Through (0, 1), hugging the x-axis: Every exponential y = aˣ (a > 0) passes through (0, 1) (because a⁰ = 1), stays positive, and has the horizontal asymptote y = 0 — it gets very close to the x-axis but never touches it.
IB-style question — key points of 2ˣ
State the y-intercept and horizontal asymptote of y = 2ˣ.
Step by step
- y-intercept: x = 0.
- As x → −∞, 2ˣ → 0.
Final answer
y-intercept (0, 1); horizontal asymptote y = 0.
Always positive: aˣ is never zero or negative, so the range is y > 0 and there's no x-intercept.
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.
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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). 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.