Comparing Linear, Quadratic, and Exponential Functions: Understanding Growth Differences

Many real-life situations, such as buying a cellphone, making an investment, or studying the population growth of a particular species, can be modeled using functions. This lesson will review some situations and identify whether they can be fitted into a linear, quadratic, or exponential function to make optimal decisions.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Linear Functions
Quadratic Functions
Exponential Functions

Challenge

Distinctive Characteristics of a Quadratic Function

The graphs of a linear function, a quadratic function, and an exponential function can be seen in the applet below.

Is it possible to name at least three characteristics of a quadratic function that neither a linear nor an exponential function has?

Example

Classifying Functions

Magdalena has just found an old printer toy she used to play with when she was a child. She discovered that the toy prints random functions as labels.

Magdalena and Linear, Quadratic, and Exponential function labels

Magdalena ran the printer and got three labels with the following functions.

y = 3 \cdot 2^{x} y = 2 x + 1 y = x^{2} + 3 x + 1

She wants to determine whether each of the functions is linear, quadratic, or exponential. Help Magdalena to accomplish this task!

Hint

Review the definitions of linear, quadratic, and exponential functions.

Solution

Consider the first given function.

y = 3 \cdot 2^{x}

Note that the independent variable

x

is written as an exponent in the function and the base

2

is a non-negative number. Therefore, this is an exponential function.

\underline{Exponential Function} y = 3 \cdot 2^{x}

Now, consider the second function.

y = 2 x + 1

This function only contains one linear term and one constant term. This means that it is a linear function.

\underline{Linear Function} y = 2 x + 1

Finally, consider the third function.

y = x^{2} + 3 x + 1

Notice that this is a polynomial function and the highest exponent of the independent variable

x

2 .

Therefore, it can be concluded that this is a quadratic equation.

\underline{Quadratic Function} y = x^{2} + 3 x + 1

This information is summarized in the following table.

Linear Function	$y = 2 x + 1$
Quadratic Function	$y = x^{2} + 3 x + 1$
Exponential Function	$y = 3 \cdot 2^{x}$

Pop Quiz

Determining if a Function Is Linear, Quadratic, or Exponential

In the following applet, determine whether the given function is linear, quadratic, or exponential.

Identify whether the given function is linear, quadratic, or exponential.

Discussion

Finding the Pattern of a Table of Values to Determine Its Model

When considering a table of values, the trend of the data can be determined by observing how the dependent variable changes over equal intervals. To do so, the following situations need to be analyzed.

If the $y -$ values have a common difference, the data follows a linear model.
If the $y -$ values have a common ratio, the data follows an exponential model.
If the second differences — the differences of the first differences — are constant, the data follows a quadratic model.

The applet below illustrates each situation.

An applet showing a linear, quadratic, and exponential model in a table of values

Example

Determining the Pattern Describing Real-Life Data Sets

Magdalena's father works as a manager for a fish farming company. This morning, Magdalena and her brother Vincenzo poked around in her father's briefcase. They found some papers containing information about the fish population in three nearby lakes.

Fish Population (Thousands)
Time (Months)	Lake Verdastro	Lake Rumoroso	Lake Mezzo
$0$	$1356$	$721$	$1207$
$1$	$6676$	$2163$	$3007$
$2$	$11996$	$6489$	$7807$
$3$	$17316$	$19467$	$15607$
$4$	$22636$	$58401$	$26407$

Magdalena told her brother that determining if the fish growth follows some pattern will help their father make better decisions for the company. However, they are not sure how to do this task. Help them determine whether the collected data for each lake can be modeled by a linear, quadratic, or exponential function.

Answer

Lake Verdastro: Linear Function
Lake Rumoroso: Exponential Function
Lake Mezzo: Quadratic Function

Hint

The $y -$ values have a common difference in a linear function, while in an exponential function, the $y -$ values have a common ratio. Furthermore, in a quadratic function, the second differences are constant.

Solution

The recorded data for each lake will be analyzed to determine if they have a common difference, a common ratio, or a constant second difference. So, the data for Lake Verdastro will be analyzed first.

Because the fish population in Lake Verdastro has a common difference of

5320,

it can be said that it follows a linear model. The data for Lake Rumoroso will be analyzed in a similar fashion.

Note that the data of Lake Rumoroso has a common ratio of

3 .

Therefore, it can be fitted into an exponential function. Finally, the model for the data of Lake Mezzo will be determined.

Since the second differences of the data of Lake Mezzo are constant, the fish population for the lake over time can be modeled with a quadratic function. With this, the growth model of the fish population for each lake has been determined.

