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Modeling, Functions, and Graphs

Katherine Yoshiwara

Section 1.1 Linear Models

Investigation 1.1. Sales on Commission.

Delbert is offered a part-time job selling restaurant equipment. He will be paid $1000 per month plus a 6% commission on his sales. The sales manager tells Delbert he can expect to sell about $8000 worth of equipment per month. To help him decide whether to accept the job, Delbert does a few calculations.
  1. Based on the sales manager’s estimate, what monthly income can Delbert expect from this job? What annual salary would that provide?
  2. What would Delbert’s monthly salary be if he sold only $5000 of equipment per month? What would his salary be if he sold $10,000 worth per month? Compute monthly incomes for each sales total shown in the table.
    Sales Income
    5000
    8000
    10,000
    12,000
    15,000
    18,000
    20,000
    25,000
    30,000
    35,000
     
     
    grid
  3. Plot your data points on a graph, using sales, \(S\text{,}\) on the horizontal axis and income, \(I\text{,}\) on the vertical axis, as shown in the figure. Connect the data points to show Delbert’s monthly income for all possible monthly sales totals.
  4. Add two new data points to the table by reading values from your graph.
  5. Write an algebraic expression for Delbert’s monthly income, \(I\text{,}\) in terms of his monthly sales, \(S\text{.}\) Use the description in the problem to help you:
    He will be paid: $1000 . . . plus a 6% commission on his sales.
    Income \(= \fillinmath{XXXXXXXXXXXXXXX}\)
  6. Test your formula from part (5) to see if it gives the same results as those you recorded in the table.
  7. Use your formula to find out what monthly sales total Delbert would need in order to have a monthly income of $2500.
  8. Each increase of $1000 in monthly sales increases Delbert’s monthly income by .
  9. Summarize the results of your work: In your own words, describe the relationship between Delbert’s monthly sales and his monthly income. Include in your discussion a description of your graph.

Subsection Tables, Graphs and Equations

The first step in creating a model is to describe relationships between variables. In Investigation 1.1, we analyzed the relationship between Delbert’s sales and his income. Starting from a verbal description, we represented the relationship in three different ways.
  1. A table of values displays specific data points with precise numerical values.
  2. A graph is a visual display of the data. It is easier to spot trends and describe the overall behavior of the variables from a graph.
  3. An algebraic equation is a compact summary of the model. It can be used to analyze the model and to make predictions
We begin our study of modeling with some examples of linear models. In the examples that follow, observe the interplay among the three modeling tools, and how each contributes to the model.

Example 1.1.

Annelise is on vacation at a seaside resort. She can rent a bicycle from her hotel for $3 an hour, plus a $5 insurance fee. (A fraction of an hour is charged as the same fraction of $3.)
  1. Make a table of values showing the cost, \(C\text{,}\) of renting a bike for various lengths of time, \(t\text{.}\)
  2. Plot the points on a graph. Draw a curve through the data points.
  3. Write an equation for \(C\) in terms of \(t\text{.}\)
Solution.
  1. To find the cost, we multiply the time by $3, and add the result to the $5 insurance fee. For example, the cost of a 1-hour bike ride is
    \begin{align*} \text{Cost}\amp=(\$5\text{ insurance fee})+(\$3\text{ per hour})\times(\alert{1}\text{ hour})\\ C\amp=5+3(\alert{1})=8 \end{align*}
    A 1-hour bike ride costs $8. We record the results in a table, as shown here:
    Length of rental
    (hours)    		 Cost of rental
    (dollars)    		 \((t,C)\)
    \(1\) \(8\) \(\quad C=5+3(\alert{1})\quad\) \((1,8)\)
    \(2\) \(11\) \(C=5+3(\alert{2})\) \((2,11)\)
    \(3\) \(14\) \(C=5+3(\alert{3})\) \((3,14)\)
  2. Each pair of values represents a point on the graph. The first value gives the horizontal coordinate of the point, and the second value gives the vertical coordinate. The points lie on a straight line, as shown in the figure. The line extends infinitely in only one direction, because negative values of \(t\) do not make sense here.
    graph
  3. To write an equation, we let \(C\) represent the cost of the rental, and we use \(t\) for the number of hours:
    \begin{align*} \text{Cost}\amp=(\$5\text{ insurance fee})+(\$3\text{ per hour})\times\text{(number of hours)}\\ C\amp=5+3\cdot t \end{align*}

Checkpoint 1.2. QuickCheck 1.

