Past Exam Questions

(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

Answer:

For

\(y=\frac{2x^{2}+kx}{x+1},\)

differentiate using the quotient rule:

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

Stationary points require

\(2x^{2}+4x+k=0.\)

For no stationary points, this quadratic must have no real roots:

\(4^{2}-4(2)(k)\lt 0.\)

Thus

\(16-8k\lt 0\quad\Rightarrow\quad k\gt 2.\)

When \(k=4\),

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

So the vertical asymptote is \(x=-1\), and the oblique asymptote is \(y=2x+2\).

The intercepts come from

\(2x^{2}+4x=2x(x+2)=0,\)

so the curve crosses the \(x\)-axis at \((-2,0)\) and \((0,0)\). The curve has one branch below the oblique asymptote for \(x\gt -1\), and one branch above it for \(x\lt -1\), with the vertical asymptote at \(x=-1\).

Answer: \(-\frac12\leqslant y\leqslant\frac92\).

The turning points are \(\left(-3,\frac92\right)\) and \(\left(\frac13,-\frac12\right)\). The asymptote is \(y=4\).

The curve intersects the coordinate axes at \((0,0)\) and \(\left(\frac34,0\right)\), and intersects the asymptote at \(\left(-\frac43,4\right)\).

Start with

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

To verify the first form, expand the numerator:

\(-\frac12+\frac{(3x-1)^2}{2(x^2+1)}=\frac{-(x^2+1)+(3x-1)^2}{2(x^2+1)}\).

Now \((3x-1)^2=9x^2-6x+1\), so

\(\frac{-(x^2+1)+9x^2-6x+1}{2(x^2+1)}=\frac{8x^2-6x}{2(x^2+1)}=\frac{4x^2-3x}{x^2+1}\).

This is the given equation of \(C\).

For the second form,

\(\frac92-\frac{(x+3)^2}{2(x^2+1)}=\frac{9(x^2+1)-(x+3)^2}{2(x^2+1)}\).

Since \((x+3)^2=x^2+6x+9\), this becomes

\(\frac{9x^2+9-(x^2+6x+9)}{2(x^2+1)}=\frac{8x^2-6x}{2(x^2+1)}=\frac{4x^2-3x}{x^2+1}\).

So this is also equal to the original expression.

Now use the two rewritten forms to find bounds.

Because \((3x-1)^2\ge 0\) and \(x^2+1\gt 0\) for all \(x\), we have

\(\frac{(3x-1)^2}{2(x^2+1)}\ge 0\), hence \(y\ge -\frac12\).

Similarly, \((x+3)^2\ge 0\) and \(x^2+1\gt 0\), so

\(\frac{(x+3)^2}{2(x^2+1)}\ge 0\), hence \(y\le \frac92\).

Therefore \(-\frac12\le y\le \frac92\).

Turning points occur when these squared terms are zero. From \((3x-1)^2=0\), we get \(x=\frac13\) and then \(y=-\frac12\). From \((x+3)^2=0\), we get \(x=-3\) and then \(y=\frac92\). So the turning points are \(\left(\frac13,-\frac12\right)\) and \(\left(-3,\frac92\right)\).

For the asymptote, divide numerator and denominator:

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

As \(x\to\pm\infty\), the fraction tends to \(0\), so the asymptote is \(y=4\).

To find the intersections with the axes, set \(y=0\):

\(4x^2-3x=x(4x-3)=0\), so \(x=0\) or \(x=\frac34\). Thus the x-intercepts are \((0,0)\) and \(\left(\frac34,0\right)\). The y-intercept is also \((0,0)\), since substituting \(x=0\) gives \(y=0\).

To find where the curve meets the asymptote, set \(y=4\):

\(\frac{4x^2-3x}{x^2+1}=4\), so \(4x^2-3x=4x^2+4\), hence \(-3x=4\) and \(x=-\frac43\). Therefore the intersection is \(\left(-\frac43,4\right)\).

