is \(y = x\).

(b) To find stationary points, differentiate \(y = \frac{x^2 + x + 9}{x + 1}\) and set \(\frac{dy}{dx} = 0\).

\(\frac{dy}{dx} = 1 - \frac{9}{(x+1)^2} = 0 \Rightarrow (x+1)^2 = 9\).

Solving gives \(x + 1 = 3\) or \(x + 1 = -3\), so \(x = 2\) or \(x = -4\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{2^2 + 2 + 9}{2 + 1} = 5\), giving point \((2, 5)\).

For \(x = -4\), \(y = \frac{(-4)^2 - 4 + 9}{-4 + 1} = -7\), giving point \((-4, -7)\).

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 \(y=-(x+2)=-x-2\).

Answer:

1. The asymptotes are \(x=-\frac12\) and \(y=\frac12x-\frac14\).
2. The stationary points are \((0,0)\) and \((-1,-1)\).
3. The sketch has vertical asymptote \(x=-\frac12\), oblique asymptote \(y=\frac12x-\frac14\), an upper branch for \(x>-\frac12\) with minimum at \((0,0)\), and a lower branch for \(x<-\frac12\) with maximum at \((-1,-1)\).

1. The vertical asymptote occurs where the denominator is zero:

   \(2x+1=0\), so \(x=-\frac12\).

   Now divide \(x^2\) by \(2x+1\):

   \(\displaystyle \frac{x^2}{2x+1}=\frac12x-\frac14+\frac{1}{4(2x+1)}\).

   As \(|x|\to\infty\), the final term tends to \(0\), so the oblique asymptote is

   \(y=\frac12x-\frac14\).

   Therefore the asymptotes are \(x=-\frac12\) and \(y=\frac12x-\frac14\).

2. Using

   \(\displaystyle y=\frac12x-\frac14+\frac{1}{4(2x+1)}\),

   differentiate:

   \(\displaystyle \frac{dy}{dx}=\frac12-\frac{1}{2(2x+1)^2}\).

   At a stationary point, \(\frac{dy}{dx}=0\), so

   \(\displaystyle \frac12-\frac{1}{2(2x+1)^2}=0\).

   Hence \((2x+1)^2=1\), giving

   \(2x+1=\pm1\).

   So \(x=0\) or \(x=-1\).

   Substituting into \(y=\frac{x^2}{2x+1}\):

   when \(x=0\), \(y=0\); when \(x=-1\), \(y=-1\).

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

3. For the sketch, draw the vertical asymptote \(x=-\frac12\) and the oblique asymptote \(y=\frac12x-\frac14\).

   Since

   \(\displaystyle y-\left(\frac12x-\frac14\right)=\frac{1}{4(2x+1)}\),

   the curve is below the oblique asymptote for \(x<-\frac12\), and above it for \(x>-\frac12\).

   As \(x\to-\frac12^-\), \(y\to-\infty\), and as \(x\to-\frac12^+\), \(y\to+\infty\).

   The left-hand branch approaches the oblique asymptote from below as \(x\to-\infty\), has a stationary point at \((-1,-1)\), and then tends to \(-\infty\) as it approaches the vertical asymptote.

   The right-hand branch tends to \(+\infty\) as it approaches the vertical asymptote from the right, has a stationary point at \((0,0)\), and then approaches the oblique asymptote from above as \(x\to+\infty\).

Answer: (i) The asymptotes are

\(x=\frac{1}{k}\) and \(y=\frac{x}{k}\).

(ii) The stationary points are

\(\left(\frac{2}{k},\,\frac{4}{k^2}\right)\) and \((0,0)\).

(iii) The sketch is a rational curve with vertical asymptote \(x=\frac{1}{k}\), oblique asymptote \(y=\frac{x}{k}\), passing through \((0,0)\) and \(\left(\frac{2}{k},\frac{4}{k^2}\right)\), with the branch for \(x<\frac{1}{k}\) lying below the slant asymptote for large negative \(x\) and tending to \(-\infty\) as \(x\to\left(\frac{1}{k}\right)^-\), while the branch for \(x>\frac{1}{k}\) tends to \(+\infty\) as \(x\to\left(\frac{1}{k}\right)^+\) and then approaches the line \(y=\frac{x}{k}\) as \(x\to\infty\).

(i) Write

\(y=\dfrac{x^2}{kx-1}\).

