Real Life Impacts of Analyzing Exponential Functions

Investment Period	Bank A	Bank B
$A (t) = 15000 + 450 t$	$B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}$
$1$ Year ( $12$ Months)	$15450$	$\approx 15379$
$2$ Years	$15900$	$\approx 15768$
$3$ Years	$16350$	$\approx 16167$
$4$ Years	$16800$	$\approx 16576$
$5$ Years	$17250$	$\approx 16995$
$6$ Years	$17700$	$\approx 17425$
$7$ Years	$18150$	$\approx 17865$
$8$ Years	$18600$	$\approx 18317$
$9$ Years	$19050$	$\approx 18780$
$10$ Years	$19500$	$\approx 19255$
$11$ Years	$19950$	$\approx 19742$
$12$ Years	$20400$	$\approx 20241$
$13$ Years	$20850$	$\approx 20753$
$14$ Years	$21300$	$\approx 21278$
$15$ Years	$21750$	$\approx 21816$
$16$ Years	$22200$	$\approx 22368$
$17$ Years	$22650$	$\approx 22934$
$18$ Years	$23100$	$\approx 23514$
$19$ Years	$23550$	$\approx 24108$
$20$ Years	$24000$	$\approx 24718$

Investment Period

Bank A

Bank B

A (t) = 15000 + 450 t

B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}

1

Year (

12

Months)

15450

\approx 15379

2

Years

15900

\approx 15768

3

Years

16350

\approx 16167

4

Years

16800

\approx 16576

5

Years

17250

\approx 16995

6

Years

17700

\approx 17425

7

Years

18150

\approx 17865

8

Years

18600

\approx 18317

9

Years

19050

\approx 18780

10

Years

19500

\approx 19255

11

Years

19950

\approx 19742

12

Years

20400

\approx 20241

13

Years

20850

\approx 20753

14

Years

21300

\approx 21278

15

Years

21750

\approx 21816

16

Years

22200

\approx 22368

17

Years

22650

\approx 22934

18

Years

23100

\approx 23514

19

Years

23550

\approx 24108

20

Years

24000

\approx 24718

Investment Period in Years	Bank A	Bank B
$A (t) = 15000 + 450 t$	$B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}$
$1$	$15450$	$\approx 15379$
$2$	$15900$	$\approx 15768$
$3$	$16350$	$\approx 16167$
$4$	$16800$	$\approx 16576$
$5$	$17250$	$\approx 16995$
$6$	$17700$	$\approx 17425$
$7$	$18150$	$\approx 17865$
$8$	$18600$	$\approx 18317$
$9$	$19050$	$\approx 18780$
$10$	$19500$	$\approx 19255$
$11$	$19950$	$\approx 19742$
$12$	$20400$	$\approx 20241$
$13$	$20850$	$\approx 20753$
$14$	$21300$	$\approx 21278$
$15$	$21750$	$\approx 21816$
$16$	$22200$	$\approx 22368$
$17$	$22650$	$\approx 22934$
$18$	$23100$	$\approx 23514$
$19$	$23550$	$\approx 24108$
$20$	$24000$	$\approx 24718$

Investment Period in Years

Bank A

Bank B

A (t) = 15000 + 450 t

B (m) = 15000 (1 + \frac{0 . 0 2 5}{1 2})^{m}

1

15450

\approx 15379

2

15900

\approx 15768

3

16350

\approx 16167

4

16800

\approx 16576

5

17250

\approx 16995

6

17700

\approx 17425

7

18150

\approx 17865

8

18600

\approx 18317

9

19050

\approx 18780

10

19500

\approx 19255

11

19950

\approx 19742

12

20400

\approx 20241

13

20850

\approx 20753

14

21300

\approx 21278

15

21750

\approx 21816

16

22200

\approx 22368

17

22650

\approx 22934

18

23100

\approx 23514

19

23550

\approx 24108

20

24000

\approx 24718

$x$	$55 x + 10$	$y_{C} = 55 x + 10$	$(x, y_{C})$
$0$	$55 (0) + 10$	$10$	$(0, 10)$
$2$	$55 (2) + 10$	$120$	$(2, 120)$
$4$	$55 (4) + 10$	$230$	$(4, 230)$
$6$	$55 (6) + 10$	$340$	$(6, 340)$
$8$	$55 (8) + 10$	$450$	$(8, 450)$

x

55 x + 10

y_{C} = 55 x + 10

(x, y_{C})

0

55 (0) + 10

10

(0, 10)

2

55 (2) + 10

120

(2, 120)

4

55 (4) + 10

230

(4, 230)

6

55 (6) + 10

340

(6, 340)

8

55 (8) + 10

450

(8, 450)

$x$	$4 \cdot 2^{x}$	$y_{T} = 4 \cdot 2^{x}$	$(x, y_{T})$
$0$	$4 \cdot 2^{0}$	$4$	$(0, 4)$
$2$	$4 \cdot 2^{2}$	$16$	$(2, 16)$
$4$	$4 \cdot 2^{4}$	$64$	$(4, 64)$
$6$	$4 \cdot 2^{6}$	$256$	$(6, 256)$
$8$	$4 \cdot 2^{8}$	$1024$	$(8, 1024)$

x

4 \cdot 2^{x}

y_{T} = 4 \cdot 2^{x}

(x, y_{T})

0

4 \cdot 2^{0}

4

(0, 4)

2

4 \cdot 2^{2}

16

(2, 16)

4

4 \cdot 2^{4}

64

(4, 64)

6

4 \cdot 2^{6}

256

(6, 256)

8

4 \cdot 2^{8}

1024

(8, 1024)

