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2.5 · Vertical Translation Test
How do you test if a table with vertical translation is exponential?
What do you add or subtract?
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2.5 · Writing the Equation
After adding/subtracting k to reveal proportional growth, how do you write f(x)?
f(x) − k = a·bx
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2.5 · Apply: Find the Equation
f(x) − 1 gives values 6, 12, 24, 48, 96. Write the equation for f(x).
Find the ratio, initial value, then add back k
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2.5 · Apply: Decay with Translation
h(x) + 1 gives values 64, 32, 16, 8, 4 for x = 0,1,2,3,4. Write h(x).
64 · (1/2)x = 64 · 2−x
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2.5 · Modeling from Two Points
g(x) = abx, g(3) = 21.54 and g(8) = 3.62. Write the two equations needed to find a and b.
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2.5 · Initial Value
For f(t) = a(b)t, how do you find the initial value a from a table?
Which input value makes this easiest?
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2.5 · Two Input-Output Pairs
F(2) = 47 and F(20) = 2602 for F(t) = a(b)t. Write the two equations.
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2.5 · Natural Base e
What is the natural base e and approximately what value does it equal?
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✓ Writing the Equation
f(x) = a·bx + k
Since f(x) − k = a·bx, add k to both sides. The initial value a = adjusted f(0), and b = common ratio of adjusted values.
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✓ Vertical Translation Test
Add or subtract a constant k so the adjusted outputs are proportional over equal-length intervals
If f(x) − k shows a constant multiplier (ratio), then f is exponential with vertical shift k.
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✓ Decay with Translation
h(x) = 64(2−x) − 1
h(x) + 1 = 64(1/2)x = 64(2)−x. Subtract 1: h(x) = 64(2)−x − 1.
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✓ Apply: Find the Equation
f(x) = 6(2)x + 1
Ratio = 2, initial adjusted value = 6, so f(x) − 1 = 6(2)x. Adding 1: f(x) = 6(2)x + 1.
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✓ Initial Value
Use x = 0: a = f(0) = a · b0 = a · 1 = a
When x = 0, b0 = 1, so f(0) = a. If x = 0 isn't in the table, use two points to solve a system.
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✓ Modeling from Two Points
ab3 = 21.54 and ab8 = 3.62
Divide: b5 = 3.62/21.54. Solve for b, then substitute back to find a. Use calculator ExpReg to verify.
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✓ Natural Base e
e ≈ 2.718 — the natural number used as the base in many real-world exponential models
Example: B(t) = 17e0.31t. Common in biology, finance, and physics models.
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✓ Two Input-Output Pairs
a(b)2 = 47 and a(b)20 = 2602
Substitute t = 2 and t = 20 into F(t) = a(b)t. Divide equations to eliminate a, then solve for b.
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2.5 · Natural Base — Initial Value
B(t) = 17e0.31t. If A(t) = a·bt is equivalent, find a.
Evaluate both at t = 0
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2.5 · Percent Growth → Base
A population grows by 13% per year. What is the base b in the exponential model?
Growth rate r% → b = ?
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2.5 · Percent Decay → Base
A substance decays by 20% per year. What is the base b in the exponential model?
Decay rate r% → b = ?
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2.5 · Changing Time Units
D(t) = 50·(1.13)t where t is in years. Rewrite D so t is in months.
1 year = 12 months
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✓ Percent Growth → Base
b = 1 + r/100 = 1.13
13% growth → b = 1 + 0.13 = 1.13. The model is D(t) = 50(1.13)t. For growth, b > 1.
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✓ Natural Base — Initial Value
a = 17
A(0) = B(0) → a·b0 = 17e0 → a·1 = 17·1 → a = 17. Evaluate both at t = 0.
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✓ Changing Time Units
D(t) = 50·(1.13)t/12
Replace t (years) with t/12 (months). The base stays 1.13 but the exponent becomes t/12 to account for monthly measurement.
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✓ Percent Decay → Base
b = 1 − r/100 = 0.80
20% decay → b = 1 − 0.20 = 0.80. For decay, 0 < b < 1. The model decreases toward 0 over time.