# 10.7: Linear, Exponential and Quadratic Models

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify functions using differences and ratios.
• Write equations for functions.
• Perform exponential and quadratic regressions with a graphing calculator.
• Solve real-world problems by comparing function models.

## Introduction

In this course we’ve learned about three types of functions, linear, quadratic and exponential.

• Linear functions take the form y=mx+b\begin{align*}y=mx+b\end{align*}.
• Quadratic functions take the form y=ax2+bx+c\begin{align*}y=ax^2+bx+c\end{align*}.
• Exponential functions take the form y=abx\begin{align*}y=a \cdot b^x\end{align*}.

In real-world applications, the function that describes some physical situation is not given; it has to be found before the problem can be solved. For example, scientific data such as observations of planetary motion are often collected as a set of measurements given in a table. Part of the scientist’s job is to figure out which function best fits the data. In this section, you’ll learn some methods that are used to identify which function describes the relationship between the variables in a problem.

## Identify Functions Using Differences or Ratios

One method for identifying functions is to look at the difference or the ratio of different values of the dependent variable. For example, if the difference between values of the dependent variable is the same each time we change the independent variable by the same amount, then the function is linear.

Example 1

Determine if the function represented by the following table of values is linear.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
2\begin{align*}-2\end{align*} –4
–1 –1
0 2
1 5
2 8

If we take the difference between consecutive y\begin{align*}y-\end{align*}values, we see that each time the x\begin{align*}x-\end{align*}value increases by one, the y\begin{align*}y-\end{align*}value always increases by 3.

Since the difference is always the same, the function is linear.

When we look at the difference of the y\begin{align*}y-\end{align*}values, we have to make sure that we examine entries for which the x\begin{align*}x-\end{align*}values increase by the same amount.

For example, examine the values in this table:

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
0 5
1 10
3 20
4 25
6 35

At first glance this function might not look linear, because the difference in the y\begin{align*}y-\end{align*}values is not always the same. But if we look closer, we can see that when the y\begin{align*}y-\end{align*}value increases by 10 instead of 5, it’s because the x\begin{align*}x-\end{align*}value increased by 2 instead of 1. Whenever the x\begin{align*}x-\end{align*}value increases by the same amount, the y\begin{align*}y-\end{align*}value does too, so the function is linear.

Another way to think of this is in mathematical notation. We can say that a function is linear if y2y1x2x1\begin{align*}\frac{y_2-y_1}{x_2-x_1}\end{align*} is always the same for any two pairs of x\begin{align*}x-\end{align*} and y\begin{align*}y-\end{align*}values. Notice that the expression we used here is simply the definition of the slope of a line.

Differences can also be used to identify quadratic functions. For a quadratic function, when we increase the x\begin{align*}x-\end{align*}values by the same amount, the difference between y\begin{align*}y-\end{align*}values will not be the same. However, the difference of the differences of the y\begin{align*}y-\end{align*}values will be the same.

Here are some examples of quadratic relationships represented by tables of values:

In this quadratic function, y=x2\begin{align*}y=x^2\end{align*}, when we increase the x\begin{align*}x-\end{align*}value by one, the value of y\begin{align*}y\end{align*} increases by different values. However, it increases at a constant rate, so the difference of the difference is always 2.

In this quadratic function, y=2x23x+1\begin{align*}y=2x^2-3x+1\end{align*}, when we increase the x\begin{align*}x-\end{align*}value by one, the value of y\begin{align*}y\end{align*} increases by different values. However, the increase is constant: the difference of the difference is always 4.

To identify exponential functions, we use ratios instead of differences. If the ratio between values of the dependent variable is the same each time we change the independent variable by the same amount, then the function is exponential.

Example 2

Determine if the function represented by each table of values is exponential.

a)

b)

a) If we take the ratio of consecutive y\begin{align*}y-\end{align*}values, we see that each time the x\begin{align*}x-\end{align*}value increases by one, the y\begin{align*}y-\end{align*}value is multiplied by 3. Since the ratio is always the same, the function is exponential.

b) If we take the ratio of consecutive y\begin{align*}y-\end{align*}values, we see that each time the x\begin{align*}x-\end{align*}value increases by one, the y\begin{align*}y-\end{align*}value is multiplied by 12\begin{align*}\frac{1}{2}\end{align*}. Since the ratio is always the same, the function is exponential.

## Write Equations for Functions

Once we identify which type of function fits the given values, we can write an equation for the function by starting with the general form for that type of function.

Example 3

Determine what type of function represents the values in the following table.

x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
0 5
1 1
2 -3
3 -7
4 -11

Solution

Let’s first check the difference of consecutive values of y\begin{align*}y\end{align*}.