Lake	Model
Verdastro	Linear Function
Rumoroso	Exponential Function
Mezzo	Quadratic Function

Note that the analysis suggests that Lake Rumoroso is the best for fish farming. The children gave this information to their father, who was delighted by this excellent analysis and took them for ice cream as a reward.

Example

Saving for a New Computer

Magdalena will start college soon, so she decided to start saving money to buy a new computer. The computer she wants costs around $$ 2000 .$ Thanks to her help with the analysis for the fish farming company, her father gave Magdalena an initial amount of $$ 1000 .$ She plans to add $$ 40$ to her savings each week.

Magdalena looking at the computer she desires

Magdalena wants to buy the computer as soon as possible. She thinks she will be able to afford it after $30$ weeks of saving the money but she isn't sure. Answer the following questions to help her discover if she will save enough money for the new computer by then.

a Can this situation be modeled by a linear, quadratic, or exponential function? Write the function that models the situation.

b How much money will Magdalena have saved after

30

weeks? Interpret the results.

Answer

a Model: Linear Function
Function:

B (x) = 40 x + 1000

b Answer:

$ 2200

Interpretation: Because the amount of money Magdalena will save after

30

weeks is greater than the computer's price, she will be able to buy the computer.

Hint

a After

x

weeks of saving money, the total amount of savings can be determined by adding the product of

x

and the amount of weekly savings to the initial amount.

b Substitute

30

for

x

into the function found in Part A. Compare the result with the price of the computer.

Solution

a Magdalena starts with the initial

$ 1000

that her father gave her. Each week she plans to add

$ 40

to this amount. Therefore, her balance

B (x)

after

x

weeks can be calculated by adding the product of

x

and

$ 40

to the initial amount of

$ 1000 .

\underline{Balance After x Weeks} B (x) = 1000 + 40 x ⇓ B (x) = 40 x + 1000

Looking at the function, it can be concluded that this situation is modeled by a linear function.

c The amount of money that Magdalena will have after

30

weeks can be found by substituting

30

for

x

in the function found in Part A.

B (x) = 40 x + 1000

Substitute

$x = 30$

B (30) = 40 (30) + 1000

Multiply

B (30) = 1200 + 1000

AddTerms

Add terms

B (30) = 2200

If everything goes according to plan, Magdalena will have

$ 2200

after

30

weeks. Recall that the price of the computer is around

$ 2000 .

The amount that Magdalena is expected to have saved is greater. Therefore, she will be able to buy the new computer and use it for her school projects by that time.

Example

Finding the Optimal Increase to Get the Maximum Revenue

Magdalena is happy with her new computer. She is also using it for her part-time job as a designer to help pay for her school expenses. She sells $50$ designs per month at $$ 59$ each. However, because creating each design is complex, she plans to increase the price of each design. She estimates that for each $$ 5$ increase, two fewer designs will be sold per month.

Magdalena wants to know what price will maximize her revenue, or the amount of money she makes. Identify the following information to help Magdalena make the best decision about her small business.

a Identify and write the function that models this situation.

b What is the maximum amount of money that Magdalena could make each month?

c What price would allow Magdalena to sell her designs so that she could make the most money each month?

Answer

a Model: Quadratic Function
Function:

R (x) = - 10 x^{2} + 132 x + 2950

$ 3385.60

$ 92

Hint

a Begin by writing a function for the new price of a design and another function for the number of designs sold.

b Remember that the axis of symmetry intersects the parabola at its vertex.

c Substitute the

x -

value of the vertex into the expression of the new price of a design.

Solution

a Magdalena has estimated how each

$ 5

increase will affect the number of designs sold. Let

x

be the number of

$ 5

price increases. This means the new price

N (x)

of a design will be the current price of

$ 59

plus

x

times

$ 5 .

\underline{New Price of a Design :} N (x) = 59 + 5 x

Increasing the price of the designs means that Magdalena will not sell as many designs. For each

$ 5

increase,

2

fewer designs will be sold. This means that the number of designs sold

S (x)

will be given by the current number of designs sold per month minus

2

times

x

\underline{Number of Designs Sold} S (x) = 50 - 2 x

Finally, the amount of money Magdalena can make each month

R (x)

can be determined by multiplying the number of designs sold

S (x)

by the new price of a design

N (x) .

\underline{Magdalena ’ s Revenue} R (x) = S (x) \cdot N (x)

Substitute the expression of each function and simplify to find the model describing the amount of money Magdalena can make.

