The Key Insight: Everything Revolves Around the Asymptotes#
The hyperbola is the function students find most confusing — but it follows simple rules. Unlike the parabola (which has a turning point) or the exponential (which has an asymptote on one side), the hyperbola has two asymptotes that completely determine its shape and position.
Once you find the asymptotes, the rest falls into place.
1. The Standard Form and Its Parameters#
$$ \boxed{y = \frac{a}{x - p} + q} $$| Parameter | What it controls | How to find it |
|---|---|---|
| $p$ | Horizontal shift → Vertical asymptote at $x = p$ | The “forbidden” $x$-value (denominator = 0) |
| $q$ | Vertical shift → Horizontal asymptote at $y = q$ | The value $y$ approaches as $x \to \pm\infty$ |
| $a > 0$ | Curves in Q1 and Q3 relative to asymptote centre | Top-right and bottom-left branches |
| $a < 0$ | Curves in Q2 and Q4 relative to asymptote centre | Top-left and bottom-right branches |
| $\|a\|$ large | Curves are further from asymptotes (wider) | The branches spread out more |
| $\|a\|$ small | Curves are closer to asymptotes (tighter) | The branches hug the asymptotes |
The point $(p;\; q)$ is the centre of the asymptotes — the “heart” of the hyperbola. Every key feature is measured relative to this point.
2. Why Do the Asymptotes Exist?#
The Vertical Asymptote ($x = p$)#
When $x = p$, the denominator $(x - p) = 0$. Division by zero is undefined, so the function doesn’t exist at this point. As $x$ gets closer and closer to $p$, the fraction $\frac{a}{x-p}$ becomes enormous (positive or negative), and the graph “blows up” — shooting toward $+\infty$ or $-\infty$.
The Horizontal Asymptote ($y = q$)#
As $x \to +\infty$ or $x \to -\infty$, the denominator $(x - p)$ becomes very large, so $\frac{a}{x-p} \to 0$. This means $y \to 0 + q = q$. The graph gets infinitely close to the line $y = q$ but never reaches it (because $\frac{a}{x-p}$ is never exactly zero).
Think of it this way: The asymptotes are invisible boundaries that the curve chases forever but never catches.
3. Sketching — The 6-Step Method#
Worked Example 1: Sketch $y = \frac{3}{x - 2} + 1$#
Step 1 — Identify parameters: $a = 3$, $p = 2$, $q = 1$
Step 2 — Draw asymptotes (dashed lines):
- Vertical: $x = 2$
- Horizontal: $y = 1$
- Centre: $(2;\; 1)$
Step 3 — Determine quadrants: $a = 3 > 0$ → curves in Q1 and Q3 relative to $(2; 1)$ (top-right and bottom-left).
Step 4 — y-intercept (let $x = 0$):
$y = \frac{3}{0 - 2} + 1 = -\frac{3}{2} + 1 = -\frac{1}{2}$
y-intercept: $(0;\; -\frac{1}{2})$ — this is in the bottom-left branch ✓
Step 5 — x-intercept (let $y = 0$):
$0 = \frac{3}{x - 2} + 1$
$\frac{3}{x - 2} = -1$
$3 = -(x - 2) = -x + 2$
$x = -1$
x-intercept: $(-1;\; 0)$ — also in the bottom-left branch ✓
Step 6 — Domain and Range:
Domain: $x \in \mathbb{R},\; x \neq 2$
Range: $y \in \mathbb{R},\; y \neq 1$
Sketch: Draw the asymptotes as dashed lines, mark the intercepts, then draw smooth curves in the correct quadrants, approaching but never touching the asymptotes.
Worked Example 2: Sketch $y = \frac{-4}{x + 1} - 2$#
Parameters: $a = -4$, $p = -1$, $q = -2$
Asymptotes: $x = -1$ and $y = -2$. Centre: $(-1;\; -2)$.
