Through (0, 1), hugging the x-axis: Every exponential y = aˣ (with a > 0) does three things.
The three features
- goes through (0, 1) — because a⁰ = 1
- stays above the x-axis (always positive)
- flattens toward y = 0 but never touches it — that line is its horizontal asymptote
IB-style question — key points & range of 2ˣ
State the y-intercept, the horizontal asymptote, and the range of y = 2ˣ.
Step by step
- y-intercept: x = 0.
- As x → −∞, 2ˣ → 0 (but never reaches it).
- 2ˣ stays positive, so every output sits above the asymptote.
Final answer
y-intercept (0, 1); asymptote y = 0; range y > 0.
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Always positive: aˣ is never zero or negative, so the range is y > 0 and there's no x-intercept.
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The base sets the direction: If the base a > 1, the graph grows (rises steeply to the right). If 0 < a < 1, it decays (falls toward the x-axis). Both still pass through (0, 1).
Growth (a > 1)
- e.g. y = 3ˣ
- rises to the right
- → ∞ as x → ∞
Decay (0 < a < 1)
- e.g. y = (½)ˣ
- falls to the right
- → 0 as x → ∞
Decay = reflected growth: (½)ˣ = 2⁻ˣ, so a decay curve is just a growth curve reflected in the y-axis.
Growth (base > 1) rises faster and faster; decay (0 < base < 1) falls fast then flattens. Each hugs the x-axis on one side — that horizontal asymptote is y = 0.
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Adding c lifts the asymptote: y = k·aˣ + c stretches by k and shifts up by c. The horizontal asymptote moves to y = c, and the y-intercept becomes k + c (at x = 0).
IB-style question — shifted exponential
State the horizontal asymptote and y-intercept of y = 2ˣ + 3.
Step by step
- Asymptote: 2ˣ → 0, leaving the + 3.
- y-intercept: x = 0.
Final answer
Horizontal asymptote y = 3; y-intercept (0, 4).
The asymptote isn't always y = 0: A + c shifts the whole curve up, so the curve now levels off at y = c, not y = 0.
Read the model: initial value × (factor)ᵗ: A model A(t) = A₀·bᵗ has initial value A₀ (at t = 0) and per-period factor b (b > 1 growth, b < 1 decay).
| From the question | Factor b |
|---|---|
| grows r% | b = 1 + r/100 |
| loses r% | b = 1 − r/100 |
Using the model
- To find a value, substitute t.
- To find a time, solve for t (logs or GDC).
IB-style question — a growth model
A population is P = 200·(1.05)ᵗ (t in years). Find (a) the initial population and (b) the population after 10 years.
Step by step
- (a) t = 0.
- (b) t = 10.
Final answer
(a) 200; (b) about 326.
IB-style question — a decay model
A laptop bought for $1200 loses 18% of its value each year. (a) Write a model V for its value after t years. (b) Find its value after 4 years.
Step by step
- Losing 18% means you keep 82% each year, so the decay factor is b = 1 − 0.18.
- (a) Start value 1200, factor 0.82.
- (b) Substitute t = 4.
Final answer
(a) V = 1200·(0.82)ᵗ; (b) about $543.
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