When you graph the data given in a table, on which axis do you show the variable in the first row of the table?
  • The linear axis
  • The horizontal axis
  • The vertical axis
  • Both axes
Answer.
\(\text{The horizontal axis}\)
Solution.
The horizontal axis

Example 1.3.

Use the equation \(C=5+3\cdot t\) you found in Example 1.1 to answer the following questions. Then show how to find the answers by using the graph.
  1. How much will it cost Annelise to rent a bicycle for 6 hours?
  2. How long can Annelise bicycle for $18.50?
Solution.
  1. We substitute \(t=\alert{6}\) into the expression for \(C\) to find
    \begin{equation*} C=5+3(\alert{6})=23 \end{equation*}
    A 6-hour bike ride will cost $23. The point \(P\) on the graph in the figure represents the cost of a 6-hour bike ride. The value on the \(C\)-axis at the same height as point \(P\) is 23, so a 6-hour bike ride costs $23.
  2. We substitute \(C=\alert{18.50}\) into the equation and solve for \(t\text{.}\)
    \begin{align*} \alert{18.50}\amp=5+3t\\ 13.50\amp=3t\\ t\amp=4.5 \end{align*}
    For $18.50 Annelise can bicycle for 4½ hours. The point \(Q\) on the graph represents an $18.50 bike ride. The value on the \(t\)-axis below point \(Q\) is 4.5, so $18.50 will buy a 4.5 hour bike ride.
    line though points

Note 1.4.

In Example 1.3, notice the different algebraic techniques we used in parts (a) and (b).
  • In part (a), we were given a value of \(t\) and we evaluated the expression \(5+3t\) to find \(C\text{.}\)
  • In part (b) we were given a value of \(C\) and we solved the equation \(C=5+3t\) to find \(t\text{.}\)

Checkpoint 1.5. Practice 1.

Frank plants a dozen corn seedlings, each 6 inches tall. With plenty of water and sunlight they will grow approximately 2 inches per day. Complete the table of values for the height, \(h\text{,}\) of the seedlings after \(t\) days.
Complete the table of values for the height, \(h\) of the seedlings after \(t\) days.
\(t\) \(0\) \(5\) \(10\) \(15\) \(20\)
\(h\)
  1. Write an equation for the height \(h\) of the seedlings in terms of the number \(t\) of days since they were planted.
    Equation:
  2. Graph the equation.
Answer 1.
\(6\)
Answer 2.
\(16\)
Answer 3.
\(26\)
Answer 4.
\(36\)
Answer 5.
\(46\)
Answer 6.
\(h-2t = 6\)
Solution.
\(t\) \(0\) \(5\) \(10\) \(15\) \(20\)
\(h\) 6 16 26 36 46
  1. \(\displaystyle h=6+2t\)
  2. The graph of seedling height vs time is shown below.
The graph of seedling height vs time for part (b) is shown below.
graph of seedlings height vs time

Checkpoint 1.6. Practice 2.

Use your equation from Practice 1 to answer the questions. Illustrate each answer on the graph.
  1. How tall is the corn after 3 weeks? Use “ft” for feet or “in” for inches.
    Answer (including units):
  2. How long will it be before the corn is 6 feet tall? Use “day” for days.
    Answer (including units):
Answer 1.
\(48\ {\rm in}\)
Answer 2.
\(33\ {\rm day}\)
Solution.
  1. 48 inches tall
  2. 33 days
A graph is below.
The graph below illustrates the answers.
graph of seedlings height vs time

Checkpoint 1.7. Pause and Reflect.

What is the difference between an expression and an equation?

Subsection Choosing Scales for the Axes

To create a useful graph, we must choose appropriate scales for the axes.
  • The axes must extend far enough to show the values of the variables.
  • The tick marks should be equally spaced.
  • Usually we should use no more than 10 or 15 tick marks.