R (x) = S (x) \cdot N (x)

SubstituteII

$S (x) = 50 - 2 x$ , $N (x) = 59 + 5 x$

R (x) = (50 - 2 x) (59 + 5 x)

▼

Simplify right-hand side

Distr

Distribute $50 - 2 x$

R (x) = (50 - 2 x) 59 + (50 - 2 x) 5 x

CommutativePropMult

Commutative Property of Multiplication

R (x) = 59 (50 - 2 x) + 5 x (50 - 2 x)

Distr

Distribute $59 & 5 x$

R (x) = 2950 - 118 x + 250 x - 10 x^{2}

AddTerms

Add terms

R (x) = 2950 + 132 x - 10 x^{2}

CommutativePropAdd

Commutative Property of Addition

R (x) = - 10 x^{2} + 132 x + 2950

Therefore, Magdalena's situation can be modeled with a quadratic function.

b Consider the quadratic function

R (x)

found in Part A.

R (x) = - 10 x^{2} + 132 x + 2950

The function will be modeled with a parabola. Since the leading coefficient is negative, the parabola opens downward and reaches its maximum at its vertex. The maximum revenue is given by the

y -

value of the vertex. First, the

x -

value of the vertex will be determined. Consider the formula for the axis of symmetry of a parabola.

x = - \frac{b}{2 a}

Since the axis of symmetry intersects the parabola at the vertex, the

x -

value of the vertex is also given by this expression. In this quadratic function,

a = - 10

and

b = 132 .

x = - \frac{b}{2 a}

SubstituteII

$a = - 10$ , $b = 132$

x = - \frac{1 3 2}{2 ( - 1 0 )}

▼

Evaluate right-hand side

MultPosNeg

$a (- b) = - a \cdot b$

x = - \frac{1 3 2}{- 2 0}

RemoveNegFracAndDenom

$- \frac{a}{- b} = \frac{a}{b}$

x = \frac{1 3 2}{2 0}

CalcQuot

Calculate quotient

x = 6.6

The axis of symmetry and the

x -

coordinate of the vertex is

x = 6.6 .

To find the

y -

coordinate of the vertex, substitute

x = 6.6

into the quadratic function and simplify.

R (x) = - 10 x^{2} + 132 x + 2950

Substitute

$x = 6.6$

R (6.6) = - 10 (6.6)^{2} + 132 (6.6) + 2950

▼

Simplify right-hand side

CalcPow

Calculate power

R (6.6) = - 10 (43.56) + 132 (6.6) + 2950

Multiply

R (6.6) = - 435.6 + 871.2 + 2950

AddTerms

Add terms

R (6.6) = 3385.60

The vertex of the parabola is

(6.6, 3385.60) .

This means that the maximum amount of money that Magdalena can expect to make each month is

$ 3385.60 .

It is worth noting that

x = 6.6

does make sense in this context, since

x

represents the number of increases of

5

needed to be made to the design price.

c It was determined that the maximum revenue

$ 3385.60

per month is reached when

x = 6.6 .

This means that Magdalena will need to make

6.6

increases of

$ 5

to the price of a design to make the most amount of money each month. Substitute this value into the function

N (x)

to find the new price for a design.

N (x) = 59 + 5 x

Substitute

$x = 6.6$

N (6.6) = 59 + 5 (6.6)

▼

Simplify right-hand side

Multiply

N (6.6) = 59 + 33

AddTerms

Add terms

N (6.6) = 92

Therefore, Magdalena should sell her designs for

$ 92

each to make the most amount of money each month.

Example

Determining the Storage Needed for a School Assignment

Vincenzo, Magdalena's little brother, loves dinosaurs. He wants to make a $10 -$ minute film for a school assignment, explaining the history of this extinguished group of reptiles. He asked Magdalena to help him make and edit the video on her computer. However, the siblings are worried because Vincenzo has a USB drive with only $16 GB$ to store the film on.

After some research, Magdalena found that any high-quality video file requires an initial storage space of $20 MB .$ This required storage space doubles for each additional minute. The children now have to investigate whether the USB drive has enough space for Vincenzo's completed film. Find the following information to help them.

a Identify the model and write a function representing the size of an

x -

minute long video.

b Does the USB have enough storage space for Vincenzo's completed video? Explain.

Answer

a Model: Exponential

Function: $f (x) = 20 \cdot 2^{x}$

b No, because the video size will be

20 GB,

which is larger than the USB storage.

Hint

a Consider that exponential functions grow by equal factors over equal intervals.

b Substitute

20

for

x

in the function found in Part A. Then divide the result by

1024

to convert it from megabytes to gigabytes.