A sketch should show the horizontal asymptote \(y=4\), the turning points \(\left(-3,\frac92\right)\) and \(\left(\frac13,-\frac12\right)\), and the axis intersections \((0,0)\) and \(\left(\frac34,0\right)\). The curve approaches \(y=4\) as \(x\to\pm\infty\), crosses it at \(\left(-\frac43,4\right)\), rises to the maximum at \((-3,\frac92)\), falls through \((0,0)\) to the minimum at \(\left(\frac13,-\frac12\right)\), then rises again through \(\left(\frac34,0\right)\) and tends back towards \(y=4\).

Answer: Either

(i) The asymptotes are the vertical lines \(x=1\) and \(x=3\), and the horizontal asymptote is \(y=1\).

(ii) The curve meets the horizontal asymptote at \(x=\frac{2a-3}{a-2}\).

(iii) The stationary points have \(x\)-coordinates satisfying \((a-2)x^{2}+(6-4a)x+(5a-6)=0\). Hence the curve has stationary points when \(1\lt a\lt 3\), with \(a\neq 2\) already given.

(iv) For \(a\gt 3\), the graph has two vertical asymptotes at \(x=1\) and \(x=3\), and a horizontal asymptote \(y=1\). For \(2\lt a\lt 3\), the same asymptotes occur, with the middle branch having a maximum between \(y=0\) and \(y=1\), and the outside branches having a minimum point.

Or

(i) \(S_{n+4}-5S_{n+2}+2S_{n+1}-S_n=0\).

(ii) \(S_2=10\) and \(S_4=54\).

(iii) \(S_3=-6\) and \(S_6=292\).

(iv) The required value is \(248\).

Either

Let \(y=\frac{(x-2)(x-a)}{(x-1)(x-3)}\).

(i) The denominator is zero when \(x=1\) or \(x=3\), so these are vertical asymptotes. Since the numerator and denominator both have degree 2 with leading coefficient 1, the horizontal asymptote is \(y=1\).

(ii) To find where the curve meets the horizontal asymptote, set \(y=1\):

\(\frac{(x-2)(x-a)}{(x-1)(x-3)}=1\).

So

\((x-2)(x-a)=(x-1)(x-3)\).

Expanding gives

\(x^{2}-(a+2)x+2a=x^{2}-4x+3\).

Hence

\((4-a)x=3-2a\),

so

\(x=\frac{2a-3}{a-2}\).

Therefore the curve meets its horizontal asymptote at \(x=\frac{2a-3}{a-2}\).

(iii) Write \(y=\frac{N}{D}\), where \(N=(x-2)(x-a)\) and \(D=(x-1)(x-3)\). Then

\(y'=\frac{N'D-ND'}{D^{2}}\).

Stationary points satisfy \(y'=0\), so

\(N'D-ND'=0\).

Now \(N= x^{2}-(a+2)x+2a\), so \(N'=2x-(a+2)\), and \(D=x^{2}-4x+3\), so \(D'=2x-4\).

Thus

\((2x-a-2)(x^{2}-4x+3)-(x^{2}-(a+2)x+2a)(2x-4)=0\).

Expanding and simplifying gives

\((a-2)x^{2}+(6-4a)x+(5a-6)=0\).

For stationary points to exist, this quadratic in \(x\) must have real roots, so its discriminant must be non-negative:

\((6-4a)^{2}-4(a-2)(5a-6)\ge 0\).

This simplifies to

\(a^{2}-4a+3\le 0\),

so

\((a-1)(a-3)\le 0\).

Hence \(1\le a\le 3\). Since \(a\neq 2\) is given, the set of values is \(1\le a\lt 2\) or \(2\lt a\le 3\).

(iv) The graph always has vertical asymptotes \(x=1\) and \(x=3\), and horizontal asymptote \(y=1\). The middle branch lies between the vertical asymptotes, and the outer branches lie to the left of \(x=1\) and to the right of \(x=3\).