If we take the difference between consecutive y\begin{align*}y-\end{align*}values, we see that each time the x\begin{align*}x-\end{align*}value increases by one, the y\begin{align*}y-\end{align*}value always decreases by 4. Since the difference is always the same, the function is linear.

To find the equation for the function, we start with the general form of a linear function: y=mx+b\begin{align*}y=mx+b\end{align*}. Since m\begin{align*}m\end{align*} is the slope of the line, it’s also the quantity by which y\begin{align*}y\end{align*} increases every time the value of x\begin{align*}x\end{align*} increases by one. The constant b\begin{align*}b\end{align*} is the value of the function when x=0\begin{align*}x = 0\end{align*}. Therefore, the function is y=4x+5\begin{align*}y=-4x+5\end{align*}.

Example 4

Determine what type of function represents the values in the following table.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 0
1 5
2 20
3 45
4 80
5 125
6 180

Solution

Here, the difference between consecutive \begin{align*}y-\end{align*}values isn’t constant, so the function is not linear. Let’s look at those differences more closely.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 0
1 5 \begin{align*}5-0=5\end{align*}
2 20 \begin{align*}20-5=15\end{align*}
3 45 \begin{align*}45-20=25\end{align*}
4 80 \begin{align*}80-45=35\end{align*}
5 125 \begin{align*}125-80=45\end{align*}
6 180 \begin{align*}180-125=55\end{align*}

When the \begin{align*}x-\end{align*}value increases by one, the difference between the \begin{align*}y-\end{align*}values increases by 10 every time. Since the difference of the differences is constant, the function describing this set of values is quadratic.

To find the equation for the function that represents these values, we start with the general form of a quadratic function: \begin{align*}y=ax^2+bx+c\end{align*}.

We need to use the values in the table to find the values of the constants: \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*}.

The value of \begin{align*}c\end{align*} represents the value of the function when \begin{align*}x = 0\end{align*}, so \begin{align*}c = 0\end{align*}.

\begin{align*}\text{Plug in the point} \ (1, 5): && 5 &= a+b\\ \text{Plug in the point} \ (2, 20): && 20 &= 4a+2b \Rightarrow 10=2a+b\\ \text{To find} \ a \ \text{and} \ b, \text{we solve the system of equations:} && 5 &= a+b\\ && 10 &= 2a+b\\ \text{Solve the first equation for} \ b: && 5 &= a+b \Rightarrow b=5-a\\ \text{Plug the first equation into the second:} && 10 &= 2a+5-a\\ \text{Solve for} \ a \ \text{and} \ b && a &= 5 \ \text{and} \ b=0\end{align*}

Therefore the equation of the quadratic function is \begin{align*}y=5x^2\end{align*}.

Example 5

Determine what type of function represents the values in the following table.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 400
1 500
2 25
3 6.25
4 1.5625

Solution

The differences between consecutive \begin{align*}y-\end{align*}values aren’t the same, and the differences between those differences aren’t the same either. So let’s check the ratios instead.

Each time the \begin{align*}x-\end{align*}value increases by one, the \begin{align*}y-\end{align*}value is multiplied by \begin{align*}\frac{1}{4}\end{align*}. Since the ratio is always the same, the function is exponential.

To find the equation for the function that represents these values, we start with the general form of an exponential function, \begin{align*}y=a \cdot b^x\end{align*}.

Here \begin{align*}b\end{align*} is the ratio between the values of \begin{align*}y\end{align*} each time \begin{align*}x\end{align*} is increased by one. The constant \begin{align*}a\end{align*} is the value of the function when \begin{align*}x = 0\end{align*}. Therefore, the function is \begin{align*}y=400 \left( \frac{1}{4} \right)^x\end{align*}.

## Perform Exponential and Quadratic Regressions with a Graphing Calculator

Earlier, you learned how to perform linear regression with a graphing calculator to find the equation of a straight line that fits a linear data set. In this section, you’ll learn how to perform exponential and quadratic regression to find equations for curves that fit non-linear data sets.

Example 6

The following table shows how many miles per gallon a car gets at different speeds.

Speed (mph) Miles per gallon
30 18
35 20
40 23
45 25
50 28
55 30
60 29
65 25
70 25

Using a graphing calculator:

a) Draw the scatterplot of the data.

b) Find the quadratic function of best fit.

c) Draw the quadratic function of best fit on the scatterplot.

d) Find the speed that maximizes the miles per gallon.

e) Predict the miles per gallon of the car if you drive at a speed of 48 mph.

Solution

Step 1: Input the data.

Press [STAT] and choose the [EDIT] option.

Input the values of \begin{align*}x\end{align*} in the first column \begin{align*}(L_1)\end{align*} and the values of \begin{align*}y\end{align*} in the second column \begin{align*}(L_2)\end{align*}. (Note: in order to clear a list, move the cursor to the top so that \begin{align*}L_1\end{align*} or \begin{align*}L_2\end{align*} is highlighted. Then press [CLEAR] and then [ENTER].)

