CMClearMathAcademy

Exponential Functions and Sequences Checkpoint

A free Algebra I lesson from the “Exponential Functions and Sequences” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint covers Exponential Functions, Exponential Growth and Decay, Arithmetic Sequences, and Geometric Sequences. Complete 25 review-practice questions, then 25 separately written scored questions. A sequence's a_n means the term in position n; a_1 is its first term, and a_(n−1) is the preceding term. An explicit rule gives a term directly from its position. A recursive definition gives an initial term and a rule using the preceding term, normally for integers n at least 2. Sequence positions start at 1 unless a question states otherwise. A time model may instead start at t = 0. For y = a * b^x as a real-input exponential model here, a is positive, b is positive and different from 1; a negative ratio is allowed for a geometric sequence at integer positions. Use a calculator for numerical powers if needed; keep intermediate values exact and round only when requested. For typed answers, enter a number or an exact fraction such as −2/3 without units or currency signs; each prompt states its units and any rounding. Select a listed option for formulas or explanations. The score samples proposed outcomes A1.9 and A1.10 and does not establish external certification. The three written tasks are unscored: use their rubrics with a teacher, tutor, or your own written review. Allow additional time for those tasks.

What you'll learn

Why it matters: Repeated percentage change models savings, depreciation, and population estimates; equal additions and repeated products also describe rows, stages, and term-by-term patterns. The time unit and first counted position determine the model.

Worked example

Problem. A sequence starts at a_1 = 12 and each next term is 1.25 times the preceding term. Write explicit and recursive rules and find term 4.

  1. The first position is 1, so term n uses n − 1 repeated products: a_n = 12 * (1.25)^(n − 1).
  2. A recursive definition also needs the starting term: a_1 = 12. Then a_n = 1.25 * a_(n−1) for integers n at least 2, where a_(n−1) is the preceding term.
  3. The terms are 12, 15, 18.75, 23.4375. The explicit calculation 12 * (1.25)^3 agrees.

Answer: a_n = 12 * (1.25)^(n − 1); a_1 = 12, a_n = 1.25 * a_(n−1) for integers n at least 2; a_4 = 23.4375

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. For f(x) = 36 * (3/2)^x, find f(−2). Enter a number.

Show solution
  1. A negative power uses the reciprocal: (3/2)^(−2) = (2/3)^2 = 4/9.
  2. The output is 36 * 4/9 = 16; a negative input does not make the output negative.

Answer: 16

2. A table lists x = 0, 2, 4 and y = 7, 63, 567. Assume y = a * b^x with a and b positive. Which rule fits all three entries?

Choices: y = 7 * 9^x · y = 7 * 3^x · y = 7 + 28x · y = 3 * 7^x

Show solution
  1. At x = 0, y = a = 7. The ratio 63/7 = 9 spans two x-units, so b^2 = 9.
  2. The positive value of b is 3. Checking x = 4 gives 7 * 3^4 = 567.

Answer: y = 7 * 3^x

3. The model M(t) = 84 * (0.6)^t gives the mass in grams after t completed days, starting at t = 0. Which interpretation is correct?

Choices: It starts at 84 g and loses 60% of its current mass each day. · It starts at 50.4 g and loses 40% of its current mass each day. · It starts at 84 g and loses 40% of its current mass each day. · It starts at 84 g and loses 0.6 g each day.

Show solution
  1. At t = 0 the power equals 1, so the starting mass is 84 g.
  2. Each day 60% remains; the lost portion is 100% − 60% = 40% of the current mass.

Answer: It starts at 84 g and loses 40% of its current mass each day.

4. Each list gives outputs at x = 0, 1, 2. Which list is consistent with a nonconstant rule y = a * b^x with positive a and b?

Choices: 6, 18, 30 · 6, 18, 54 · 6, 12, 24.5 · 6, 6, 6

Show solution
  1. Equal input steps require the same ratio of consecutive outputs for this model.
  2. Only 18/6 = 54/18 = 3 is constant and different from 1. The repeated 6s describe a constant function.

Answer: 6, 18, 54

5. The plot shows part of y = 9 * (1/3)^x. For the same rule at all real inputs x, which description is correct?

Choices: It stays positive, increases, and approaches 0 as x decreases. · It becomes negative once x is larger than 3. · It stays positive, decreases, and approaches 0 as x increases. · It decreases by exactly 6 for every one-unit step in x.

Show solution
  1. The values at x = 0, 1, 2, 3 are 9, 3, 1, 1/3. The decreases are not equal amounts.
  2. Every real power of the positive base 1/3 is positive. Repeatedly taking a third makes the values approach zero without reaching it.

Answer: It stays positive, decreases, and approaches 0 as x increases.