For \(a\gt 3\), the curve does not have stationary points. The sketch therefore shows the three branches approaching the asymptotes, with the middle branch passing through the horizontal asymptote once.

For \(2\lt a\lt 3\), the curve has stationary points. The middle branch has a maximum with value between \(0\) and \(1\), and each outer branch has a minimum point.

Or

Let the roots of \(x^{4}-5x^{2}+2x-1=0\) be \(\alpha,\beta,\gamma,\delta\), and define \(S_n=\alpha^n+\beta^n+\gamma^n+\delta^n\).

(i) If \(r\) is any root, then

\(r^{4}-5r^{2}+2r-1=0\).

Multiplying by \(r^n\) gives

\(r^{n+4}-5r^{n+2}+2r^{n+1}-r^n=0\).

Applying this to each root and adding,

\(S_{n+4}-5S_{n+2}+2S_{n+1}-S_n=0\).

(ii) By Vieta's formulae, the sum of the roots is zero, so \(S_1=0\), and the sum of pairwise products is \(-5\).

Now

\(S_2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\cdots)=0^2-2(-5)=10\).

Using the recurrence with \(n=0\),

\(S_4-5S_2+2S_1-S_0=0\).

Since \(S_0=4\), this gives

\(S_4=5(10)-2(0)+4=54\).

(iii) Use the recurrence with \(n=-1\):

\(S_3-5S_1+2S_0-S_{-1}=0\).

To find \(S_{-1}\), let \(y=1/x\). Then the roots \(1/\alpha,1/\beta,1/\gamma,1/\delta\) satisfy

\(1-5y^2+2y^3-y^4=0\), or equivalently \(y^4-2y^3+5y^2-1=0\).

Its roots sum to \(2\), so \(S_{-1}=2\). Therefore

\(S_3-5(0)+2(4)-2=0\),

so \(S_3=-6\).

Now use the recurrence with \(n=2\):

\(S_6-5S_4+2S_3-S_2=0\).

Hence

\(S_6=5(54)-2(-6)+10=292\).

(iv) Consider

\(\alpha^2(\beta^4+\gamma^4+\delta^4)+\beta^2(\gamma^4+\delta^4+\alpha^4)+\gamma^2(\delta^4+\alpha^4+\beta^4)+\delta^2(\alpha^4+\beta^4+\gamma^4)\).

When expanded, this is the sum of all terms \(r^2s^4\) with distinct roots \(r\neq s\). So it equals

\((\alpha^2+\beta^2+\gamma^2+\delta^2)(\alpha^4+\beta^4+\gamma^4+\delta^4)-\left(\alpha^6+\beta^6+\gamma^6+\delta^6\right)\).

That is, \(S_2S_4-S_6\).

Therefore the required value is

\(10\cdot 54-292=248\).

(a) Vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by comparing the degrees of the numerator and denominator, giving \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplify and solve \(4x^2 + 16x - 20 = 0\) to find \(x = -5\) and \(x = 1\). Substitute back to find \(y\)-coordinates: \((-5, \frac{25}{7})\) and \((1, \frac{1}{3})\).

(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2}, 0)\).

(d) Sketch the curve \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) with correct shape and position.

(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\). This leads to solving \(16x^2 - 38x + 4 = 0\) and \(20x^2 - 52x - 4 = 0\). The critical points are \(x = \frac{19}{16} - \frac{3}{16} \sqrt{33}, \ x = \frac{19}{16} + \frac{3}{16} \sqrt{33}, \ x = \frac{13}{10} - \frac{3}{10} \sqrt{21}, \ x = \frac{13}{10} + \frac{3}{10} \sqrt{21}\). The solution is \(\frac{13}{10} - \frac{3}{10} \sqrt{21} < x < \frac{19}{16} - \frac{3}{16} \sqrt{33}\) or \(\frac{19}{16} + \frac{3}{16} \sqrt{33} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).

