Analyze growth and decay models, interpret the base and initial value, and solve exponential problems.
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The equation is already y = abx. Name the starting value and the percent change.
y = 250(1.15)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 250
For a percent increase, start at 100% and add the percent.
100% + 15% = 115%
115% = 1.15
The base 1.15 means a 15% increase each year.
A 20% decrease works the same way from 100%. 100% − 20% = 80%, and 80% = 0.8, so the base is 0.8.
Answer: starting value 250, 15% increase
y = 80(1.25)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 80
For a percent increase, start at 100% and add the percent.
100% + 25% = 125%
125% = 1.25
The base 1.25 means a 25% increase each year.
Answer: starting value 80, 25% increase
y = 640(0.9)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 640
For a percent decrease, start at 100% and subtract the percent.
100% − 10% = 90%
90% = 0.9
The base 0.9 means a 10% decrease each year.
Answer: starting value 640, 10% decrease
y = 15(1.04)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 15
For a percent increase, start at 100% and add the percent.
100% + 4% = 104%
104% = 1.04
The base 1.04 means a 4% increase each hour.
Answer: starting value 15, 4% increase
y = 1,200(0.75)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 1,200
For a percent decrease, start at 100% and subtract the percent.
100% − 25% = 75%
75% = 0.75
The base 0.75 means a 25% decrease each year.
Answer: starting value 1,200, 25% decrease
y = 48(1.5)x
The starting value is a, the y-intercept. It is the value of y when x = 0.
a = 48
For a percent increase, start at 100% and add the percent.
100% + 50% = 150%
150% = 1.5
The base 1.5 means a 50% increase each year.
Answer: starting value 48, 50% increase
Use the starting amount for a. Build the base from 100% and the percent change.
y = a(b)x
a = 18
For a percent increase, start at 100% and add the percent.
100% + 12% = 112%
112% = 1.12
b = 1.12
y = 18(1.12)x
Answer: y = 18(1.12)x
y = a(b)x
a = 2,400
For a percent increase, start at 100% and add the percent.
100% + 8% = 108%
108% = 1.08
b = 1.08
y = 2,400(1.08)x
Answer: y = 2,400(1.08)x
y = a(b)x
a = 18,000
For a percent decrease, start at 100% and subtract the percent.
100% − 15% = 85%
85% = 0.85
b = 0.85
y = 18,000(0.85)x
Answer: y = 18,000(0.85)x
y = a(b)x
a = 500
For a percent increase, start at 100% and add the percent.
100% + 40% = 140%
140% = 1.4
b = 1.4
y = 500(1.4)x
Answer: y = 500(1.4)x
y = a(b)x
a = 900
For a percent decrease, start at 100% and subtract the percent.
100% − 30% = 70%
70% = 0.7
b = 0.7
y = 900(0.7)x
Answer: y = 900(0.7)x
y = a(b)x
a = 3,500
For a percent increase, start at 100% and add the percent.
100% + 6% = 106%
106% = 1.06
b = 1.06
y = 3,500(1.06)x
Answer: y = 3,500(1.06)x
The starting amount is a. The factor it is multiplied by each year is b.
y = a(b)x
a = 1,200
The population doubles every year, so b = 2.
y = 1,200(2)x
Answer: y = 1,200(2)x
y = a(b)x
a = 45
The population triples every year, so b = 3.
y = 45(3)x
Answer: y = 45(3)x
y = a(b)x
a = 800
The amount is halved every year, so b = 12.
y = 800(12)x
Answer: y = 800(12)x
y = a(b)x
a = 16
The count quadruples every year, so b = 4.
y = 16(4)x
Answer: y = 16(4)x
y = a(b)x
a = 250
The number is multiplied by 5 every year, so b = 5.
y = 250(5)x
Answer: y = 250(5)x
y = a(b)x
a = 60
The count becomes 10 times as large every year, so b = 10.
y = 60(10)x
Answer: y = 60(10)x
t is the number of years after the year in the story. Read a or b from that year.
t is the number of years after 2005.
When t = 0, the year is 2005.
P(t) = 18,000(1.02)t
P(0) = 18,000(1.02)0
P(0) = 18,000 × 1
P(0) = 18,000
18,000 is the population in 2005.
Year 0 is not the starting year in this story.
Answer: A. The population in 2005.
t is the number of years after 1998.
When t = 0, the year is 1998.
B(t) = 5,400(1.03)t
B(0) = 5,400(1.03)0
B(0) = 5,400 × 1
B(0) = 5,400
5,400 is the balance in 1998.
Year 0 is not the starting year in this story.
Answer: C. The balance in 1998.
t is the number of years after 2012.
The base is b.
b = 1.04
100% + 4% = 104%
104% = 1.04
Each year after 2012, the value is multiplied by 1.04. That is a 4% increase.
The multiplying starts after 2012. It does not start in year 0.
Answer: A. Each year after 2012, the value is multiplied by 1.04.
t is the number of years after 2020.