A vertical asymptote occurs when the denominator is zero:

\(kx-1=0 \implies x=\dfrac{1}{k}.\)

For the oblique asymptote, divide \(x^2\) by \(kx-1\):

\(\dfrac{x^2}{kx-1}=\dfrac{1}{k}x+\dfrac{1}{k^2}+\dfrac{1/k^2}{kx-1}\).

Since the final fraction tends to 0 as \(|x|\to\infty\), the slant asymptote is

\(y=\dfrac{x}{k}+\dfrac{1}{k^2}.\)

However, it is cleaner to check by long division carefully:

\(x^2=(kx-1)\left(\dfrac{x}{k}+\dfrac{1}{k^2}\right)+\dfrac{1}{k^2}\),

so

\(y=\dfrac{x}{k}+\dfrac{1}{k^2}+\dfrac{1/k^2}{kx-1}.\)

Hence the oblique asymptote is \(y=\dfrac{x}{k}+\dfrac{1}{k^2}\).

(ii) Differentiate using the quotient rule:

\(y' = \dfrac{(2x)(kx-1)-x^2(k)}{(kx-1)^2}.\)

Simplify the numerator:

\(2x(kx-1)-kx^2 = 2kx^2-2x-kx^2 = kx^2-2x = x(kx-2).\)

So

\(y' = \dfrac{x(kx-2)}{(kx-1)^2}.\)

Stationary points occur when \(y'=0\), so

\(x(kx-2)=0\), giving \(x=0\) or \(x=\dfrac{2}{k}.\)

Now find the corresponding \(y\)-values:

For \(x=0\), \(y=0\).

For \(x=\dfrac{2}{k}\),

\(y=\dfrac{(2/k)^2}{k(2/k)-1}=\dfrac{4/k^2}{2-1}=\dfrac{4}{k^2}.\)

So the stationary points are \((0,0)\) and \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).

(iii) To sketch the curve, note the key features:

• Vertical asymptote: \(x=\dfrac{1}{k}\).
• Oblique asymptote: \(y=\dfrac{x}{k}+\dfrac{1}{k^2}\).
• Intercept and stationary point: the curve passes through \((0,0)\), which is a stationary point.
• Second stationary point: \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).

For \(x<\dfrac{1}{k}\), the denominator is negative, so \(y\le 0\) except at \(x=0\) where it is 0. As \(x\to \left(\dfrac{1}{k}\right)^-\), \(y\to -\infty\).

For \(x>\dfrac{1}{k}\), the denominator is positive and the curve goes to \(+\infty\) as \(x\to \left(\dfrac{1}{k}\right)^+\), then turns at \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\), and tends to the oblique asymptote as \(x\to\infty\).

So the sketch consists of two branches separated by the vertical asymptote, with the left branch crossing the origin and the right branch having a turning point at \(\left(\dfrac{2}{k},\dfrac{4}{k^2}\right)\).

Answer: (i) Comparing the given curve with the asymptote tells us that for large values of \(x\),

\(\displaystyle \frac{x^2+1}{ax+b} \sim 2x+1.\)

Now divide \(x^2+1\) by \(ax+b\):

\(\displaystyle \frac{x^2+1}{ax+b}=\frac{1}{a}x-\frac{b}{a^2}+\frac{\text{remainder}}{ax+b}.\)

So the slant asymptote is \(y=\frac{1}{a}x-\frac{b}{a^2}\). Matching this with \(y=2x+1\) gives

\(\frac{1}{a}=2 \Rightarrow a=\frac12,\)

and

\(-\frac{b}{a^2}=1 \Rightarrow -\frac{b}{(1/2)^2}=1 \Rightarrow -4b=1 \Rightarrow b=-\frac14.\)

So \(a=\frac12\) and \(b=-\frac14\).

(ii) The other asymptote is where the denominator is zero:

\(\frac12 x-\frac14=0 \Rightarrow x=\frac12.\)

So the other asymptote is \(x=\frac12\).