(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x - 4 = 0\). Solving gives \(x = -\frac{1}{2}\) and \(x = 4\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x - 4)(4x - 5) - (2x^2 - 5x)(4x - 7)}{(2x^2 - 7x - 4)^2}\). Simplifying gives \(4x^2 + 16x - 20 = 0\), leading to points \((-5, \frac{25}{3})\) and \((1, \frac{1}{3})\).

(c) The curve intersects the axes at \((0,0)\) and \((\frac{5}{2},0)\).

(d) The sketch of \(y = \left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right|\) reflects the negative parts of the curve above the x-axis.

(e) Solve \(\left| \frac{2x^2 - 5x}{2x^2 - 7x - 4} \right| < \frac{1}{9}\) by considering \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} < \frac{1}{9}\) and \(\frac{2x^2 - 5x}{2x^2 - 7x - 4} > -\frac{1}{9}\). Solving these inequalities gives the intervals \(\frac{13}{10} < x < \frac{19}{16}\) and \(\frac{\sqrt{33}}{16} < x < \frac{13}{10} + \frac{3}{10} \sqrt{21}\).

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

(b) Differentiate \(y = \frac{x^2 + a}{x + a}\) to find stationary points: \(\frac{dy}{dx} = \frac{(x+a)^2 - (x^2+a)}{(x+a)^2} = \frac{-a^2}{(x+a)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x = -a \pm \sqrt{a^2 + a}\).

(c) Sketch the curve showing the asymptotes \(x = -a\) and \(y = x - a\). The curve intersects the y-axis at \((0, 1)\).

(d) Sketch the absolute value of the curve from (c), reflecting any negative parts above the x-axis.

(e) Solve \(\left| \frac{x^2 + a}{x + a} \right| = a\). This leads to \(x^2 - ax + a - a^2 = 0\). The discriminant \(a^2 - 4(a-a^2) = 5a^2 - 4a > 0\) gives \(a > \frac{3}{4}\).

(a) The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\). The leading coefficients of the numerator and denominator are equal, so \(y = 1\) is the horizontal asymptote.

(b) Consider \(y = \frac{x^2 + x - 4}{x^2 + x + 2}\). Rearrange to form a quadratic in \(x\): \((y-1)x^2 + (y-1)x + (2y + 4) = 0\). The discriminant \(\Delta = (y-1)^2 - 4(y-1)(2y+4) \geq 0\) gives \((y-1)(7y + 17) \leq 0\). Solving this inequality gives \(-\frac{17}{7} \leq y < 1\).

(c) Find stationary points by setting \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{12x + 6}{(x^2 + x + 2)^2} = 0\). Solving gives \(x = -\frac{1}{2}\). Substitute back to find \(y = -\frac{17}{7}\). Stationary point is \(\left(-\frac{1}{2}, -\frac{17}{7}\right)\).

(d) Sketch the curve, noting the asymptote \(y = 1\) and intersections at \((0, -2), \left(\frac{1}{2}(1-\sqrt{7}), 0\right), \left(\frac{1}{2}(1+\sqrt{7}), 0\right)\).

(e) For \(y = \frac{|x|^2 + |x| - 4}{|x|^2 + |x| + 2} < -\frac{1}{2}\), solve \(3|x|^2 + 3|x| - 6 = 0\). Critical points are \(|x| = 1\). Thus, \(-1 < x < 1\).

(a) To find the vertical asymptotes, set the denominator equal to zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = \frac{4}{2} = 2\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\).

Set \(\frac{dy}{dx} = 0\) and solve \(-3x^2 + 2x + 1 = 0\) to find \(x = -\frac{1}{3}\) and \(x = 1\). Substitute back to find \(y\) values: \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) and \((1, -3)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve intersects the y-axis at \((0, \frac{1}{3})\).

(d) For \(\left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, \(k > 3\).

