(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 \(\frac{5}{2} < x < \frac{11}{2}\).

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 3\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + 2x + 1}{x - 3}\):

\(y = \frac{(x-3)(x+5) + 16}{x-3} = x + 5 + \frac{16}{x-3}\)

Thus, the oblique asymptote is \(y = x + 5\).

(b) Differentiate \(y = \frac{x^2 + 2x + 1}{x - 3}\) to find turning points:

\(\frac{dy}{dx} = 1 - \frac{16}{(x-3)^2}\)

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

\(1 - \frac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16\)

Solving gives \(x = -1\) and \(x = 7\). Substitute back to find \(y\)-coordinates: \((-1, 0)\) and \((7, 16)\).

(c) Sketch the curve using the asymptotes and turning points found.

(d) For \(y = \left| \frac{x^2 + 2x + 1}{x - 3} \right|\), reflect the negative parts of the curve above the x-axis. For \(y^2 = \frac{x^2 + 2x + 1}{x - 3}\), consider both positive and negative square roots, sketching both branches.

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + x + 1 = 0\). The discriminant is \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\), which is less than zero, so no real roots exist. Therefore, there are no vertical asymptotes. The horizontal asymptote is found by considering the degrees of the polynomials. As \(x \to \infty\), \(y \to \frac{2x^2}{x^2} = 2\). Thus, the horizontal asymptote is \(y = 2\).

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

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

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(3x^2 + 6x = 0\).

Solving gives \(x = 0\) or \(x = -2\). Substituting back into the original equation gives points \((0, -1)\) and \((-2, 3)\).

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

\(2x^2 - x - 1 = 0\).

Solving gives \(x = 1\) and \(x = -\frac{1}{2}\). The curve intersects the y-axis where \(x = 0\), giving \((0, -1)\).

(d) For \(\left| \frac{2x^2 - x - 1}{x^2 + x + 1} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at 4 points. This occurs when \(0 < k < 1\).

(a) The curve \(y = \frac{x+1}{x-1}\) has a vertical asymptote at \(x = 1\) and a horizontal asymptote at \(y = 1\). The branches are in the first and third quadrants.

(b) The curve \(y = \frac{|x|+1}{|x|-1}\) is symmetrical about \(x = 0\). The critical points are found by solving \(\frac{|x|+1}{|x|-1} = -2\), leading to \(|x| = \frac{1}{3}\). Thus, the set of values of \(x\) for which \(\frac{|x|+1}{|x|-1} < -2\) is \(-1 < x < -\frac{1}{3}, \frac{1}{3} < x < 1\).

(a) The vertical asymptote occurs where the denominator is zero, so \(x = 1\).

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

\(y = (x - 1)(ax + a + 1) + a\) gives \(y = ax + a + 1 + \frac{a}{x - 1}\).

Thus, the oblique asymptote is \(y = ax + a + 1\).

(b) Rewrite the equation as \(ax^2 + x - 1 = y(x - 1)\), leading to \(ax^2 + x - 1 = yx - y\).

This simplifies to \(ax^2 - (y - 1)x + (y - 1) = 0\).

The discriminant \((y - 1)^2 - 4a(y - 1)\) must be less than zero for no real roots, giving \(1 < y < 1 + 4a\).

Thus, there is no point on \(C\) for which \(1 < y < 1 + 4a\).

(c) The sketch shows the curve with the vertical asymptote at \(x = 1\) and the oblique asymptote \(y = ax + a + 1\). The curve has two branches, one in each of the regions determined by the asymptotes.

(a) The vertical asymptote occurs where the denominator is zero: \(x + 1 = 0 \Rightarrow x = -1\).
The oblique asymptote is found by dividing \(x^2 - x\) by \(x + 1\), giving \(y = x - 2\).

(b) Differentiate \(y = \frac{x^2 - x}{x + 1}\) using the quotient rule:
\(\frac{dy}{dx} = \frac{(2x - 1)(x + 1) - (x^2 - x)}{(x + 1)^2} = \frac{1 - 2(x + 1)^2}{(x + 1)^2}\).
Set \(\frac{dy}{dx} = 0\) to find stationary points:
\(1 - 2(x + 1)^2 = 0 \Rightarrow (x + 1)^2 = \frac{1}{2}\).
Solving gives \(x = -1 \pm \sqrt{2}\).
Substitute back to find \(y\):
\(y = \frac{(-1 + \sqrt{2})^2 - (-1 + \sqrt{2})}{-1 + \sqrt{2} + 1} = -3 + 2\sqrt{2}\),
\(y = \frac{(-1 - \sqrt{2})^2 - (-1 - \sqrt{2})}{-1 - \sqrt{2} + 1} = -3 - 2\sqrt{2}\).