When t = 0, the year is 2020.
H(t) = 75(0.96)t
H(0) = 75(0.96)0
H(0) = 75 × 1
H(0) = 75
75 is the height in 2020.
Year 0 is not the starting year in this story.
Answer: B. The height in 2020.
t is the number of years after 2016.
The base is b.
b = 0.92
100% − 8% = 92%
92% = 0.92
Each year after 2016, the population is multiplied by 0.92. That is an 8% decrease.
The multiplying starts after 2016. It does not start in year 0.
Answer: D. Each year after 2016, the population is 0.92 times the year before.
t is the number of years after 2008.
When t = 0, the year is 2008.
F(t) = 12,750(1.05)t
F(0) = 12,750(1.05)0
F(0) = 12,750 × 1
F(0) = 12,750
12,750 is the value of the fund in 2008.
Year 0 is not the starting year in this story.
Answer: C. The value of the fund in 2008.
No equation is given. The amount doubles, triples, or quadruples on a steady schedule.
y = a(b)x
a = 3
b = 2
The amount is multiplied every hour, and 10 hours have passed, so the exponent is 10.
y = 3(2)10
210 = 1,024
y = 3 × 1,024
y = 3,072
Answer: 3,072 bacteria
y = a(b)x
a = 6
b = 2
The amount is multiplied every hour, and 12 hours have passed, so the exponent is 12.
y = 6(2)12
212 = 4,096
y = 6 × 4,096
y = 24,576
Answer: 24,576 subscribers
y = a(b)x
a = 2
b = 3
The amount is multiplied every hour, and 10 hours have passed, so the exponent is 10.
y = 2(3)10
310 = 59,049
y = 2 × 59,049
y = 118,098
Answer: 118,098 cells
y = a(b)x
a = 3
b = 4
The amount is multiplied every hour, and 10 hours have passed, so the exponent is 10.
y = 3(4)10
410 = 1,048,576
y = 3 × 1,048,576
y = 3,145,728
Answer: 3,145,728 stamps
y = a(b)x
a = 80
b = 2
The amount is multiplied every 2 years. The exponent is the number of those periods in 12 years.
12 ÷ 2 = 6
y = 80(2)6
26 = 64
y = 80 × 64
y = 5,120
Answer: 5,120 trees
y = a(b)x
a = 25
b = 3
The amount is multiplied every 3 years. The exponent is the number of those periods in 12 years.
12 ÷ 3 = 4
y = 25(3)4
34 = 81
y = 25 × 81
y = 2,025
Answer: 2,025 members
Add the percent increase to 100% before you write the base.
y = a(b)x
a = 40
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 150% = 250%
250% = 2.5
b = 2.5
The base is not 1.5. That number uses the percent increase and leaves off the original 100%.
y = 40(2.5)x
Answer: y = 40(2.5)x
y = a(b)x
a = 200
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 200% = 300%
300% = 3
b = 3
The base is not 2. That number uses the percent increase and leaves off the original 100%.
y = 200(3)x
Answer: y = 200(3)x
y = a(b)x
a = 75
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 120% = 220%
220% = 2.2
b = 2.2
The base is not 1.2. That number uses the percent increase and leaves off the original 100%.
y = 75(2.2)x
Answer: y = 75(2.2)x
y = a(b)x
a = 500
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 250% = 350%
350% = 3.5
b = 3.5
The base is not 2.5. That number uses the percent increase and leaves off the original 100%.
y = 500(3.5)x
Answer: y = 500(3.5)x
y = a(b)x
a = 16
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 300% = 400%
400% = 4
b = 4
The base is not 3. That number uses the percent increase and leaves off the original 100%.
y = 16(4)x
Answer: y = 16(4)x
y = a(b)x
a = 90
This percent increase is larger than 100%. Start at 100%, then add the percent increase.
100% + 180% = 280%
280% = 2.8
b = 2.8
The base is not 1.8. That number uses the percent increase and leaves off the original 100%.
y = 90(2.8)x
Answer: y = 90(2.8)x
t counts years. One year is 12 months, so replace t with m over 12.
12 months = 1 year.
t = m12
P(t) = 500(1.04)t
Replace t with the number of years.
\(P(m) = 500(1.04)^{\frac{m}{12}}\)
Answer: \(P(m) = 500(1.04)^{\frac{m}{12}}\)
12 months = 1 year.
t = m12
V(t) = 2,400(0.85)t
Replace t with the number of years.
\(V(m) = 2{,}400(0.85)^{\frac{m}{12}}\)
Answer: \(V(m) = 2{,}400(0.85)^{\frac{m}{12}}\)
12 months = 1 year.
t = m12
A(t) = 150(1.2)t
Replace t with the number of years.
\(A(m) = 150(1.2)^{\frac{m}{12}}\)
Answer: \(A(m) = 150(1.2)^{\frac{m}{12}}\)
12 months = 1 year.
t = m12
B(t) = 60(2)t
Replace t with the number of years.