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 1 = 0\). This has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 1\).

(b) Rewrite \(y = 1 + \frac{2}{x^2 + 1}\). Since \(\frac{2}{x^2 + 1} > 0\), it follows that \(y > 1\). The maximum value of \(y\) occurs when \(\frac{2}{x^2 + 1}\) is maximized, which is 2, giving \(y = 3\). Thus, \(1 < y \leq 3\).

(c) Differentiate \(y = \frac{x^2 + 3}{x^2 + 1}\) to find stationary points: \(\frac{d}{dx} \left( \frac{x^2 + 3}{x^2 + 1} \right) = \frac{(x^2 + 1)(2x) - (x^2 + 3)(2x)}{(x^2 + 1)^2} = 0\). Solving gives \(x = 0\), and substituting back gives \(y = 3\). Thus, the stationary point is \((0, 3)\).

(d) The sketch shows the curve with a horizontal asymptote at \(y = 1\) and a stationary point at \((0, 3)\). The curve intersects the y-axis at \((0, 3)\).

(e) For \(y = \frac{x^2 + 1}{x^2 + 3} < \frac{1}{2}\), solve \(2(x^2 + 1) < x^2 + 3\), giving \(x^2 < 1\). Thus, \(-1 < x < 1\).

(a) The vertical asymptotes occur where the denominator is zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 2\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\). Simplifying the numerator gives \(-3x^2 + 2x + 1 = 0\). Solving this quadratic gives the stationary points \(\left( -\frac{1}{3}, \frac{1}{3} \right)\) and \((1, -3)\).

(c) The sketch should show the asymptotes at \(x = \frac{1}{2}, x = 3\), and \(y = 2\). The curve intersects the y-axis at \((0, \frac{1}{3})\).

(d) For \(y = \left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at four points. This occurs when \(k > 3\).

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 3 = 0\). This equation has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limit as \(x\) approaches infinity: \(y = \frac{x+1}{x^2+3} \to 0\). Thus, the horizontal asymptote is \(y = 0\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\): \(\frac{dy}{dx} = \frac{(x^2+3) - (x+1)(2x)}{(x^2+3)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x^2 + 2x - 3 = 0\). Solving this quadratic equation gives \(x = -3\) and \(x = 1\). Substituting back into the original equation gives the stationary points \((-3, -\frac{1}{6})\) and \((1, \frac{1}{2})\).

(c) The intersections with the axes occur when \(y = 0\) and \(x = 0\). For \(y = 0\), solve \(x+1 = 0\) to get \(x = -1\), so the intersection is \((-1, 0)\). For \(x = 0\), \(y = \frac{1}{3}\), so the intersection is \((0, \frac{1}{3})\).

(d) For \(y^2 = \frac{x+1}{x^2+3}\), the intersections with the axes are the same as in part (c), but \(y\) can be positive or negative. Thus, the intersections are \((-1, 0)\) and \((0, \pm \frac{1}{3})\). The stationary points are \((1, \pm \frac{1}{2})\).

(a) The vertical asymptotes occur where the denominator is zero: \(x^2 - x - 2 = 0\). Solving gives \(x = -1\) and \(x = 2\). The horizontal asymptote is found by considering the limits as \(x\) approaches infinity, giving \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes at \(x = -1\), \(x = 2\), and \(y = 1\). The curve intersects the y-axis at \((0, -1)\).

(d) The sketch of \(y = \frac{1}{f(x)}\) is derived from the sketch in (c), showing the reciprocal relationship.

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution gives the intervals \(-4 < x < -1\), \(0 < x < \frac{1}{2}\), and \(x > 2\).