Example 1.8.

In 1990, the median price of a home in the US was $92,000. The median price increased by about $4700 per year over the next decade.
  1. Make a table of values showing the median price of a house in 1990, 1994, 1998, and 2000.
  2. Choose suitable scales for the axes and plot the values you found in part (a) on a graph. Use \(t\text{,}\) the number of years since 1990, on the horizontal axis and the price of the house, \(P\text{,}\) on the vertical axis. Draw a curve through the points.
  3. Write an equation that expresses \(P\) in terms of \(t\text{.}\)
  4. How much did the price of the house increase from 1990 to 1996? Illustrate the increase on your graph.
Solution.
  1. In 1990 the median price was $92,000. Four years later, in 1994, the price had increased by \(\alert{4}(4700)=18,800\) dollars, so
    \begin{equation*} P=92,000+\alert{4}(4700)=110,800 \end{equation*}
    In 1998 the price had increased by \(\alert{8}(4700)=37,600\) dollars, so
    \begin{equation*} P=92,000+\alert{8}(4700)=129,600 \end{equation*}
    You can verify the price of the house in 2000 by a similar calculation.
    Year Price of House) \((t,P)\)
    \(1990\) \(92,000\) \((0,\, 92,000)\)
    \(1994\) \(110,800\) \((4,\, 110,800)\)
    \(1998\) \(129,600\) \((8,\, 129,600)\)
    \(2000\) \(139,000\) \((10,\, 139,000)\)
  2. We let \(t\) stand for the number of years since 1990, so that \(t=0\) in 1990, \(t=4\) in 1994, and so on. To choose scales for the axes, we look at the values in the table. For this graph we scale the horizontal axis, or \(t\)-axis, in 1-year intervals and the vertical axis, or \(P\)-axis, for $90,000 to $140,000 in intervals of $5,000. The points lie on a straight line, as shown in the figure.
    grid
  3. Look back at the calculations in part (a). The price of the house started at $92,000 in 1990 and increased by \(t \times 4700\) dollars after \(t\) years. Thus,
    \begin{equation*} P=92,000+4700t \end{equation*}
  4. We find the points on the graph for 1990 and 1996. These points lie above \(t=0\) and \(t=6\) on the \(t\)-axis. Next we find the values on the \(P\)-axis corresponding to the two points. The values are \(P=92,000\) in 1990 and \(P=120,200\) in 1996. The increase in price is the difference of the two \(P\)-values.
    \begin{align*} \text{increase in price}\amp=120,200-92,000\\ \amp=28,200 \end{align*}
    The price of the home increased $28,200 between 1990 and 1996. This increase is indicated by the arrows in the figure.

Checkpoint 1.9. QuickCheck 2.

If \(C\) is expressed in terms of \(H\text{,}\) which variable goes on the horizontal axis?
  • \(\displaystyle C\)
  • \(\displaystyle H\)
  • \(\displaystyle x\)
  • The smaller one
Answer.
\(\text{Choice 2}\)
Solution.
\(H\)

Note 1.10.

The graphs in the preceding examples are increasing graphs . As we move along the graph from left to right (in the direction of increasing \(t\) ), the second coordinate increases as well. Try Checkpoint 1.11, which illustrates a decreasing graph.

Checkpoint 1.11. Practice 3.