\(B(m) = 60(2)^{\frac{m}{12}}\)
Answer: \(B(m) = 60(2)^{\frac{m}{12}}\)
12 months = 1 year.
t = m12
C(t) = 900(1.08)t
Replace t with the number of years.
\(C(m) = 900(1.08)^{\frac{m}{12}}\)
Answer: \(C(m) = 900(1.08)^{\frac{m}{12}}\)
12 months = 1 year.
t = m12
D(t) = 75(0.5)t
Replace t with the number of years.
\(D(m) = 75(0.5)^{\frac{m}{12}}\)
Answer: \(D(m) = 75(0.5)^{\frac{m}{12}}\)
The first term is term number 1. The exponent on that term is 0.
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 6, so y = 6 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 6(4)n−1
Check the first term.
n = 1
y = 6(4)0
y = 6 × 1
y = 6
An answer with exponent n gives 24 when n = 1. That skips the first term and starts too high. The multiplier is 4.
Answer: B. y = 6(4)n−1
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 5, so y = 5 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 5(3)n−1
Check the first term.
n = 1
y = 5(3)0
y = 5 × 1
y = 5
An answer with exponent n gives 15 when n = 1. That skips the first term and starts too high. The multiplier is 3.
Answer: C. y = 5(3)n−1
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 12, so y = 12 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 12(2)n−1
Check the first term.
n = 1
y = 12(2)0
y = 12 × 1
y = 12
An answer with exponent n gives 24 when n = 1. That skips the first term and starts too high. The multiplier is 2.
Answer: A. y = 12(2)n−1
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 7, so y = 7 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 7(10)n−1
Check the first term.
n = 1
y = 7(10)0
y = 7 × 1
y = 7
An answer with exponent n gives 70 when n = 1. That skips the first term and starts too high. The multiplier is 10.
Answer: D. y = 7(10)n−1
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 20, so y = 20 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 20(12)n−1
Check the first term.
n = 1
y = 20(12)0
y = 20 × 1
y = 20
An answer with exponent n gives 10 when n = 1. That skips the first term and starts too high. The multiplier is 1/2.
Answer: B. y = 20(12)n−1
A geometric sequence starts at n = 1. The first term is term number 1, not term number 0.
The first term is 9, so y = 9 when n = 1.
y = a(b)n
A power of 0 is 1, so the exponent on the first term has to be 0.
The exponent that is 0 when n = 1 is n − 1.
y = 9(5)n−1
Check the first term.
n = 1
y = 9(5)0
y = 9 × 1
y = 9
An answer with exponent n gives 45 when n = 1. That skips the first term and starts too high. The multiplier is 5.
Answer: C. y = 9(5)n−1
Enter the table in Desmos, then type y1 ~ abx1. Read a and b.
| x | y |
|---|---|
| 0 | 80 |
| 1 | 120 |
| 2 | 180 |
y1 ~ abx1
a = 80
b = 1.5
y = 80(1.5)x
The table includes the starting value, the output when x = 0.
In Desmos, enter the three rows, then type the regression.
y1 ~ abx1
Desmos gives a = 80 and b = 1.5.
y = a(b)x
y = 80(1.5)x
Shortcut: the y-intercept is 80, and the outputs increase, so the base is greater than 1. On a multiple-choice question, the y-intercept and the increase or decrease often leave one answer.
Answer: y = 80(1.5)x
| x | y |
|---|---|
| 0 | 50 |
| 1 | 40 |
| 2 | 32 |
The table includes the starting value, the output when x = 0.
In Desmos, enter the three rows, then type the regression.
y1 ~ abx1
Desmos gives a = 50 and b = 0.8.
y = a(b)x
y = 50(0.8)x
Shortcut: the y-intercept is 50, and the outputs decrease, so the base is between 0 and 1. On a multiple-choice question, the y-intercept and the increase or decrease often leave one answer.
Answer: y = 50(0.8)x
| x | y |
|---|---|
| 0 | 200 |
| 1 | 300 |
| 2 | 450 |
The table includes the starting value, the output when x = 0.
In Desmos, enter the three rows, then type the regression.
y1 ~ abx1
Desmos gives a = 200 and b = 1.5.
y = a(b)x
y = 200(1.5)x
Shortcut: the y-intercept is 200, and the outputs increase, so the base is greater than 1. On a multiple-choice question, the y-intercept and the increase or decrease often leave one answer.
Answer: y = 200(1.5)x
| x | y |
|---|---|
| 0 | 12 |
| 1 | 36 |
| 2 | 108 |
The table includes the starting value, the output when x = 0.
In Desmos, enter the three rows, then type the regression.
y1 ~ abx1
Desmos gives a = 12 and b = 3.
y = a(b)x
y = 12(3)x
Shortcut: the y-intercept is 12, and the outputs increase, so the base is greater than 1. On a multiple-choice question, the y-intercept and the increase or decrease often leave one answer.
Answer: y = 12(3)x
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