Question. Curve $C$: $y = \dfrac{ax^2}{bx - 2}$ has asymptote $y = 4x + 8$. Find $a$ and $b$.

Solution.

Long division:

$$y = \dfrac{a}{b}x + \dfrac{2a}{b^2} + \dfrac{4a/b^2}{bx - 2}$$

So the oblique asymptote is $y = \dfrac{a}{b}x + \dfrac{2a}{b^2}$.

Compare with $y = 4x + 8$:

• Gradient: $\dfrac{a}{b} = 4 \Rightarrow a = 4b$ ...(1)
• Intercept: $\dfrac{2a}{b^2} = 8 \Rightarrow a = 4b^2$ ...(2)

Setting (1) = (2): $4b = 4b^2 \Rightarrow b(b-1) = 0$.

Since $b \neq 0$ (otherwise no curve), $b = 1$, so $a = 4$.

Answer: $a = 4$, $b = 1$.

Question. Sketch $y^2 + 8y - 2x^2 + 4x - 4 = 0$, stating centre and asymptotes.

Solution — complete the square.

Group $y$ and $x$ terms: $y^2 + 8y - 2(x^2 - 2x) - 4 = 0$

$(y + 4)^2 - 16 - 2[(x-1)^2 - 1] - 4 = 0$

$(y+4)^2 - 2(x-1)^2 = 18$

$$\dfrac{(y+4)^2}{18} - \dfrac{(x-1)^2}{9} = 1$$

Read off:

• $y^2$ term positive $\Rightarrow$ opens up/down.
• Centre: $(1, -4)$.
• $y$-distance from centre: $\sqrt{18} = 3\sqrt{2}$.
• $x$-distance: $\sqrt{9} = 3$.

Asymptotes:

$$y + 4 = \pm \dfrac{\sqrt{18}}{3}(x - 1) = \pm\sqrt{2}\,(x - 1)$$ $$\Rightarrow y = \pm\sqrt{2}\,(x - 1) - 4$$

Question. Given $y = f(x)$, sketch $y = 3 - f(2x)$. Describe the sequence.

Solution — work inside-out, one substitution at a time:

Step 1. Replace $x$ with $2x$ — scale along $x$-axis by factor $\tfrac{1}{2}$.

$$y = f(2x)$$

Step 2. Replace $y$ with $-y$ — reflect in the $x$-axis.

$$-y = f(2x) \;\Longrightarrow\; y = -f(2x)$$

Step 3. Replace $y$ with $y - 3$ — translate $+3$ along $y$-axis.

$$y - 3 = -f(2x) \;\Longrightarrow\; y = 3 - f(2x)$$

Sequence: $x$-scale by $\tfrac{1}{2}$ → reflect in $x$-axis → translate up by 3.

Question. $y = f(x)$ has min point at $\left(1, -\tfrac{1}{2}\right)$, cuts the $x$-axis at $(-2, 0)$, $(0, 0)$, $(2, 0)$, and has asymptotes $x = -1$, $x = 3$, $y = -3$. Find the number of distinct real roots of $|f(x)| - \tfrac{1}{2} = 0$.

Solution.

Equivalent to $|f(x)| = \tfrac{1}{2}$.

Step 1. Sketch $y = |f(x)|$:

• Reflect the parts of $y = f(x)$ below the $x$-axis to above.
• All $x$-intercepts $(-2, 0), (0, 0), (2, 0)$ unchanged (become cusps).
• Min point $(1, -\tfrac{1}{2})$ reflects to max point $(1, \tfrac{1}{2})$.
• VAs $x = -1$ and $x = 3$ unchanged.
• HA $y = -3$ becomes $y = 3$.

Step 2. Add the horizontal line $y = \tfrac{1}{2}$ to the same axes.

Step 3. Count intersections — the line cuts $|f(x)|$ at 5 distinct points.

Answer: 5 roots.

Question. Sketch $y = f(|3 - x|)$ from $y = f(x)$.

Solution — handle the modulus first, then transform the outside.

Step 1. Replace $x$ with $|x|$ — apply the $f(|x|)$ method:

• Remove the left half of $y = f(x)$.
• Reflect the right half in the $y$-axis.
• Result: $y = f(|x|)$ (symmetric about $y$-axis).

Step 2. Replace $x$ with $x + 3$ — translate $-3$ along $x$-axis:

• Result: $y = f(|x + 3|) = f(|3 + x|)$ (symmetric about $x = -3$).