Silver Lake has been polluted by industrial waste products. The concentration of toxic chemicals in the water is currently 285 parts per million (ppm). Environmental officials would like to reduce the concentration by 15 ppm each year.
  1. Complete the table of values showing the desired concentration, \(C\text{,}\) of toxic chemicals \(t\) years from now. For each \(t\)-value, calculate the corresponding value for \(C\text{.}\) Write your answers as ordered pairs.
    \(t\) \(C\) \((t,C)\)
    \(0\) \(C=285-15(\alert{0})\) (0, )
    \(5\) \(C=285-15(\alert{5})\) (5, )
    \(10\) \(C=285-15(\alert{10})\) (10, )
    \(15\) \(C=285-15(\alert{15})\) (15, )
  2. To choose scales for the axes, notice that the value of \(C\) starts at 285 and decreases from there. We’ll scale the vertical axis up to 300, and use 10 tick marks at intervals of 30. Graph the ordered pairs on the grid, and connect them with a straight line.
  3. Write an equation for the concentration, \(C\text{,}\) of toxic chemicals \(t\) years from now.
    Equation:
Hint.
For part (c): The concentration is initially 285 ppm, and we subtract 15 ppm for each year that passes, or \(15 \times t\text{.}\)
Answer 1.
\(285\)
Answer 2.
\(210\)
Answer 3.
\(135\)
Answer 4.
\(60\)
Answer 5.
\(C+15t = 285\)
Solution.
  1. \((t,C)\)
    \((0,285)\)
    \((5,210)\)
    \((10,135)\)
    \((15,60)\)
  2. The graph is shown below.
  3. \(\displaystyle C=285-15t\)
The graph for part(b):
decreasing graph

Note 1.12.

In the previous Exercise, we extend the graph until it reaches the horizontal axis, but no farther. Points with negative \(C\)-coordinates have no meaning for the problem.

Checkpoint 1.13. QuickCheck 3.

If \(x \gt 5\text{,}\) what is true about \(-2x\text{?}\)
  • It is greater than 3
  • It is less than 3
  • It is greater than 10
  • It is less than -10
Answer.
\(\text{It is less than -10}\)
Solution.
\(-2x\) is less than \(-10\text{.}\)

Technology 1.14. Graphing an Equation.

We can use a graphing utililty to graph an equation. On most utilities, we follow three steps.
To Graph an Equation:
  1. Press Y= and enter the equation you wish to graph.
  2. Press WINDOW and select a suitable graphing window.
  3. Press GRAPH

Example 1.15. Using a Graphing Utility.

In Example 1.8, we found the equation
\begin{equation*} P = 92,000 + 4700t \end{equation*}
for the median price of a house \(t\) years after 1990. Graph this equation with technology.
Solution.
To begin, we press Y= and enter
\begin{equation*} Y1 = 92,000 + 4700X \end{equation*}
For this graph, we’ll use the grid in Example 1.8 for our window settings, so we press WINDOW and enter
Xmin\(=0\) Xmax\(=10\)
Ymin\(=90,000\) Ymax\(=140,000\)
Finally, we press GRAPH. The graph is shown in the figure.
grid

Checkpoint 1.16. Practice 4.

  1. Solve the equation \(2y-1575=45x\) for \(y\) in terms of \(x\text{.}\)
    \(y=\)
  2. Graph the equation with a graphing utility. Use the window
    Xmin\(=-50\) Xmax\(=50\) Xscl\(=5\)
    Ymin\(=-500\) Ymax\(=1000\) Yscl\(=100\)
  3. Sketch the graph on paper. Use the window settings to choose appropriate scales for the axes.
Answer.
\(\frac{1575+45x}{2}\)
Solution.
  1. \(\displaystyle y=(1575+45x)/2\)
  2. The calculator graph is shown below.
  3. The graph is shown below.
The graph for (b):
calculator graph
c.
graph

Checkpoint 1.17. QuickCheck 4.

bad grid
What is wrong with the grid above?
  • The grid lines on the \(x\)-axis are not evenly spaced.
  • The scale on the \(y\)-axis does not start at 0.
  • The axes are not labeled with the variables.
  • All of the above.
Answer.
\(\text{Choice 4}\)

Subsection Linear Equations

All the models in the preceding examples have equations with a similar form:
\begin{equation*} \blert{y=\text{(starting value)}+\text{(rate of change)}\cdot x} \end{equation*}
(We’ll talk more about rate of change in Section 1.4.) Their graphs were all portions of straight lines. For this reason such equations are called linear equations. The order of the terms in the equation does not matter. For example, the equation in Example 1.1,
\begin{equation*} C=5+3t \end{equation*}
can be written equivalently as
\begin{equation*} -3t+C=5 \end{equation*}
and the equation in Example 1.8,
\begin{equation*} P=92,000+4700t \end{equation*}
can be written as
\begin{equation*} -4700t +P=92,000 \end{equation*}
This form of a linear equation, \(Ax+By=C \text{,}\) is called the general form.