Quadrants: $a < 0$ → Q2 and Q4 relative to centre (top-left and bottom-right).
y-intercept (let $x = 0$):
$y = \frac{-4}{0 + 1} - 2 = -4 - 2 = -6$
y-intercept: $(0;\; -6)$ — bottom-right of centre ✓
x-intercept (let $y = 0$):
$0 = \frac{-4}{x + 1} - 2$
$\frac{-4}{x + 1} = 2$
$-4 = 2(x + 1) = 2x + 2$
$x = -3$
x-intercept: $(-3;\; 0)$ — top-left of centre ✓
Domain: $x \in \mathbb{R},\; x \neq -1$
Range: $y \in \mathbb{R},\; y \neq -2$
Worked Example 3: Sketch $y = \frac{2}{x} + 3$#
Parameters: $a = 2$, $p = 0$, $q = 3$
This is a hyperbola centred at the y-axis.
Asymptotes: $x = 0$ (the y-axis itself!) and $y = 3$
y-intercept: None! The vertical asymptote IS the y-axis, so the graph never crosses it.
x-intercept (let $y = 0$):
$0 = \frac{2}{x} + 3 \Rightarrow \frac{2}{x} = -3 \Rightarrow x = -\frac{2}{3}$
x-intercept: $(-\frac{2}{3};\; 0)$
Domain: $x \in \mathbb{R},\; x \neq 0$
Range: $y \in \mathbb{R},\; y \neq 3$
4. Finding the Equation from a Graph#
This is a very common exam question. You need to identify $a$, $p$, and $q$.
The Method#
- Read $p$ and $q$ from the asymptotes (dashed lines on the graph)
- Substitute one known point to find $a$
Worked Example 4#
Asymptotes: $x = -1$ and $y = 3$. The graph passes through $(1;\; 5)$.
Step 1: $p = -1$, $q = 3$
$$y = \frac{a}{x - (-1)} + 3 = \frac{a}{x + 1} + 3$$Step 2: Substitute $(1;\; 5)$:
$5 = \frac{a}{1 + 1} + 3 = \frac{a}{2} + 3$
$\frac{a}{2} = 2 \Rightarrow a = 4$
$$\boxed{y = \frac{4}{x + 1} + 3}$$Worked Example 5#
Asymptotes: $x = 3$ and $y = -1$. The graph passes through $(5;\; 1)$.
$p = 3$, $q = -1$
$y = \frac{a}{x - 3} - 1$
Substitute $(5;\; 1)$: $1 = \frac{a}{5 - 3} - 1 = \frac{a}{2} - 1$
$\frac{a}{2} = 2 \Rightarrow a = 4$
$$y = \frac{4}{x - 3} - 1$$Worked Example 6 — Using x-intercept and y-intercept#
The graph has asymptotes at $x = 2$ and $y = -3$. The y-intercept is $(0;\; -4)$. Find the equation.
$p = 2$, $q = -3$
$y = \frac{a}{x - 2} - 3$
Substitute $(0;\; -4)$: $-4 = \frac{a}{0 - 2} - 3 = \frac{a}{-2} - 3$
$-1 = -\frac{a}{2} \Rightarrow a = 2$
$$y = \frac{2}{x - 2} - 3$$5. Lines of Symmetry#
Every hyperbola has two lines of symmetry that pass through the centre $(p;\; q)$ at $45°$ angles:
$$y = (x - p) + q \qquad \text{(gradient } +1\text{)}$$$$y = -(x - p) + q \qquad \text{(gradient } -1\text{)}$$These are the lines along which you could fold the graph and the two branches would lie on top of each other.
Worked Example 7#
Find the lines of symmetry of $y = \frac{3}{x - 2} + 1$.
Centre: $(2;\; 1)$
Line 1: $y = (x - 2) + 1 = x - 1$
Line 2: $y = -(x - 2) + 1 = -x + 3$
Worked Example 8 — Exam-Style#
The line $y = x + k$ is an axis of symmetry of $y = \frac{a}{x - 1} + 2$. Find $k$.
Centre: $(1;\; 2)$. The axis of symmetry with gradient $+1$ is:
$y = (x - 1) + 2 = x + 1$
Comparing with $y = x + k$: $k = 1$.
6. Determining the Quadrants — The $a$-Sign Rule#
The asymptotes divide the plane into four regions (like quadrants around the centre). The sign of $a$ determines which two regions contain the branches:
| $a > 0$ | $a < 0$ |
|---|---|
| Top-right and bottom-left | Top-left and bottom-right |
| (Q1 and Q3 relative to centre) | (Q2 and Q4 relative to centre) |
The “Test Point” Method#
If you’re unsure, substitute an $x$-value far to the right of the vertical asymptote (e.g., $x = 100$). Calculate $y$:
- If $y > q$: the right branch is above the horizontal asymptote (Q1 relative to centre)
- If $y < q$: the right branch is below the horizontal asymptote (Q4 relative to centre)
The other branch is always in the diagonally opposite quadrant.
