Key Idea: Two mirror-image curves that model growth and decay — and you'll be asked to sketch them and read off their key features (intercept, asymptote, domain) on Paper 1, then evaluate a model with the GDC on Paper 2.
📈 The graph of y = aˣ
| Feature of y = aˣ (a > 0) | Value |
|---|---|
| Passes through | (0, 1) — because a⁰ = 1 |
| Range | y > 0 (always positive, no x-intercept) |
| Horizontal asymptote | y = 0 — hugs the x-axis, never touches |
| Shape | a > 1 grows ↗; 0 < a < 1 decays ↘ |
For y = k·aˣ + c, the curve levels off at y = c, not y = 0, and the y-intercept becomes k + c. And (½)ˣ = 2⁻ˣ, so a decay curve is just a growth curve reflected in the y-axis.
📉 The graph of y = logₐx (its inverse)
| y = aˣ | y = logₐx (the inverse) |
|---|---|
| through (0, 1) | through (1, 0) |
| asymptote y = 0 (horizontal) | asymptote x = 0 (vertical) |
| domain all x, range y > 0 | domain x > 0, range all y |
| — | no y-intercept (can't log 0 or a negative) |
y = logₐx is the inverse of y = aˣ, so each is the other reflected in the line y = x — their domain and range swap. For y = logₐ(x − h) + k the vertical asymptote moves to x = h and the domain becomes x > h (the inside must stay positive).
✏️ IB-style worked examples
IB-style question — y-intercept and asymptote of an exponential
State the y-intercept and horizontal asymptote of y = 4ˣ.
Step by step:
y-intercept: put x = 0.
As x → −∞ the curve hugs the x-axis.
y-intercept (0, 1); horizontal asymptote y = 0.
IB-style question — transformed exponential
State the horizontal asymptote and y-intercept of y = 3ˣ + 2.
Step by step:
As x → −∞, 3ˣ → 0, leaving the + 2.
y-intercept: put x = 0.
Horizontal asymptote y = 2; y-intercept (0, 3).
IB-style question — read and use an exponential model
A colony of bacteria is modelled by N = 150·(1.08)ᵗ (t in hours). Find (a) the initial number and (b) the number after 12 hours.
Step by step:
(a) Initial value is at t = 0.
(b) Substitute t = 12 (GDC).
(a) 150; (b) about 378.
IB-style question — domain and asymptote of a log
State the vertical asymptote and domain of y = log x.
Step by step:
Only positive inputs are allowed.
As x → 0⁺, log x → −∞.
Vertical asymptote x = 0; domain x > 0.
IB-style question — transformed logarithm
State the vertical asymptote and domain of y = log(x − 3).
Step by step:
The inside must stay positive.
Asymptote where the inside is 0.
Vertical asymptote x = 3; domain x > 3.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: For a plain y = aˣ the asymptote is y = 0, but + c shifts it to y = c — read the constant. For logs the asymptote is the vertical line where the inside is 0 (x = h), and you can't take the log of 0 or a negative, so the domain is x > h.
Tap each card to reveal the answer.
y-intercept of y = 5ˣ (0, 1) — every aˣ passes through (0, 1) since a⁰ = 1.
Asymptote of y = 2ˣ − 4 y = −4 — the − 4 shifts the whole curve down, so it levels off at y = −4.
Does y = 3ˣ grow or decay? Grows — the base 3 > 1. (0 < a < 1 would decay.)
Domain of y = log x x > 0 — you can only log positive numbers; there's no y-intercept.
y = logₐx is the inverse of which graph? y = aˣ — reflect aˣ in the line y = x to get logₐx.
Vertical asymptote of y = log(x + 1) x = −1 — the inside is 0 when x = −1, and the domain is x > −1.
Exam Tips
- y = aˣ: through (0, 1), range y > 0, asymptote y = 0 — a > 1 grows, 0 < a < 1 decays.
- y = logₐx: through (1, 0), asymptote x = 0, domain x > 0 — the inverse of aˣ (reflect in y = x).
- A + c on an exponential lifts the horizontal asymptote to y = c; logₐ(x − h) moves the vertical asymptote to x = h.
- For models A₀·bᵗ: A₀ is the value at t = 0; substitute t to find a value, solve for t to find a time.
- Paper 2: just type the model into the GDC and read off the value — no algebra required.