Solution

a It is given that the size of the video grows by a factor of

2

for every minute of the video. This means this situation is represented by an exponential growth and it can be modeled with an exponential function. Consider the general form of an exponential function.

f (x) = a \cdot b^{x}

In this form,

b

represents the growth factor and

a

is the

y -

intercept, which is sometimes referred to as the initial value. Since the video file needs an initial storage space of

20 MB

and it doubles for every additional minute, substituting

20

for

a

and

2

for

b

will give the expression representing the video size after

x

minutes.

f (x) = a \cdot b^{x} \Rightarrow f (x) = 20 \cdot 2^{x}

b To determine if the USB storage is large enough for Vincenzo's

10 -

minute film, evaluate the expression written in Part A for

x = 10 .

f (x) = 20 \cdot 2^{x}

Substitute

$x = 10$

f (10) = 20 \cdot 2^{10}

CalcPow

Calculate power

f (10) = 20 \cdot 1024

Multiply

f (10) = 20480

The size of a

10 -

minute video is

20480 MB .

To compare this value with the total capacity of the USB drive, it needs to be converted to from megabytes to gigabytes. Because

1 GB

equals

1024 MB,

dividing the size of the video by

1024

will give the size in gigabytes.

\underline{Megabytes to Gigabytes} \frac{2 0 4 8 0}{1 0 2 4} = 20 GB

Therefore, the

10 -

minute film will have a size of

20 GB .

This is larger than the total storage of the

16 GB

USB drive. Unfortunately, the children cannot save the file onto the drive and have to buy a new one with storage of at least

20 GB .

Closure

Identifying Some Remarkable Characteristics of Quadratic Functions

Many situations involving linear, quadratic, and exponential functions have been covered throughout this lesson. It is now time to point out some remarkable characteristics of quadratic functions. Consider the applet presented at the beginning of the lesson.

Is it possible to name at least three characteristics of a quadratic function that neither a linear nor an exponential function has?

Answer

See solution.

Hint

Begin by analyzing the behavior of the functions. Do all functions have a turning point? What is the maximum number of $x -$ intercepts each function could have?

Solution

Three distinctive characteristics of quadratic functions will be identified one at a time.

First Characteristic

To identify the first characteristic, look at the behavior of the graphs of linear, quadratic, and exponential functions.

In the applet above, it can be seen that both linear and exponential functions increase throughout the graph. Conversely, the quadratic function is a parabola that shifts from decreasing to increasing at the vertex. Now consider another three functions.

This time both the linear and exponential functions only decrease. The parabola of the quadratic function opens downward now but it still changes directions, this time from increasing to decreasing. With this information, the following conclusion can be made.

Linear and exponential functions either only increase or only decrease. Conversely, quadratic functions always decrease and increase. The order depends on the direction in which the parabola opens.

Second Characteristic

A second unique characteristic can be determined directly from the first. The vertex of a parabola is its turning point. At this point, the function reaches its absolute maximum or minimum. If the parabola opens upward, the vertex represents the absolute minimum. If it opens downward, the vertex is the absolute maximum.

Conversely, linear and exponential functions approach positive infinity or negative infinity since they only decrease or increase, meaning that they do not have an absolute maximum or minimum. The second characteristic for quadratic functions can now be noted.

Because a parabola either opens upward or downward, there is always one point that is the absolute minimum or absolute maximum of the function. This point is called the vertex.

Third Characteristic

The

x -

intercepts of a function are known as the zeros of the function. They are found by setting the function rule equal to zero and simplifying.

f (x) = 0

Because a quadratic equation has at most

2

solutions, the quadratic function can have one, two, or no

x -

intercepts.

A linear functions whose graph is not horizontal has exactly one

x -

intercept. Exponential functions can have one or no

x -

intercepts.

Using this information, a third distinctive characteristic of quadratic functions can be determined.

While non-horizontal linear functions and exponential functions have at most one $x -$ intercept, quadratic functions can have up to two $x -$ intercepts.

Conclusion

Only three distinctive characteristics of quadratic functions are presented here. Please note that there are a few others.

Three Distinctive Characteristics of Quadratic Functions
Linear and exponential functions either only increase or only decrease. Conversely, quadratic functions always decrease and increase. The order depends on the direction in which the parabola opens.
Because a parabola either opens upward or downward, there is always one point that is the absolute minimum or absolute maximum of the function. This point is called the vertex.
While non-horizontal linear functions and exponential functions have at most one $x -$ intercept, quadratic functions can have up to two $x -$ intercepts.

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Catch-Up and Review

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

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Answer

Hint

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Answer

Hint

Solution

First Characteristic

Second Characteristic

Third Characteristic

Conclusion