(iii) The curve is

\(\displaystyle y=\frac{x^2+1}{\frac12 x-\frac14}=\frac{4(x^2+1)}{2x-1}.\)

Useful features for the sketch:

• Vertical asymptote: \(x=\frac12\)
• Oblique asymptote: \(y=2x+1\)
• \(y\)-intercept: when \(x=0\), \(y=\frac{1}{-1/4}=-4\), so the curve crosses the \(y\)-axis at \((0,-4)\)

Also, since \(x^2+1>0\) for all real \(x\), the sign of \(y\) is determined by the denominator \(\frac12 x-\frac14\). Hence the curve is below the \(x\)-axis for \(x<\frac12\) and above it for \(x>\frac12\). As \(x\to \frac12^-\), \(y\to -\infty\), and as \(x\to \frac12^+\), \(y\to +\infty\). For large positive \(x\), the curve approaches \(y=2x+1\) from above; for large negative \(x\), it approaches the same line from below.

A correct sketch should show a vertical asymptote at \(x=\frac12\), the slant asymptote \(y=2x+1\), and the point \((0,-4)\).

We are given that the line \(y=2x+1\) is an asymptote of \(C\), where \(C\) has equation \(\displaystyle y=\frac{x^2+1}{ax+b}\).

Since the numerator has degree 2 and the denominator has degree 1, the curve has a slant asymptote found by division.

Divide \(x^2+1\) by \(ax+b\). The leading term of the quotient must be \(\frac{1}{a}x\), because

\(\displaystyle \frac{x^2}{ax}=\frac{1}{a}x.\)

So the asymptote has form

\(\displaystyle y=\frac{1}{a}x+c\)

for some constant \(c\). Since the asymptote is given as \(y=2x+1\), we match coefficients:

\(\displaystyle \frac{1}{a}=2 \Rightarrow a=\frac12.\)

Now find \(b\). With \(a=\frac12\), write the division explicitly:

\(\displaystyle \frac{x^2+1}{\frac12 x+b}.\)

Take the next term in the quotient as \(2x\), because

\(\displaystyle 2x\left(\frac12 x+b\right)=x^2+2bx.\)

Subtracting from \(x^2+1\) leaves a remainder of \(-2bx+1\). The next constant term in the quotient is chosen so that the linear term of the remainder matches the asymptote constant. More simply, using the general result for division of \(x^2+1\) by \(ax+b\), the asymptote is

\(\displaystyle y=\frac{1}{a}x-\frac{b}{a^2}.\)

Matching with \(y=2x+1\), and using \(a=\frac12\), gives

\(\displaystyle -\frac{b}{(1/2)^2}=1\)

so

\(\displaystyle -4b=1 \Rightarrow b=-\frac14.\)

Therefore

\(\displaystyle a=\frac12,\quad b=-\frac14.\)

(ii) The other asymptote is the vertical asymptote where the denominator is zero:

\(\displaystyle \frac12 x-\frac14=0 \Rightarrow x=\frac12.\)

So the other asymptote is \(\boxed{x=\frac12}\).

(iii) For the sketch, substitute the values of \(a\) and \(b\):

\(\displaystyle y=\frac{x^2+1}{\frac12 x-\frac14}=\frac{4(x^2+1)}{2x-1}.\)

Key features:

• Vertical asymptote: \(x=\frac12\)
• Oblique asymptote: \(y=2x+1\)
• \(y\)-intercept: at \(x=0\), \(y=\frac{1}{-1/4}=-4\), so the curve passes through \((0,-4)\)

Since \(x^2+1>0\) for all real \(x\), the sign of \(y\) depends on \(2x-1\). Hence:

• for \(x<\frac12\), \(y<0\),
• for \(x>\frac12\), \(y>0\).

As \(x\to \frac12^-\), \(y\to -\infty\), and as \(x\to \frac12^+\), \(y\to +\infty\).

Also, because the asymptote is \(y=2x+1\), the curve approaches this straight line for large \(|x|\). A correct sketch should therefore show two branches separated by the vertical asymptote, with the left-hand branch passing through \((0,-4)\) and tending to \(-\infty\) near \(x=\frac12\), and the right-hand branch tending to \(+\infty\) near \(x=\frac12\) and then approaching the line \(y=2x+1\).

Answer: (i) The asymptotes are \(x=-b\) and \(y=x-b\).

(ii) The curve does not intersect the \(x\)-axis because \(x^2+b=0\) has no real solution when \(b>0\).

(iii) There are \(2\) stationary points on \(C\).

(iv) The curve crosses the \(y\)-axis at \((0,1)\). It has a vertical asymptote \(x=-b\) and an oblique asymptote \(y=x-b\), does not meet the \(x\)-axis, and has two stationary points.