General Form for a Linear Equation.

The graph of any equation
\begin{equation*} Ax+By=C \end{equation*}
where \(A\) and \(B\) are not both equal to zero, is a straight line.

Example 1.18.

The manager at Albert’s Appliances has $3000 to spend on advertising for the next fiscal quarter. A 30-second spot on television costs $150 per broadcast, and a 30-second radio ad costs $50.
  1. The manager decides to buy \(x\) television ads and \(y\) radio ads. Write an equation relating \(x\) and \(y\text{.}\)
  2. Make a table of values showing several choices for \(x\) and \(y\text{.}\)
  3. Plot the points from your table, and graph the equation.
Solution.
  1. Each television ad costs $150, so \(x\) ads will cost $\(150x\text{.}\) Similarly, \(y\) radio ads will cost $\(50y\text{.}\) The manager has $3000 to spend, so the sum of the costs must be $3000. Thus,
    \begin{equation*} 150x+50y=3000 \end{equation*}
  2. We choose some values of \(x\text{,}\) and solve the equation for the corresponding value of \(y\text{.}\) For example, if \(x=\alert{10}\) then
    \begin{align*} 150(\alert{10})+50y\amp=300\\ 1500+50y\amp=3000\\ 50y\amp=1500\\ y\amp=30 \end{align*}
    If the manager buys 10 television ads, she can also buy 30 radio ads. You can verify the other entries in the table.
    \(x\) \(8\) \(10\) \(12\) \(14\)
    \(y\) \(36\) \(30\) \(24\) \(18\)
  3. We plot the points from the table. All the solutions lie on a straight line, as shown in the figure.
    grid

Checkpoint 1.19. Practice 5.

In central Nebraska, each acre of corn requires 25 acre-inches of water per year, and each acre of winter wheat requires 18 acre-inches of water. (An acre-inch is the amount of water needed to cover one acre of land to a depth of one inch.) A farmer can count on 9000 acre-inches of water for the coming year. (Source: Institute of Agriculture and Natural Resources, University of Nebraska)
  1. Write an equation relating the number of acres of corn, \(x\text{,}\) and the number of acres of wheat, \(y\text{,}\) that the farmer can plant.
  2. Complete the table. Round your answers to tenths.
    \(x\) \(50\) \(100\) \(150\) \(200\)
    \(y\)
Answer 1.
\(25x+18y = 9000\)
Answer 2.
\(430.556\)
Answer 3.
\(361.111\)
Answer 4.
\(291.667\)
Answer 5.
\(222.222\)
Solution.
  1. \(\displaystyle 25x+18y=9000\)
  2. \(x\) \(50\) \(100\) \(150\) \(200\)
    \(y\) \(430.6\) \(361.1\) \(291.7\) \(222.2\)

Checkpoint 1.20. Pause and Reflect.

Write down two different equation forms for linear models. Which do you think is easier to use?

Subsection Intercepts

intercepts
Consider the graph of the equation
\begin{equation*} 3x-4y=12 \end{equation*}
shown at left. The points where the graph crosses the axes are called the intercepts of the graph. The coordinates of these points are easy to find.
The \(y\)-coordinate of the \(x\)-intercept is zero, so we set \(y=\alert{0}\) in the equation to get
\begin{align*} 3(\alert{0})-4x\amp=12\\ x=-3 \end{align*}
The \(x\)-intercept is the point \((-3,0)\text{.}\) Also, the \(x\)-coordinate of the \(y\)-intercept is zero, so we set \(x=\alert{0}\) in the equation to get
\begin{gather*} 3y-4(\alert{0})=12\\ y=4 \end{gather*}
The \(y\)-intercept is \((0,4)\text{.}\)

Checkpoint 1.21. QuickCheck 5.

What is the \(y\)-coordinate of any point on the \(x\)-axis?
  • \(\displaystyle (x,y)\)
  • \(\displaystyle Y\)
  • It depends on the value of \(x\)
  • 0
Answer.
\(\text{Choice 4}\)
Solution.
\(0\)

Intercepts of a Graph.

