Algebra 2 Exponential Growth and Decay Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Exponential Growth and Decay – Maths PDF

Cover of the book Algebra 2 Math Worksheets with Answers (PDF)

Download as a PDFThe Algebra 2 worksheets book with answersPrintable PDFSee the book ›

Sommaire

This algebra 2 exponential functions answers for Algebra 2 covers exponential growth and decay (Common Core F-LE.A.2, F-LE.B.5, A-SSE.B.3c, F-IF.C.8b), with a free PDF to print.

Every worksheet problem on exponential growth and decay is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where using the rate instead of the growth factor can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: F-LE.A.2, F-LE.B.5, A-SSE.B.3c, F-IF.C.8b

Algebra 2 exponential functions answers: exercises 1 to 10

Exercise 1 – Growth or decay

Growth if $b > 1$, decay if $0 < b < 1$.

Growth (b = 1.4); decay (b = 0.8); decay (b = 1/3)

Exercise 2 – Evaluate

$f(0) = 3$, $f(1) = 6$, $f(2) = 12$, $f(3) = 24$, $f(4) = 48$.

3, 6, 12, 24, 48

Exercise 3 – Write a function from a table

Constant ratio 2.

The initial value is $f(0) = 5$. Consecutive outputs have ratio 2.

$f(x) = 5 \cdot 2^{x}$

Exercise 4 – Two points

Divide: $\frac{ab^{3}}{ab} = \frac{54}{6}$, so $b^{2} = 9$ and $b = 3$.

Then $a(3) = 6$, so $a = 2$.

$f(x) = 2 \cdot 3^{x}$

Exercise 5 – Population growth

$P(t) = 12{,}000(1.035)^{t}$. $P(5) = 12{,}000(1.035)^{5} \approx 14,252.24$.

About 14,252 people

Exercise 6 – Depreciation

$V(t) = 24{,}000(0.85)^{t}$. $V(4) = 24{,}000(0.85)^{4} \approx 12,528.15$.

About $12,528.15

Exercise 7 – Half-life

$A(t) = 80\left(\frac{1}{2}\right)^{t/3}$. At $t = 9$: $\left(\frac{1}{2}\right)^{3} = \frac{1}{8}$, so $80 \cdot \frac{1}{8} = 10$.

10 mg

Exercise 8 – Doubling time

$B(t) = 50 \cdot 2^{t/20}$. At 60 minutes, $2^{3} = 8$, so $50 \cdot 8 = 400$.

400 bacteria

Exercise 9 – Compound interest

Compound interest repeats a multiplication.

$A = 2000\left(1 + \frac{0.04}{4}\right)^{4 \cdot 5} = 2000(1.01)^{20} \approx 2,440.38$.

About $2,440.38

Exercise 10 – Compounding frequency

Annual: $5000(1.03)^{10} \approx 6,719.58$.

Monthly: $5000\left(1 + \frac{0.03}{12}\right)^{120} \approx 6,746.77$.

Monthly compounding earns about $27.19 more.

Annual $6,719.58; monthly $6,746.77

Exercises 11 and up

Exercise 11 – Continuous compounding

$A = 1500e^{0.03 \cdot 8} = 1500e^{0.24} \approx 1,906.87$.

About $1,906.87

Exercise 12 – Rewrite exponents

$2^{3x} = (2^{3})^{x} = 8^{x}$. $\left(\frac{1}{2}\right)^{x} = (2^{-1})^{x} = 2^{-x}$.

$8^{x}$ and $2^{-x}$

Exercise 13 – Effective annual rate

$1.02^{12t} = (1.02^{12})^{t}$. $1.02^{12} \approx 1.268$.

b about 1.268: an annual rate of about 26.8%

Exercise 14 – Monthly equivalent

$1.07^{t} = (1.07^{1/12})^{12t}$. $1.07^{1/12} \approx 1.00565$.

About 0.565% per month

Exercise 15 – Graph features

Intercept (0, 2); asymptote y = 0.

$f(0) = 2$. As $x$ decreases, $f(x)$ approaches 0 but never reaches it.

As $x$ increases, $f(x)$ grows without bound.

y-intercept (0, 2); horizontal asymptote y = 0; increasing

Exercise 16 – Shifted exponential

Shifted up 3.

$g(0) = 2^{0} + 3 = 4$. The whole graph of $2^{x}$ is shifted up 3, so the asymptote moves from $y = 0$ to $y = 3$.

y-intercept (0, 4); asymptote y = 3

Exercise 17 – Linear or exponential

Values of both plans.

Plan A: $200 + 50w$. Plan B: $20 \cdot 2^{w}$.

Week 4: $400$ versus $320$. Week 5: $450$ versus $640$.

Week 5

Exercise 18 – Interpret parameters

250 is the initial concentration. 0.92 means 8% of the drug leaves each hour. $f(10) = 250(0.92)^{10} \approx 108.60$.

250 mg/L initially; 8% decrease per hour; about 108.60 mg/L

Exercise 19 – When does it exceed?

Balance by year.

Year 8: about $1,477.46, below 1,500. Year 9: about $1,551.33, above 1,500.

Year 9

Exercise 20 – Find the error

A 5% increase means the new amount is 100% + 5% = 105% of the old amount.

$b = 1.05$

Exercise 21 – Tripling

$V(m) = 4 \cdot 3^{m}$. $V(5) = 4 \cdot 243 = 972$.

972 views

Key facts behind this algebra 2 exponential functions answers

Key facts

f(x) = a·b^x: a is the initial value, b the growth factor.
Growth factor b = 1 + r (growth) or 1 – r (decay).
Equal steps in x multiply the output by b.
Compound interest: A = P(1 + r/n)^(nt); continuous: A = Pe^(rt).
Rewrite b^x expressions to find equivalent rates.

How to use this algebra 2 exponential functions answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the algebra 2 exponential functions answers as a PDF with the button below and keep it with your notes.

All Algebra 2 chapters: Algebra 2 math.

Voter.. post

Free printable PDF