A.11(F)·Algebra II·Exponential & Logarithmic FunctionsReadiness

Exponential Growth Models

Use exponential functions to model growth over time.

Students write exponential growth models of the form $f(t)=a(1+r)^t$ , converting percent rates to growth factors and matching the time variable to the rate's period.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Algebra II key. Algebra II is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

Practice the released item Ask the tutor All Algebra II released practice50 minutes

Learn the concept

Identify the parts

An exponential growth model is $f(t)=a(1+r)^t$ . The initial value a is the amount when $t=0$ . The growth rate r is a decimal, and $1+r$ is the factor that multiplies the amount each period.

Match the period

If a rate is monthly, t counts months; if it is yearly, t counts years. “Months after January” means January is t = 0, so the initial amount is January's value.

Exponential versus linear

A linear model adds a constant amount: $f(t)=a+dt$ . An exponential model multiplies by a constant factor. Over long periods, steady percent growth eventually exceeds constant additive growth.

Worked example

A town has 12,000 residents and grows 2.5% per year. Write a model and estimate the population after 4 years.

The initial value is 12,000 and the growth factor is 1 + 0.025 = 1.025.
Write $P(t)=12000(1.025)^t$ , with t in years.
Evaluate: $P(4)=12000(1.025)^4\approx 12000(1.1038)\approx 13246$ .

About 13,246 residents

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Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Algebra 2 Solving Exponential Equations

UT IPSI · Kid2Kid / Texas Education Agency

Solve exponential equations that arise from growth models.

Watch on YouTube

Texas public school video

Algebra 1 Writing Exponential Functions from Real-world Situations

Five Minute Math · Aaron Daffern, Dallas ISD

Translate a starting amount and percent growth rate into an exponential function.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

1. A channel starts with 850 subscribers and grows 6% per month. Write a model.
Show answer and reasoning
$S(m)=850(1.06)^m$
The initial value is 850. Adding 0.06 to 1 gives the monthly growth factor, and m counts months.
2. After 10 years, which is larger: 1,000 growing 5% per year or 1,000 growing by 50 per year?
Show answer and reasoning
The 5% exponential model
$1000(1.05)^{10}\approx 1629$ , while $1000+50(10)=1500$ .

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 7 · 1 point

There were 417 cell phones sold at an electronics store in January. Since then, cell phone sales at this store have increased at a rate of 3.75% per month. At this rate of growth, which function can be used to determine the monthly cell phone sales $m$ months after January?

$p(m) = 417(0.0375)^m$
$p(m) = 417(1.0375)^m$
$p(m) = 417(0.9625)^{m + 1}$
$p(m) = 417(0.0375)^{m + 1}$

Show STAAR answer and explanation

$p(m) = 417(1.0375)^m$

Each month retains 100% of the previous sales and adds 3.75%, so the growth factor is $1+0.0375=1.0375$ . The initial value is 417 and m is the number of monthly growth periods.

$p(m) = 417(0.0375)^m$ : 0.0375 is the growth rate, not the factor that retains the previous month's sales and adds growth.

$p(m) = 417(0.9625)^{m + 1}$ : 0.9625 would represent a 3.75% decrease, and January is m = 0 rather than m = 1.

$p(m) = 417(0.0375)^{m + 1}$ : This uses the rate as the base and shifts the exponent, so it gives neither 417 initial sales nor monthly growth.

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Archived released STAAR Algebra II EOC · 2013 · Item 26 · 1 point

An antibiotic is introduced into a colony of 12,000 bacteria during a laboratory experiment. The function $n(m) = 12{,}000(0.851)^{\frac{m}{10}}$ can be used to model the number of bacteria in the colony after $m$ minutes. Which value is closest to the amount of time needed for the population of the colony to drop to 8,000 bacteria?

11 min
25 min
8 min
42 min

Show STAAR answer and explanation

25 min

Solve $8000=12000(0.851)^{m/10}$ . Divide: $(0.851)^{m/10}=\frac23$ . Take logarithms: $\frac{m}{10}=\frac{\log(2/3)}{\log 0.851}\approx2.51$ , so $m\approx25$ minutes.

11 min: $n(11)=12000(0.851)^{1.1}\approx10{,}050$ ; the population is still above 8,000.

8 min: $n(8)=12000(0.851)^{0.8}\approx10{,}550$ ; the population is still above 8,000.

42 min: $n(42)=12000(0.851)^{4.2}\approx6{,}090$ ; the population has already dropped well below 8,000.

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Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

Identify the initial value and percent growth rate in a context.
Convert a percent growth rate r to the growth factor $1+r$ .
Evaluate and compare exponential and linear models over time.

Instructional sequence

1
Engage
5 min
Compare adding 40 subscribers each month with increasing subscribers by 4% each month. Ask which grows faster over many months.
2
Direct instruction
10 min
Label a as the starting amount and $1+r$ as the repeated multiplier. Practice converting 3.75%, 12%, and 0.5% to growth factors.
3
Guided practice
13 min
Build a month-by-month table by repeated multiplication, then write the matching exponential rule.
4
Compare models
14 min
Pairs compare linear and exponential models for the same starting value and explain when each is appropriate.
5
Closure
8 min
Students label the parts of a model and complete the exit ticket.

Misconceptions to address

““A 3.75% growth rate uses a base of 3.75 or 0.0375.””

Teacher move: Convert 3.75% to 0.0375 and add it to 1. The growth factor is 1.0375 because each period keeps 100% and adds 3.75%.

““Percent growth adds the same amount each period.””

Teacher move: The percent acts on the current amount, which keeps changing. Use a table to show the increases getting larger.

Materials

•Population, savings, and sales scenarios
•Growth-factor conversion table
•Scientific calculator

Teaching tips

Have students write “starts at ___, multiplies by ___ each ___” before writing any rule.
Check the period: monthly rates need t measured in months.
Use a three-row table to verify a model before evaluating far into the future.

Exit ticket · 3–5 min

A video has 2,000 views on day 0, and views grow 12% per day. Write a model and estimate the views on day 3.

Show expected answer

$V(d)=2000(1.12)^d$ . $V(3)=2000(1.404928)\approx 2810$ views.

Sources & alignment

TEA 2013 Algebra II released test TEA 2013 Algebra II answer key and standards Texas Gateway, home of the UT IPSI Kid2Kid lessons

Alignment follows the standard TEA assigned to each 2013 released item. Video playback depends on YouTube access; written instruction remains available here.

Learn the concept

Identify the parts

Match the period

Exponential versus linear

Worked example

Watch the lesson

Algebra 2 Solving Exponential Equations

Algebra 1 Writing Exponential Functions from Real-world Situations

Try it yourself

1. A channel starts with 850 subscribers and grows 6% per month. Write a model.

2. After 10 years, which is larger: 1,000 growing 5% per year or 1,000 growing by 50 per year?

Released STAAR practice

Archived released STAAR Algebra II EOC · 2013 · Item 7 · 1 point

Archived released STAAR Algebra II EOC · 2013 · Item 26 · 1 point

Learning objectives

Instructional sequence

Engage

Direct instruction

Guided practice

Compare models

Closure

Misconceptions to address