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Compare adding with repeated percentage growth

Two modeled balances begin at $160. At each completed year, plan A adds $40 and plan B increases its current balance by 25%. No other changes occur. Write both rules for nonnegative integer years t starting at 0, make a table for years 0 through 4, and plot those points with labeled axes and units. Explain when the balances match in that interval, which plan is larger afterward in that interval, and why the first year's change alone cannot distinguish the models.

Review rubric and solution
  • Writes A(t) = 160 + 40t and B(t) = 160 * (1.25)^t with integer years starting at 0.
  • Lists A: 160, 200, 240, 280, 320 and B: 160, 200, 250, 312.5, 390.625; rounds money only if stated.
  • Plots separate points for completed years 0–4, labels years and dollars, and distinguishes the two series.
  • Explains equality at years 0 and 1, then B is larger at years 2–4 because its increases grow while A's stay $40.
  • States that agreement on a single change does not establish how later values change.
  1. The formulas are A(t) = 160 + 40t and B(t) = 160 * (1.25)^t for whole years t starting at 0.
  2. Plot (0,160), (1,200), (2,240), (3,280), (4,320) for A and (0,160), (1,200), (2,250), (3,312.5), (4,390.625) for B. Dollars may be displayed as 312.50 and 390.63 if rounded to cents, but retain exact values for calculation.
  3. They match initially and after one year. On the remaining years in this table B is higher; both first increases are $40, but B's later increases are $50, $62.50, and $78.125.

Give complete recursive definitions and connect them to graphs

Sequence U begins 31, 26, 21, ... with the same difference at each step. Sequence V begins 20, 10, 5, ... with the same ratio at each step. For each sequence, define the notation, write an explicit rule and a recursive definition including the initial term and valid indices, and find term 6. Plot the first four terms of each on separate labeled graphs. Explain why a line or curve connecting the points would include inputs outside these sequences.

Review rubric and solution
  • Defines U_n and V_n as terms at positive integer position n and U_(n−1), V_(n−1) as preceding terms.
  • Writes U_n = 31 − 5(n − 1); U_1 = 31 and U_n = U_(n−1) − 5 for integer n at least 2; U_6 = 6.
  • Writes V_n = 20 * (1/2)^(n − 1); V_1 = 20 and V_n = (1/2) * V_(n−1) for integer n at least 2; V_6 = 5/8.
  • Plots U at (1,31), (2,26), (3,21), (4,16) and V at (1,20), (2,10), (3,5), (4,2.5), labeling term index and term value.
  • Explains that only positive integer indices belong to either sequence, although an extended function could also be defined.
  1. Each subscript n denotes a positive integer position, and n − 1 denotes the preceding position. U adds −5 each time: U_n = 31 − 5(n − 1), or recursively U_1 = 31 and U_n = U_(n−1) − 5 for n at least 2.
  2. V repeatedly takes half the preceding term: V_n = 20 * (1/2)^(n − 1), or recursively V_1 = 20 and V_n = (1/2) * V_(n−1) for n at least 2.
  3. The sixth terms are 6 and 20/32 = 5/8. Both graphs contain separate points; a noninteger position such as 1.5 is not an input of either sequence.

Audit a decay model and its time units

A filter leaves 70% of the dissolved material after each completed cleaning cycle. It starts with 500 mg before any cycles. A student proposes M(t) = 500 * (0.30)^t and says three cycles leave 13.5 mg. Correct the model and the prediction, and explain the error. Then number the recorded amounts as a sequence whose term 1 is the amount before any cycles. Give explicit and recursive rules for that sequence, with notation and valid indices. State one assumption needed to use the model for future cycles.

Review rubric and solution
  • Explains that 70% remains and 30% is removed, so M(t) = 500 * (0.70)^t for nonnegative integer cycles t.
  • Finds M(3) = 171.5 mg; identifies 13.5 mg as the result of repeatedly retaining the removed percentage.
  • Defines a_n as the recorded amount at position n; writes a_n = 500 * (0.70)^(n − 1) for positive integer n.
  • Gives a_1 = 500 and a_n = 0.70 * a_(n−1) for integer n at least 2; identifies three cycles as term 4.
  • States a relevant assumption, such as the filter retaining the same percentage each cycle and no material being added.
  1. The correct retention ratio is 0.70. Three cycles leave 500 * (0.70)^3 = 171.5 mg. The student's calculation 500 * (0.30)^3 = 13.5 mg is arithmetically correct for the wrong retention assumption.
  2. With term 1 recorded before any cleaning, a_n = 500 * (0.70)^(n − 1). Recursively, a_1 = 500 and a_n = 0.70 * a_(n−1) for integer n at least 2. Term 4 is the amount after three cycles.
  3. The prediction assumes a fixed retention percentage each cycle with no new material added; changing conditions can make later predictions inaccurate.

Practice this interactively with instant feedback and an AI tutor.

Practice Exponential Functions and Sequences Checkpoint Take the free placement check

More Algebra I lessons