Step 2: Draw the scatterplot.

First press [Y=] and clear any function on the screen by pressing [CLEAR] when the old function is highlighted.

Press [STATPLOT] [STAT] and [Y=] and choose option 1.

Choose the ON option; after TYPE, choose the first graph type (scatterplot) and make sure that the Xlist and Ylist names match the names on top of the columns in the input table.

Press [GRAPH] and make sure that the window is set so you see all the points in the scatterplot. In this case, the settings should be \begin{align*}30 \le x \le 80\end{align*} and \begin{align*}0 \le y \le 40\end{align*}. You can set the window size by pressing the [WINDOW] key at the top.

Press [STAT] and use the right arrow to choose [CALC].

Choose Option 5 (QuadReg) and press [ENTER]. You will see “QuadReg” on the screen.

Type in \begin{align*}L_1, L_2\end{align*} after ‘QuadReg’ and press [ENTER]. The calculator shows the quadratic function: \begin{align*}y=-0.017x^2+1.9x-25\end{align*}

Step 4: Graph the function.

Press [Y=] and input the function you just found.

Press [GRAPH] and you will see the curve fit drawn over the data points.

To find the speed that maximizes the miles per gallon, use [TRACE] and move the cursor to the top of the parabola. You can also use [CALC] [2nd] [TRACE] and option 4:Maximum, for a more accurate answer. The speed that maximizes miles per gallon is 56 mph.

Finally, plug \begin{align*}x = 48\end{align*} into the equation you found: \begin{align*}y=-0.017(48)^2+1.9(48)-25=27.032 \ miles \ per \ gallon\end{align*}.

Note: The image above shows our function plotted on the same graph as the data points from the table. One thing that is clear from this graph is that predictions made with this function won’t make sense for all values of \begin{align*}x\end{align*}. For example, if \begin{align*}x < 15\end{align*}, this graph predicts that we will get negative mileage, which is impossible. Part of the skill of using regression on your calculator is being aware of the strengths and limitations of this method of fitting functions to data.

Example 7

The following table shows the amount of money an investor has in an account each year for 10 years.

Year Value of account
1996 $5000 1997$5400
1998 $5800 1999$6300
2000 $6800 2001$7300
2002 $7900 2003$8600
2004 $9300 2005$10000
2006 11000 Using a graphing calculator: a) Draw a scatterplot of the value of the account as the dependent variable, and the number of years since 1996 as the independent variable. b) Find the exponential function that fits the data. c) Draw the exponential function on the scatterplot. d) What will be the value of the account in 2020? Solution Step 1: Input the data. Press [STAT] and choose the [EDIT] option. Input the values of \begin{align*}x\end{align*} in the first column \begin{align*}(L_1)\end{align*} and the values of \begin{align*}y\end{align*} in the second column \begin{align*}(L_2)\end{align*}. Step 2: Draw the scatterplot. First press [Y=] and clear any function on the screen. Press [GRAPH] and choose Option 1. Choose the ON option and make sure that the Xlist and Ylist names match the names on top of the columns in the input table. Press [GRAPH] make sure that the window is set so you see all the points in the scatterplot. In this case the settings should be \begin{align*}0 \le x \le 10\end{align*} and \begin{align*}0 \le y \le 11000\end{align*}. Step 3: Perform exponential regression. Press [STAT] and use the right arrow to choose [CALC]. Choose Option 0 and press [ENTER]. You will see “ExpReg” on the screen. Press [ENTER]. The calculator shows the exponential function: \begin{align*}y=4975.7(1.08)^x\end{align*} Step 4: Graph the function. Press [Y=] and input the function you just found. Press [GRAPH]. Finally, plug \begin{align*}x = 2020 - 1996 = 24\end{align*} into the function: \begin{align*}y=4975.7(1.08)^{24}=\underline{\31551.81}\end{align*} In 2020, the account will have a value of31551.81.

Note: The function above is the curve that comes closest to all the data points. It won’t return \begin{align*}y-\end{align*}values that are exactly the same as in the data table, but they will be close. It is actually more accurate to use the curve fit values than the data points.

If you don’t have a graphing calculator, there are resources available on the Internet for finding lines and curves of best fit. For example, the applet at http://science.kennesaw.edu/~plaval/applets/LRegression.html does linear regression on a set of data points; the one at http://science.kennesaw.edu/~plaval/applets/QRegression.html does quadratic regression; and the one at http://science.kennesaw.edu/~plaval/applets/ERegression.html does exponential regression. Also, programs like Microsoft Office or OpenOffice have the ability to create graphs and charts that include lines and curves of best fit.