The points where a graph crosses the axes are called the intercepts of the graph.
  1. To find the \(y\)-intercept, set \(x=0\) and solve for \(y\text{.}\)
  2. To find the \(x\)-intercept, set \(y=0\) and solve for \(x\)
The intercepts of a graph tell us something about the situation it models.

Example 1.22.

  1. Find the intercepts of the graph in Checkpoint 1.11, about the pollution in Silver Lake.
  2. What do the intercepts tell us about the problem?
Solution.
  1. An equation for the concentration of toxic chemicals is
    \begin{equation*} C=285-15t \end{equation*}
    To find the \(C\)-intercept, set \(t\) equal to zero.
    \begin{equation*} C=285-15(\alert{0})=285 \end{equation*}
    The \(C\)-intercept is the point \((0, 285)\text{,}\) or simply 285.
    To find the \(t\)-intercept, set \(C\) equal to zero and solve for \(t\text{.}\)
    \begin{equation*} \begin{aligned}[t] \alert{0}\amp =285-15t \amp \amp \blert{\text{Add }15t \text{ to both sides.}}\\ 15t\amp =285 \amp \amp \blert{\text{Divide both sides by 15.}}\\ t\amp =19 \end{aligned} \end{equation*}
    The \(t\)-intercept is the point \((19,0)\text{,}\) or simply \(19\text{.}\)
  2. The \(C\)-intercept represents the concentration of toxic chemicals in Silver Lake now: When \(t=0\text{,}\) \(C=285\text{,}\) so the concentration is currently \(285\) ppm.
    The \(t\)-intercept represents the number of years it will take for the concentration of toxic chemicals to drop to zero: When \(C=0\text{,}\) \(t=19\text{,}\) so it will take \(19\) years for the pollution to be eliminated entirely.

Checkpoint 1.23. QuickCheck 6.

Delbert says that the intercepts of the line \(3x+5y=30\) are \((10,6)\text{.}\) What is wrong with his answer?
  • \((10,6)\) is not on the \(x\)-axis.
  • An intercept must have a 0 coordinate.
  • The line has two intercepts.
  • All of the above
Answer.
\(\text{Choice 4}\)
Solution.
All of the above

Checkpoint 1.24. Practice 6.

Find the intercepts of the graph in Example 1.18, about the advertising budget for Albert’s Appliances: \(150x + 50y = 3000\text{.}\)
  1. Enter each intercept as an ordered pair.
    The \(x\)-intercept is .
    The \(y\)-intercept is .
  2. What do the intercepts tell us about the problem?
    The \(x\)-intercept tells us:
    • A) the cost of the TV ads.
    • B) the best number of TV ads.
    • C) the most TV ads she can buy for the money.
    • D) the average number of TV ads.
    The \(y\)-intercept tells us:
    • A) the cost of the radio ads.
    • B) the number of radio ads she bought
    • C) the number of radio ads if no TV ads are bought.
    • D) the average cost per ad.
Answer 1.
\(\left(20,0\right)\)
Answer 2.
\(\left(0,60\right)\)
Answer 3.
\(\text{C) the ... the money.}\)
Answer 4.
\(\text{C) the ... are bought.}\)
Solution.
  1. We find the \(x\)-intercept by setting \(y=0\) and solving for \(x\) to learn that \(x=20\text{,}\) so the \(x\)-intercept is \({\left(20,0\right)}\text{.}\)
    We find the \(y\)-intercept by setting \(x=0\) and solving for \(y\) to learn that \(y=60\text{,}\) so the \(y\)-intercept is \({\left(0,60\right)}\text{.}\)
  2. The \(x\)-intercept has \(y=0\text{,}\) that is, it corresponds to when there are zero radio ads. The \(y\)-intercept has \(x=0\text{,}\) that is, it corresponds to when there are zero tv ads.

Checkpoint 1.25. Pause and Reflect.

Explain how the words intercept and intersect are related, and how they are different.

Subsection Intercept Method for Graphing Lines

Because we really only need two points to graph a linear equation, we might as well find the intercepts first and use them to draw the graph. The values of the intercepts will also help us choose suitable scales for the axes. It is always a good idea to find a third point as a check.