We first rewrite the function by division:

\(y=\dfrac{x^2+b}{x+b}=x-b+\dfrac{b(1+b)}{x+b}\).

This form is very useful for the asymptotes and for differentiation.

(i) As \(x\to -b\), the denominator of the last term tends to \(0\), so there is a vertical asymptote at \(x=-b\).

As \(|x|\to\infty\), the term \(\dfrac{b(1+b)}{x+b}\to 0\), so the curve approaches the straight line \(y=x-b\). Hence the oblique asymptote is \(y=x-b\).

(ii) To find any \(x\)-axis intersections, set \(y=0\):

\(\dfrac{x^2+b}{x+b}=0\).

This requires \(x^2+b=0\), so \(x^2=-b\). But \(b>0\), so the right-hand side is negative and there are no real solutions. Therefore \(C\) does not intersect the \(x\)-axis.

(iii) Differentiate the rewritten form:

\(y=x-b+\dfrac{b(1+b)}{x+b}\).

So

\(\dfrac{dy}{dx}=1-\dfrac{b(1+b)}{(x+b)^2}.\)

Stationary points occur when \(\dfrac{dy}{dx}=0\):

\(1-\dfrac{b(1+b)}{(x+b)^2}=0\)

\(\Rightarrow (x+b)^2=b(1+b).\)

Since \(b>0\), \(b(1+b)>0\), so this equation has two distinct real solutions:

\(x=-b\pm \sqrt{b(1+b)}\).

Hence the curve has \(2\) stationary points.

(iv) For the sketch, note:

• vertical asymptote \(x=-b\),
• slant asymptote \(y=x-b\),
• no \(x\)-axis intercepts,
• \(y\)-axis intercept found by setting \(x=0\): \(y=\dfrac{b}{b}=1\), so the point is \((0,1)\),
• two stationary points, one on each side of the vertical asymptote.

A correct sketch should show two branches separated by \(x=-b\), approaching the line \(y=x-b\) for large \(|x|\), passing through \((0,1)\), and never crossing the \(x\)-axis.

Answer: (i) The points of intersection with the axes are \((-1,0)\), \((-6,0)\) and \((0,-3)\).

(ii) The asymptotes are \(x=2\) and \(y=x+9\).

(iii) A sketch of \(C\) should show two branches, with vertical asymptote \(x=2\), oblique asymptote \(y=x+9\), crossing the axes at \((-6,0)\), \((-1,0)\) and \((0,-3)\).

(i) To find where the curve meets the axes, we use:

• the x-axis: set \(y=0\)
• the y-axis: set \(x=0\)

For the \(x\)-axis,

\(0=\frac{x^{2}+7x+6}{x-2}\).

A fraction is zero when its numerator is zero, provided the denominator is not zero. So solve

\(x^{2}+7x+6=0\).

Factorising gives

\((x+1)(x+6)=0\).

Hence \(x=-1\) or \(x=-6\).

So the curve meets the \(x\)-axis at \((-1,0)\) and \((-6,0)\).

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

\(y=\frac{0^2+7(0)+6}{0-2}=\frac{6}{-2}=-3\).

So the curve meets the \(y\)-axis at \((0,-3)\).

(ii) The denominator is zero when \(x=2\), and there is no factor of \(x-2\) in the numerator to cancel it, so there is a vertical asymptote

\(x=2\).

To find the oblique asymptote, divide \(x^2+7x+6\) by \(x-2\):

\(x^2+7x+6=(x-2)(x+9)+24\).

Therefore

\(y= x+9+\frac{24}{x-2}\).

As \(x\to\pm\infty\), the fraction tends to 0, so the oblique asymptote is

\(y=x+9\).

(iii) For the sketch, draw the asymptotes \(x=2\) and \(y=x+9\) first.

Then plot the intercepts \((-6,0)\), \((-1,0)\) and \((0,-3)\).

The curve has two separate branches, one on each side of the vertical asymptote \(x=2\). From

\(y=x+9+\frac{24}{x-2}\),

when \(x>2\), the term \(\frac{24}{x-2}\) is positive, so the right-hand branch lies above the slant asymptote and rises to \(+\infty\) as \(x\to2^+\).

When \(x<2\), the term \(\frac{24}{x-2}\) is negative, so the left-hand branch lies below the slant asymptote and falls to \(-\infty\) as \(x\to2^-\).

A correct sketch should show these features and pass through the intercepts found in part (i).