	Catfish	Tilapia
Growth Function	$y_{C} = 55 x + 10$	$y_{T} = 4 \cdot 2^{x}$
Growth After $6 \frac{1}{2}$ Weeks	$y_{C} = 55 (6.5) + 10$	$y_{T} = 4 \cdot 2^{6.5}$
Evaluate and Approximate	$y_{C} \approx 368$	$y_{T} \approx 362$

Catfish

Tilapia

Growth Function

y_{C} = 55 x + 10

y_{T} = 4 \cdot 2^{x}

Growth After

6 \frac{1}{2}

Weeks

y_{C} = 55 (6.5) + 10

y_{T} = 4 \cdot 2^{6.5}

Evaluate and Approximate

y_{C} \approx 368

y_{T} \approx 362

User Growth per Week on Each Social Media (in Thousands)
Time (in Weeks)	Option 1	Option 2
$0$	$18$	$15$
$1$	$21$	$18$
$2$	$24$	$21.5$
$3$	$27$	$26$
$4$	$30$	$31$
$5$	$33$	$37.5$

User Growth per Week on Each Social Media (in Thousands)

Time (in Weeks)

Option 1

Option 2

0

18

15

1

21

18

2

24

21.5

3

27

26

4

30

31

5

33

37.5

For an experiment in biology class, the teacher has brought in a fast growing bacterial culture on a Petri dish that quadruples in size every quarter of an hour.

The teacher has also given the class a formula for calculating the number of bacteria after

x

number of

15

minute intervals. Ali, however, is

15

minutes late to class so he gets the following revised formula from his teacher. This equation differs from what his classmates originally received.

y = 120 \cdot 4^{x - 1}

What formula would Ali have gotten, like his classmates, if he was on time?

From the given information, we know that the bacteria quadruples every 15 minutes. Therefore, the 4 in the formula is the multiplier by which the bacteria grows. Since the bacteria grows 4 its size every 15 minutes, x has to be the number of 15 minute intervals since the experiment started. Let's look at the formula again. y =120* 4^(x-1) The second thing we notice is that the exponent contains x-1. The -1 must represent the fact that Ali was one 15-minute interval late. However many such intervals passed for the rest of the class, one less of those will have passed for Ali. y =120* 4^(x - 1) The coefficient 120 is the number of bacteria in the petri dish as Ali enters the classroom. To get the number of bacteria 15 minutes before Ali came in, we substitute x=0 and evaluate.

Now we see that the number of bacteria at the beginning of the class was 30. If Ali was on time, the initial value would be be 30 rather than 120, and -1 would not be part of the exponent. y = 30* 4^x

This special $16 K$ television with a vintage case is worth a whopping $$ 20000 .$ Five years later, it is worth $$ 12000 .$

Write an exponential function that describes the value of the

16 K

television after

x

years if the depreciation is the same percentage every year. Give an exact answer.

The general format of an exponential function is written on the following format. y=ab^x In this equation, a is the initial value and b is the change factor. We need to determine both of these values. From the exercise, we know that the special TV was worth $20 000 when first purchased. Therefore, the initial value is a= 20 000. Let's substitute this into the equation. y= 20 000* b^x To determine b, we need a second point that satisfies the equation. From the exercise, we are given that after 5 years the value is $12 000. Therefore, the second point becomes (5,12 000). Let's substitute this point into the equation.

When we know the value of b, we can complete the function. y=20 000* (sqrt(3/5))^x

A town in Ohio has seen their population leave for bigger cities over time. The four graphs model the population of the town given a depopulation rate of $7.5 %,$ $10 %,$ $12.5 %,$ and $15 %$ annually.

a graph showing two models describing the decline in value of a new computer

What is the town's current population?

What is the approximate population after ten years if the population decreases at a rate of

10 %

annually? Answer without finding the exponential equation.

Approximately how many additional years does it take for the population to halve in size when it decreases by

7.5 %

rather than

15 % ?

Answer without finding the exponential equation.

The current population is given by the graph's value when t=0. We see from the graph that it starts at y = 10 000. Therefore, the current population of the city is 10 000 people.

First, we must figure out which graph represents the one where the population decreases at a rate of 10 % annually. Notice that the greater the rate of decrease, the faster the graph will drop in the coordinate plane.

Therefore, the flattest graph must correspond to the lowest rate of decrease of 7.5 % and the second flattest graph corresponds to a 10 % depopulation and so on.

Now that we identified the correct graph, we will draw a vertical line along x=10 and identify the corresponding y-value where this line intersects the graph.

As we can see, the graph has a y-value of about 3500 when x=10.

The population has halved in size when it is 5000. In Part A we already identified which depreciation corresponds to which graph, so we can identify the graphs with depreciation of 7.5% and 15%.

Now we want to compare the time it takes for the most extreme graphs to reduce to a population of 5000. To do this, we will draw a horizontal line along 5000 and identify the corresponding x-value where this line intersects each graph.

When the population decreases at a rate of 7.5 %, it takes nearly 9 years for the population to halve. When the population decreases by 15 % it takes around 4.2 years for the population to halve. Let's calculate the difference of the years. 9-4.2=4.8 We see that it takes 4.8 additional years to halve in size when the population decreases at 7.5 % instead of 15 %.

Applications of Exponential Functions

Catch-Up and Review

Predicting Fish Growth to Win a Contest

Identifying if a Real-Life Situation Represents an Exponential or a Linear Trend

Answer

Hint

Solution

Which Account Will Have Higher Balance at the End of the Investment Period?

Answer

Hint

Solution

Bank A

Bank B

Comparison

Graphing Linear and Exponential Models to Identify Patterns

Answer

Hint

Solution

Extra

Determining User Growth on Social Media

Answer

Hint

Solution

Linear Function

Exponential Function

Fish Population Growth: Describing the Parameters

Hint

Solution

Applications of Exponential Functions

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