Example 1.26.

  1. Find the \(x\)- and \(y\)-intercepts of the graph of \(150x - 180y = 9000\text{.}\)
  2. Use the intercepts to graph the equation. Find a third point as a check.
Solution.
  1. To find the \(x\)-intercept, we set \(y = \alert{0}\text{.}\)
    \begin{equation*} \begin{aligned}[t] 150x-18(\alert{0})\amp =9000 \amp \amp \blert{\text{Simpify.}}\\ 150x\amp =9000 \amp \amp \blert{\text{Divide both sides by 150.}}\\ x\amp =60 \amp \amp \end{aligned} \end{equation*}
    The \(x\)-intercept is the point \((60, 0)\text{.}\) To find the \(y\)-intercept, we set \(x = \alert{0}\text{.}\)
    \begin{equation*} \begin{aligned}[t] 150(\alert{0})-18y\amp =9000 \amp \amp \blert{\text{Simpify.}}\\ -180y\amp =9000 \amp \amp \blert{\text{Divide both sides by } -180\text{.}}\\ y\amp =-50 \amp \amp \end{aligned} \end{equation*}
    The \(y\)-intercept is the point \((0, -50)\text{.}\)
  2. We scale both axes in intervals of 10 and then plot the two intercepts, \((60, 0)\) and \((0, -50)\text{.}\) We draw the line through them, as shown below. Finally, we find another point and check that it lies on this line. We choose \(x = \alert{20}\) and solve for \(y\text{.}\)
    \begin{align*} 150(\alert{20}) -180y \amp = 9000\\ 3000 -180y \amp = 9000\\ -180y \amp = 6000\\ y \amp =-33.\overline{3} \end{align*}
    We plot the point \((20, -33\frac{1}{3})\text{.}\) Because this point lies on the line, we can be reasonably confident that our graph is correct.
    grid

Checkpoint 1.27. QuickCheck 7.

Is it possible for the \(x\)-intercept and the \(y\)-intercept of a line to be the same point?
  • No
  • Yes
  • Only for a vertical line
  • They are always the same point.
Answer.
\(\text{Yes}\)
Solution.
Yes

To Graph a Line Using the Intercept Method:.

  1. Find the intercepts of the line.
    1. To find the \(x\)-intercept, set \(y=0\) and solve for \(x\text{.}\)
    2. To find the \(y\)-intercept, set \(x=0\) and solve for \(y\text{.}\)
  2. Plot the intercepts.
  3. Choose a value for \(x\) and find a third point on the line.
  4. Draw a line through the points.

Checkpoint 1.28. QuickCheck 8.

How many points do you need to graph a linear equation?
  • Two
  • Three
  • One in each quadrant
  • It depends on the equation
Answer.
\(\text{Two}\)
Solution.
Two

Technology 1.29. Choosing a Graphing Window.

Knowing the intercepts can also help us choose a suitable window on a graphing utility. We would like the window to be large enough to show the intercepts. For the graph in the example above, we can enter the equation
\begin{equation*} Y = (9000 -150X)/{-180} \end{equation*}
in the window
Xmin\(=-20\) Xmax\(=70\)
Ymin\(=-70\) Ymax\(=30\)

Checkpoint 1.30. Practice 7.