## Solve Real-World Problems by Comparing Function Models

Example 8

The following table shows the number of students enrolled in public elementary schools in the US (source: US Census Bureau). Make a scatterplot with the number of students as the dependent variable, and the number of years since 1990 as the independent variable. Find which curve fits this data the best and predict the school enrollment in the year 2007.

Year Number of students (millions)
1990 26.6
1991 26.6
1992 27.1
1993 27.7
1994 28.1
1995 28.4
1996 28.1
1997 29.1
1998 29.3
2003 32.5

Solution

We need to perform linear, quadratic and exponential regression on this data set to see which function represents the values in the table the best.

Step 1: Input the data.

Input the values of \begin{align*}x\end{align*} in the first column \begin{align*}(L_1)\end{align*} and the values of \begin{align*}y\end{align*} in the second column \begin{align*}(L_2)\end{align*}.

Step 2: Draw the scatterplot.

Set the window size: \begin{align*}0 \le x \le 10\end{align*} and \begin{align*}20 \le y \le 40\end{align*}.

Here is the scatterplot:

Step 3: Perform Regression.

Linear Regression

The function of the line of best fit is \begin{align*}y=0.51x+26.1\end{align*}. Here is the graph of the function on the scatterplot:

The quadratic function of best fit is \begin{align*}y=0.064x^2-.067x+26.84\end{align*}. Here is the graph of the function on the scatterplot:

Exponential Regression

The exponential function of best fit is \begin{align*}y=26.2(1.018)^x\end{align*}. Here is the graph of the function on the scatterplot:

From the graphs, it looks like the quadratic function is the best fit for this data set. We’ll use this function to predict school enrollment in 2007.

Plug in \begin{align*}x = 2007 - 1990 = 17: y=0.064(17)^2-.067(17)+26.84= 44.2 \ million \ students\end{align*}.

## Review Questions

Determine whether the data in the following tables can be represented by a linear function.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-4\end{align*} 10
-3 7
-2 4
-1 1
0 -2
1 -5
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-2\end{align*} 4
-1 3
0 2
1 3
2 6
3 11
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 50
1 75
2 100
3 125
4 150
5 175

Determine whether the data in the following tables can be represented by a quadratic function.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-10\end{align*} 10
-5 2.5
0 0
5 2.5
10 10
15 22.5
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
1 4
2 6
3 6
4 4
5 0
6 -6
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-3\end{align*} -27
-2 -8
-1 -1
0 0
1 1
2 8
3 27

Determine whether the data in the following tables can be represented by an exponential function.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 200
1 300
2 1800
3 8300
4 25800
5 62700
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 120
1 180
2 270
3 405
4 607.5
5 911.25
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 4000
1 2400
2 1440
3 864
4 518.4
5 311.04

Determine what type of function represents the values in the following tables and find the equation of each function.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 400
1 500
2 625
3 781.25
4 976.5625
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-9\end{align*} -3
-7 -2
-5 -1
-3 0
-1 1
1 2
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-3\end{align*} 14
-2 4
-1 -2
0 -4
1 -2
2 4
3 14
1. As a ball bounces up and down, the maximum height that the ball reaches continually decreases from one bounce to the next. For a given bounce, this table shows the height of the ball with respect to time:
Time (seconds) Height (inches)
2 2
2.2 16
2.4 24
2.6 33
2.8 38
3.0 42
3.2 36
3.4 30
3.6 28
3.8 14
4.0 6

Using a graphing calculator:

a) Draw the scatterplot of the data

b) Find the quadratic function of best fit

c) Draw the quadratic function of best fit on the scatterplot

d) Find the maximum height the ball reaches on the bounce

e) Predict how high the ball is at time \begin{align*}t = 2.5 \ seconds\end{align*}

1. A chemist has a 250 gram sample of a radioactive material. She records the amount of radioactive material remaining in the sample every day for a week and obtains the data in the following table:
Day Weight (grams)
0 250
1 208
2 158
3 130
4 102
5 80
6 65
7 50

Using a graphing calculator:

a) Draw a scatterplot of the data

b) Find the exponential function of best fit

c) Draw the exponential function of best fit on the scatterplot

d) Predict the amount of material after 10 days.

1. The following table shows the rate of pregnancies (per 1000) for US women aged 15 to 19. (source: US Census Bureau).
1. Make a scatterplot with the rate of pregnancies as the dependent variable and the number of years since 1990 as the independent variable.
2. Find which type of curve fits this data best.
3. Predict the rate of teen pregnancies in the year 2010.
Year Rate of pregnancy (per 1000)
1990 116.9
1991 115.3
1992 111.0
1993 108.0
1994 104.6
1995 99.6
1996 95.6
1997 91.4
1998 88.7
1999 85.7
2000 83.6
2001 79.5
2002 75.4

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