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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: F-LE.A.1a, F-LE.A.1c, F-LE.A.2, F-LE.A.3, F-IF.C.8b, F-LE.B.5 · Level: Grade 9 Algebra 1
Problem 1 Spot the Pattern, Table One
Ratios: 10 ÷ 5 = 2, 20 ÷ 10 = 2, 40 ÷ 20 = 2. The differences (5, 10, 20) are not equal.
The table is exponential with factor 2.
Problem 2 Spot the Pattern, Table Two
Differences: 12 − 7 = 5, 17 − 12 = 5, 22 − 17 = 5. The ratios are not equal (12/7 ≠ 17/12).
The table is linear with a constant difference of 5.
Problem 3 Name the Parts of a Growth Model
Compare with y = a · bx: a = 6 and b = 1.4. Since 1.4 > 1 the function grows. Rate: 1.4 − 1 = 0.40.
Initial value 6, factor 1.4, growth of 40% per step.
Problem 4 Name the Parts of a Decay Model
a = 250 and b = 0.8. Since 0 < 0.8 < 1 the function decays. Rate: 1 − 0.8 = 0.20.
Initial value 250, factor 0.8, decay of 20% per step.
Problem 5 Evaluate an Exponential Function
f(0) = 5(1) = 5. f(1) = 5(4) = 20. f(2) = 5(16) = 80. f(3) = 5(64) = 320.
5, 20, 80, 320
Problem 6 Write the Rule from a Table
Initial value a = 8. Factor: 12 ÷ 8 = 1.5, 18 ÷ 12 = 1.5, 27 ÷ 18 = 1.5.
Check x = 3: 8(1.5)3 = 8(3.375) = 27 ✓.
y = 8(1.5)x
Problem 7 Read an Exponential Graph
The y-intercept is 3, so a = 3. Factor: 6 ÷ 3 = 2, 12 ÷ 6 = 2, 24 ÷ 12 = 2, so b = 2.
Equation: y = 3(2)x. At x = 4: 3(16) = 48.
y = 3(2)x; y = 48 when x = 4
Problem 8 Table Without the Starting Value
Factor: 40 ÷ 80 = 0.5, 20 ÷ 40 = 0.5, 10 ÷ 20 = 0.5, so b = 0.5 (decay).
Step back one: at x = 0 the value is 80 ÷ 0.5 = 160.
Check x = 3: 160(0.5)3 = 20 ✓.
y = 160(0.5)x
Problem 9 A Bank Rate Model
b = 1.035 = 1 + 0.035, so the rate is 3.5%.
y = 1200(1.035)2 = 1200(1.071225) = 1285.47.
3.5% per year; $1,285.47 after 2 years
Problem 10 Line Versus Curve
Table: x = 4: f = 85, g = 48. x = 5: f = 95, g = 96.
For x = 0 to 4, g < f (3, 6, 12, 24, 48 against 45, 55, 65, 75, 85). At x = 5 the curve is just higher.
x = 5
Problem 11 Annual Compound Interest
A = 2400(1.03)6. Since 1.033 = 1.092727, 1.036 = 1.0927272 = 1.1940523…
A = 2400(1.1940523) = 2865.73.
$2,865.73
Problem 12 Depreciating Truck
Decay factor: 1 − 0.12 = 0.88, so V = 18500(0.88)t.
0.883 = 0.681472, and 18500 × 0.681472 = 12607.232.
V = 18,500(0.88)t; about $12,607.23
Problem 13 Bacteria in a Lab Dish
N = 300(2)h. Table: h = 5 gives 300 × 32 = 9,600; h = 6 gives 300 × 64 = 19,200.
9,600 < 10,000 < 19,200.
N = 300(2)h; after 6 hours
Problem 14 Interpret a Follower Model
1,800 is the starting number of followers (t = 0). 1.06 means the count grows by 6% each week.
F(10) = 1800(1.06)10 = 1800(1.790848…) = 3223.5…, which rounds to 3,224.
1,800 starting followers; 6% weekly growth; about 3,224 followers
Problem 15 Half-Life of a Medicine
Halve repeatedly: 640 → 320 (6 h) → 160 (12 h) → 80 (18 h) → 40 (24 h) → 20 (30 h) → 10 (36 h) → 5 (42 h).
After 18 hours: 80 mg. At 36 hours the amount is exactly 10 mg, not less than 10 mg, so the first time below 10 mg is 42 hours.
80 mg; 42 hours
Problem 16 Compounding Four Times a Year
Quarterly rate: 6% ÷ 4 = 1.5%. Number of quarters: 4 × 2 = 8.
A = 5000(1.015)8 = 5000(1.126493…) = 5632.46.
$5,632.46
Problem 17 Find the Growth Rate from Data
8,400 ÷ 8,000 = 1.05 and 8,820 ÷ 8,400 = 1.05, so the rate is 5%.
P = 8000(1.05)t. P(4) = 8000(1.21550625) = 9724.05.
5% per year; P = 8,000(1.05)t; about 9,724 people
Problem 18 Two Savings Plans
Plan A: A(n) = 100 + 25n. Plan B: B(n) = 100(1.2)n.
n = 3: A = 175, B = 172.80. n = 4: A = 200, B = 207.36.
At month 3 Plan A is ahead; at month 4 Plan B is ahead.
Month 4
Problem 19 Which Amount Is Larger?
A(n) = 25 + 15n; B(n) = 2n.
Day 6: A = 115, B = 64. Day 7: A = 130, B = 128. Day 8: A = 145, B = 256.
Day 7 is still lower for B (128 < 130).
Day 8
Problem 20 Phone Value Over Time
V = 900(0.85)t. Half the price is $450.
0.854 = 0.52200625 → $469.81. 0.855 = 0.4437… → $399.33.
At 4 years the value is above $450; at 5 years it is below.
V = 900(0.85)t; after 5 years