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

(d) Solve \(\left| \frac{x^2 - x}{x + 1} \right| < 6\):
\(\frac{x^2 - x}{x + 1} < 6\) or \(\frac{x^2 - x}{x + 1} > -6\).
For \(\frac{x^2 - x}{x + 1} < 6\):
\(x^2 - 7x - 6 = 0 \Rightarrow x = -3, -2\).
For \(\frac{x^2 - x}{x + 1} > -6\):
\(x^2 + 5x + 6 = 0 \Rightarrow x = -2, -3\).
Thus, \(-3 < x < -2 \text{ and } \frac{1}{2} - \frac{1}{2}\sqrt{73} < x < \frac{1}{2} + \frac{1}{2}\sqrt{73}\).

(a) The vertical asymptotes occur where the denominator is zero: \(1 + x - x^2 = 0\). Solving \(x^2 - x - 1 = 0\) gives \(x = \frac{1 \pm \sqrt{5}}{2}\). The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = -1\).

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Using the quotient rule, \(\frac{dy}{dx} = \frac{(1 + x - x^2)(2x - 1) - (x^2 - x - 3)(1 - 2x)}{(1 + x - x^2)^2}\). Simplifying gives \((2x - 1)(x - 2) = 0\), so \(x = \frac{1}{2}\). Substituting back, \(y = -\frac{13}{9}\), giving the point \(\left( \frac{1}{2}, -\frac{13}{9} \right)\).

(c) The intersections with the axes are found by setting \(y = 0\) and \(x = 0\). Solving \(x^2 - x - 3 = 0\) gives \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{13}\). For \(x = 0\), \(y = -3\). Thus, intersections are \(\left( \frac{1}{2} + \frac{1}{2}\sqrt{13}, 0 \right), \left( \frac{1}{2} - \frac{1}{2}\sqrt{13}, 0 \right), (0, -3)\).

(d) For \(\left| \frac{x^2 - x - 3}{1 + x - x^2} \right| < 3\), solve \(\frac{x^2 - x - 3}{1 + x - x^2} < 3\) and \(\frac{x^2 - x - 3}{1 + x - x^2} > -3\). Solving gives critical points \(x = \frac{1}{2} \pm \frac{1}{2}\sqrt{7}, x = 0, x = 1\). The solution is \(x < \frac{1}{2} - \frac{1}{2}\sqrt{7}, \; 0 < x < 1, \; x > \frac{1}{2} + \frac{1}{2}\sqrt{7}\).

(a) The vertical asymptotes occur where the denominator is zero: \(4 - 4x^2 = 0\), giving \(x = 1\) and \(x = -1\). The horizontal asymptote is \(y = 0\) as \(x\) approaches infinity.

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{(4-4x^2)(4) - (4x+5)(-8x)}{(4-4x^2)^2}\). Simplify and solve \(16x^2 + 40x + 16 = 0\) to find \((-2, \frac{1}{4})\) and \(-\frac{1}{2}, 1\).

(c) The curve intersects the x-axis where \(y = 0\), giving \(x = -\frac{5}{4}\). It intersects the y-axis where \(x = 0\), giving \(y = \frac{5}{4}\).

(d) For \(y = \left| \frac{4x+5}{4-4x^2} \right|\), solve \(4|4x+5| > 5|4-4x^2|\). This gives two cases: \(\frac{4x+5}{4-4x^2} > \frac{5}{4}\) and \(\frac{4x+5}{4-4x^2} < -\frac{5}{4}\). Solving these inequalities gives the ranges \(\frac{2}{3} - \frac{1}{3}\sqrt{6} < x < -\frac{4}{3}\) and \(0 < x < \frac{2}{3} + \frac{1}{3}\sqrt{6}\), excluding \(x = \pm 1\).

(a) To find the vertical asymptote, set the denominator equal to zero: \(x - 3 = 0\), so \(x = 3\).
For the oblique asymptote, perform polynomial long division on \(\frac{x^2}{x-3}\):

\(x^2 \div (x-3) = x + 3 + \frac{9}{x-3}\).
As \(x \to \infty\), \(\frac{9}{x-3} \to 0\), so the oblique asymptote is \(y = x + 3\).