7. Intersection with Other Graphs#
Exam questions often ask where a hyperbola intersects a straight line or another function.
Worked Example 9#
Find the intersection of $y = \frac{4}{x - 1} + 2$ and $y = x + 1$.
Set equal: $\frac{4}{x - 1} + 2 = x + 1$
$\frac{4}{x - 1} = x - 1$
$4 = (x - 1)^2$
$x - 1 = \pm 2$
$x = 3$ or $x = -1$
For $x = 3$: $y = 3 + 1 = 4$ → point $(3;\; 4)$
For $x = -1$: $y = -1 + 1 = 0$ → point $(-1;\; 0)$
Intersection points: $(3;\; 4)$ and $(-1;\; 0)$
8. Does the Hyperbola Always Have Intercepts?#
x-intercept#
Set $y = 0$: $0 = \frac{a}{x - p} + q \Rightarrow x = p - \frac{a}{q}$
This gives a valid x-intercept only if $q \neq 0$. If $q = 0$, the horizontal asymptote IS the x-axis, so the graph never crosses it — no x-intercept.
y-intercept#
Set $x = 0$: $y = \frac{a}{0 - p} + q = -\frac{a}{p} + q$
This gives a valid y-intercept only if $p \neq 0$. If $p = 0$, the vertical asymptote IS the y-axis — no y-intercept.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Not drawing asymptotes as dashed lines | Exams deduct marks for missing or solid asymptotes | Always draw dashed lines and label with equations ($x = p$ and $y = q$) |
| The sign of $p$ | $\frac{a}{x - 3}$ → $p = +3$. But $\frac{a}{x + 2} = \frac{a}{x - (-2)}$ → $p = -2$ | Same sign trap as the parabola: the sign you see is the opposite |
| Drawing curves that touch asymptotes | The graph gets infinitely close but never reaches the asymptote | Leave a visible gap near asymptotes |
| Wrong quadrants for the branches | The quadrants are relative to the asymptote centre $(p; q)$, not the origin | Plot the centre first, then decide which quadrants based on the sign of $a$ |
| Forgetting domain and range | Domain must exclude $x = p$; range must exclude $y = q$ | Always state: “$x \in \mathbb{R},\; x \neq p$” and “$y \in \mathbb{R},\; y \neq q$” |
| Assuming there’s always a y-intercept | If $p = 0$, the y-axis IS the vertical asymptote — no y-intercept exists | Check whether $p = 0$ before computing the y-intercept |
| Lines of symmetry through the origin | The symmetry lines pass through $(p; q)$, NOT the origin | Use $y = \pm(x - p) + q$ |
💡 Pro Tips for Exams#
1. The Sketching Checklist#
For every hyperbola sketch, mark these 6 things:
- ✅ Vertical asymptote (dashed, with equation)
- ✅ Horizontal asymptote (dashed, with equation)
- ✅ y-intercept (if it exists)
- ✅ x-intercept (if it exists)
- ✅ Domain and range
- ✅ Two smooth curves in the correct quadrants
2. Quick Domain/Range Template#
For $y = \frac{a}{x - p} + q$:
Domain: $x \in \mathbb{R},\; x \neq p$
Range: $y \in \mathbb{R},\; y \neq q$
This is always true — no calculation needed.
3. Reading the Graph Backwards#
Given a graph, find the equation:
- Read asymptotes → gives $p$ and $q$
- Check which quadrants the branches are in → determines sign of $a$
- Read one clear point → substitute to find $|a|$
🔗 Related Grade 11 topics:
- The Parabola & The Exponential — the other two functions you must sketch
- Quadratic Equations — solving equations with fractions often produces hyperbola-related expressions
📌 Grade 10 foundation: Sketching Graphs — the basic hyperbola $y = \frac{a}{x} + q$
📌 Grade 12 extension: Hyperbola Function & Inverse — all parameters and finding inverses.
⏮️ Parabola | 🏠 Back to Functions | ⏭️ Exponential