Step 3. Replace $x$ with $-x$ — reflect in the $y$-axis:

• Result: $y = f(|3 - x|)$ (symmetric about $x = 3$).

Key principle: for $y = f(|ax + b|)$, do the modulus step first, then handle the outer translation / scaling / reflection.

Question. Solve $\dfrac{x^2 - 16}{x - 2} \leq -\sqrt{2 - x^2}$.

Solution — sketch both sides.

LHS $y = \dfrac{x^2 - 16}{x - 2}$. Long division: $y = x + 2 - \dfrac{12}{x - 2}$.

• VA: $x = 2$.
• Oblique asymptote: $y = x + 2$.
• $y$-intercept: $\dfrac{-16}{-2} = 8$.
• $x$-intercepts: $x = \pm 4$.

RHS $y = -\sqrt{2 - x^2}$ is the lower half of the circle $x^2 + y^2 = 2$ (centre origin, radius $\sqrt{2}$). Domain: $-\sqrt{2} \leq x \leq \sqrt{2}$.

(Squaring gives $y^2 = 2 - x^2$, i.e. $x^2 + y^2 = 2$, with $y \leq 0$.)

Read off the graph — LHS lies below the lower semicircle when:

$$\boxed{-\sqrt{2} \leq x \leq \sqrt{2}}$$

Key insight: $y = \pm\sqrt{r^2 - x^2}$ are the upper/lower halves of a circle of radius $r$ centred at the origin.

Question. $y = \dfrac{2x^2 - 7x + 3}{2 - x}$ has asymptotes $x = 2$ and $y = -2x + 3$, meeting at $(2, -1)$. Find the range of $k$ such that

$$\dfrac{2x^2 - 7x + 3}{2 - x} = k(2 - x) - 1$$

has no real roots.

Solution — interpret the RHS as a family of lines.

$y = k(2 - x) - 1$ is a family of lines passing through $(2, -1)$ with gradient $-k$. Note $(2, -1)$ is exactly the intersection of the curve's asymptotes — a perfect pivot.

Method:

1. Sketch the curve and mark the pivot $(2, -1)$.
2. Rotate the line about the pivot. For each rotation, count intersections with the curve.
3. No real roots $\Leftrightarrow$ the line misses the curve entirely.

From the graph, this happens when the line is steep enough (in the right direction) so that both branches of the curve sit on the same side of the line. Reading off the critical gradients:

$$\boxed{k \geq 2}$$

Why this trick works: when the pivot is the intersection of the curve's asymptotes, the line "lives between the branches" for a clean range of gradients.

Question. Curve $C$: $x = \cos^3 t$, $y = \sin^3 t$ for $0 \leq t \leq \tfrac{\pi}{2}$. Find $\dfrac{dy}{dx}$. Show that the tangent at $P\!\left(\cos^3 p,\, \sin^3 p\right)$ is

$$x \sin p + y \cos p = \sin p \cos p,$$

and hence find the area of the triangle formed with the coordinate axes.

Solution.

Parametric gradient:

$$\dfrac{dx}{dt} = -3\cos^2 t \sin t, \quad \dfrac{dy}{dt} = 3\sin^2 t \cos t$$ $$\dfrac{dy}{dx} = \dfrac{3\sin^2 t \cos t}{-3\cos^2 t \sin t} = -\tan t$$

Tangent at $P$: gradient $-\tan p$, passing through $(\cos^3 p, \sin^3 p)$:

$$y - \sin^3 p = -\tan p \,(x - \cos^3 p)$$

Multiply by $\cos p$ and simplify (using $\sin^2 p + \cos^2 p = 1$):

$$x \sin p + y \cos p = \sin p \cos p \quad \checkmark$$

Axial intercepts:

• Setting $y = 0$: $x \sin p = \sin p \cos p \Rightarrow x = \cos p$, so $Q = (\cos p, 0)$.
• Setting $x = 0$: $y \cos p = \sin p \cos p \Rightarrow y = \sin p$, so $R = (0, \sin p)$.

Area of $\triangle OQR$:

$$[\triangle OQR] = \tfrac{1}{2}|OQ||OR| = \tfrac{1}{2} \sin p \cos p = \tfrac{1}{4}\sin(2p)$$

Answer: area $= \tfrac{1}{4}\sin(2p)$ (maximum $\tfrac{1}{4}$ at $p = \tfrac{\pi}{4}$).