(b) Substitute \(y = \frac{x^2}{x-3}\) into \(yx - 3y = x^2\) to form a quadratic: \(x^2 - yx + 3y = 0\).
The discriminant \(b^2 - 4ac = y^2 - 4(3y) = y^2 - 12y\) must be negative for no real solutions.
\(y^2 - 12y < 0\) implies \(0 < y < 12\), showing no points exist for \(0 < y < 12\).

(c) Sketch the curve showing the vertical asymptote at \(x = 3\) and the oblique asymptote \(y = x + 3\). The curve approaches these asymptotes.

(d)(i) Sketch \(y = \left| \frac{x^2}{x-3} \right|\) and \(y = |x| - 3\).
Find intercepts: \(y = \left| \frac{x^2}{x-3} \right|\) intersects the y-axis at \((0,0)\).
\(y = |x| - 3\) intersects the y-axis at \((0,-3)\) and x-axis at \((3,0)\) and \((-3,0)\).

(d)(ii) From the sketch, \(\left| \frac{x^2}{x-3} \right| \leq |x| + c\) has no solution when \(c < -3\).

Answer: (i) The asymptotes of \(C_1\) are \(x=-5\) and \(y=a\).

(ii) The oblique asymptote of \(C_2\) is \(y=x+a+5\).

(iii) The curves do not intersect.

(iv) The stationary points of \(C_2\) are \((-5,-2)\) and \((-1,a+1)\).

(v) A suitable sketch shows \(C_1\) with asymptotes \(x=-5\), \(y=a\), and a horizontal branch passing through \((0,0)\), and \(C_2\) with vertical asymptote \(x=-5\), oblique asymptote \(y=x+a+5\), stationary points at \((-5,-2)\) and \((-1,a+1)\), and no intersections with \(C_1\).

(i) For \(C_1\),

\(y=\dfrac{ax}{x+5}\).

As \(x\to -5\), the denominator tends to \(0\), so there is a vertical asymptote

\(x=-5\).

Also, dividing numerator and denominator by \(x\),

\(y=\dfrac{a}{1+5/x}\),

so as \(x\to \pm\infty\), \(y\to a\). Hence the horizontal asymptote is

\(y=a\).

(ii) For \(C_2\), divide the numerator by \(x+5\):

\(x^2+(a+10)x+5a+26 = (x+5)(x+a+5)+1\).

So

\(y=\dfrac{(x+5)(x+a+5)+1}{x+5}=x+a+5+\dfrac{1}{x+5}.\)

Therefore the oblique asymptote is

\(y=x+a+5\).

(iii) To find intersections, set the two expressions for \(y\) equal:

\(\dfrac{ax}{x+5}=\dfrac{x^2+(a+10)x+5a+26}{x+5}.\)

Since both have the same denominator, multiply through by \(x+5\):

\(ax=x^2+(a+10)x+5a+26.\)

Rearranging gives

\(0=x^2+10x+5a+26.\)

The discriminant is

\(\Delta=10^2-4(1)(5a+26)=100-20a-104=-20a-4.\)

Because \(a>2\), we have \(-20a-4<0\). So there are no real solutions, and hence \(C_1\) and \(C_2\) do not intersect.

(iv) Write \(C_2\) as

\(y=x+a+5+\dfrac{1}{x+5}.\)

Differentiate:

\(\dfrac{dy}{dx}=1-\dfrac{1}{(x+5)^2}.\)

Set this equal to zero for stationary points:

\(1-\dfrac{1}{(x+5)^2}=0\)

\(\Rightarrow (x+5)^2=1\)

\(\Rightarrow x=-4 \text{ or } x=-6.\)

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

When \(x=-4\),

\(y=-4+a+5+\dfrac{1}{1}=a+2.\)

When \(x=-6\),

\(y=-6+a+5+\dfrac{1}{-1}=a-2.\)

So the stationary points are \((-4,a+2)\) and \((-6,a-2)\).

(v) For the sketch:

• \(C_1\) has vertical asymptote \(x=-5\) and horizontal asymptote \(y=a\).
• Since \(y=\dfrac{ax}{x+5}\), it passes through \((0,0)\).
• For \(x>-5\), the curve goes through the origin and approaches \(y=a\) as \(x\to\infty\), and \(y\to +\infty\) as \(x\to -5^+\).
• For \(x<-5\), the branch lies above \(y=a\) and tends to \(-\infty\) as \(x\to -5^-\).
• \(C_2\) has vertical asymptote \(x=-5\) and oblique asymptote \(y=x+a+5\).
• It has stationary points at \((-6,a-2)\) and \((-4,a+2)\).
• It does not intersect \(C_1\).

This gives the required combined sketch.