(a) The vertical asymptotes occur where the denominator is zero: \(x^2 - x - 2 = 0\). Solving gives \(x = -1\) and \(x = 2\). The horizontal asymptote is found by considering the limits as \(x\) approaches infinity, giving \(y = 1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). The derivative is \(\frac{dy}{dx} = \frac{(x^2 - x - 2)(2x) - (x^2 + 2)(2x - 1)}{(x^2 - x - 2)^2}\). Solving \(x^2 + 8x - 2 = 0\) gives the stationary points \((-8.2, 0.9)\) and \((0.2, -0.9)\).

(c) The sketch shows the curve with asymptotes and intersection at \((0, -1)\).

(d) The curve \(y = \frac{1}{f(x)}\) is sketched by inverting the values of \(f(x)\) from part (c).

(e) Solve \(\frac{x^2 + 2}{x^2 - x - 2} = 1\) and \(\frac{x^2 + 2}{x^2 - x - 2} = -1\) to find critical points. The solution is \(-4 < x < -1, \ 0 < x < \frac{1}{2}, \ x > 2\).

(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 2 = 0 \Rightarrow x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x + 4)(x - 2) + 7 \Rightarrow y = x + 4\).

(b) Differentiate \(y = \frac{x^2 + 2x - 15}{x - 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x-2)(2x+2) - (x^2+2x-15)}{(x-2)^2}\). Simplify to get \(\frac{dy}{dx} = \frac{x^2 - 4x + 11}{(x-2)^2}\). The quadratic \(x^2 - 4x + 11\) has no real roots (discriminant \(16 - 44 < 0\)), so no stationary points.

(c) Intersections: Set \(y = 0\) to find \(x\)-intercepts: \(x^2 + 2x - 15 = 0 \Rightarrow (x+5)(x-3) = 0 \Rightarrow x = -5, 3\). For \(y\)-intercept, set \(x = 0\): \(y = \frac{-15}{-2} = 7.5\).

(d) Sketch the curve \(y = \left| \frac{x^2 - 2x - 15}{x - 2} \right|\) using the shape from part (c) and reflecting any negative parts above the \(x\)-axis.

(e) Solve \(\left| \frac{2x^2 + 4x - 30}{x - 2} \right| < 15\). Consider \(\frac{2x^2 + 4x - 30}{x - 2} = 15\) and \(\frac{2x^2 + 4x - 30}{x - 2} = -15\). Solve \(2x^2 + 4x - 30 = 15(x-2)\) and \(2x^2 + 4x - 30 = -15(x-2)\) to find critical points \(x = 0, \frac{11}{2}, -12, \frac{5}{2}\). Test intervals to find \(-12 < x < 0\) or \(\frac{5}{2} < x < \frac{11}{2}\).

(a) The vertical asymptote occurs where the denominator is zero: \(x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x - 2)(x + 4) - 7\), giving \(y = x + 4\).

(b) Differentiate \(y\) to find stationary points: \(\frac{dy}{dx} = \frac{(x - 2)(2x + 2) - (x^2 + 2x - 15)}{(x - 2)^2} = \frac{x^2 - 4x + 11}{(x - 2)^2}\). The quadratic \(x^2 - 4x + 11 = 0\) has no real roots (discriminant \(4^2 - 4 \times 1 \times 11 = -28 < 0\)), so no stationary points.

(c) The curve intersects the y-axis at \(x = 0\): \(y = \frac{0^2 + 2 \times 0 - 15}{0 - 2} = 7.5\). Intersects the x-axis where \(y = 0\): \(x^2 + 2x - 15 = 0\), giving roots \(x = -5\) and \(x = 3\).

(d) Sketch the absolute value function based on the sketch from (c), reflecting the negative parts above the x-axis.

(e) Solve \(\left| \frac{2x^2 + 4x - 30}{x - 2} \right| < 15\). Critical points from \(2x^2 + 4x - 30 = 15(x - 2)\) and \(2x^2 + 4x - 30 = -15(x - 2)\) give \(x = 0, \frac{11}{2}, -12, \frac{5}{2}\). Test intervals to find \(-12 < x < 0\) or \(\frac{5}{2} < x < \frac{11}{2}\).