In Checkpoint 1.19 you wrote an equation about crops in Nebraska.
  1. Find the intercepts of the graph.
    Note: Enter each intercept as an ordered pair.
    The \(x\)-intercept is .
    The \(y\)-intercept is .
  2. Use the intercepts to help you choose appropriate scales for the axes, and then graph the equation.
  3. What do the intercepts tell us about the problem?
    The \(x\)-intercept tells us:
    • A) the total water usage of corn.
    • B) the cost of the corn if no wheat is planted.
    • C) the optimal number of acres of corn.
    • D) the total acres if all corn is planted.
    The \(y\)-intercept tells us:
    • A) the largest possible number of acres of wheat.
    • B) the amount of water used by the wheat.
    • C) the total cost of the wheat.
    • D) how many acres are left after the wheat is planted.
Answer 1.
\(\left(360,0\right)\)
Answer 2.
\(\left(0,500\right)\)
Answer 3.
\(\text{D) the ... is planted.}\)
Answer 4.
\(\text{A) the ... of wheat.}\)
Solution.
  1. We find the \(x\)-intercept by setting \(y=0\) and solving for \(x\) to learn that \(x=360\text{,}\) so the \(x\)-intercept is \({\left(360,0\right)}\text{.}\)
    We find the \(y\)-intercept by setting \(x=0\) and solving for \(y\) to learn that \(y=500\text{,}\) so the \(y\)-intercept is \({\left(0,500\right)}\text{.}\)
  2. The \(x\)-intercept has \(y=0\text{,}\) that is, it corresponds to when there are zero radio ads. The \(y\)-intercept has \(x=0\text{,}\) that is, it corresponds to when there are zero tv ads.
A graph is shown below.
graph of acres of wheat vs acres of corn

Note 1.31.

The examples in this section model simple linear relationships between two variables. Such relationships, in which the value of one variable is determined by the value of the other, are called functions. We will study various kinds of functions throughout the course.

Checkpoint 1.32. Pause and Reflect.

What was the most difficult part of this section to understand? Write a question whose answer would help you understand it.

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.
  • Variable
  • Solve an equation
  • Evaluate an expression
  • Linear equation
  • Increasing graph
  • Decreasing graph
  • Intercept
  • Mathematical model

Subsubsection CONCEPTS

  1. We can describe a relationship between variables with a table of values, a graph, or an equation.
  2. Linear models have equations of the following form:
    \begin{equation*} y = (\text{starting value}) + (\text{rate of change})\cdot x \end{equation*}
  3. To make a useful graph, we must choose appropriate scales for the axes.
  4. General Form for a Linear Equation.
    The graph of any equation
    \begin{equation*} Ax+By=C \end{equation*}
    where \(A\) and \(B\) are not both equal to zero, is a straight line.
  5. The intercepts of a graph are the points where the graph crosses the axes.
  6. We can use the intercepts to graph a line.
    To Graph a Line Using the Intercept Method:.
    1. Find the intercepts of the line.
      1. To find the \(x\)-intercept, set \(y=0\) and solve for \(x\text{.}\)
      2. To find the \(y\)-intercept, set \(x=0\) and solve for \(y\text{.}\)
    2. Plot the intercepts.
    3. Choose a value for \(x\) and find a third point on the line.
    4. Draw a line through the points.
  7. The intercepts are also useful for interpreting a model.

Subsubsection STUDY QUESTIONS

  1. Name three ways to represent a relationship between two variables.
  2. If \(C\) is expressed in terms of \(H\text{,}\) which variable goes on the horizontal axis?
  3. Explain the difference between evaluating an expression and solving an equation.
  4. How many points do you need to graph a linear equation?
  5. Explain how the words intercept and intersect are related; explain how they are different.
  6. Delbert says that the intercepts of the line \(3x + 5y = 30\) are \((10, 6)\text{.}\) What is wrong with his answer?

Subsubsection SKILLS

Practice each skill in the Homework problems listed.
  1. Make a table of values: #1–4, 7 and 8
  2. Plot points and draw a graph: #1–4, 7 and 8
  3. Choose appropriate scales for the axes: #5–12
  4. Write a linear model of the form \(y = (\text{starting value}) + (\text{rate of change})\cdot x\text{:}\) #1–8
  5. Write a linear model in general form: #25–28, 33–36
  6. Evaluate a linear expression, algebraically and graphically: #1–4
  7. Solve a linear equation, algebraically and graphically: #1–4
  8. Find the intercepts of a graph: #5 and 6, 13–24, 45–52
  9. Graph a line by the intercept method: #5 and 6, 13–24
  10. Interpret the meaning of the intercepts: #5 and 6, 25–28
  11. Use a graphing calculator to graph a line: #37–52
  12. Sketch on paper a graph obtained on a calculator: #37–44

Exercises Homework 1.1

1.

The temperature in the desert at 6 a.m., just before sunrise, was \(65\degree\)F. The temperature rose \(5\) degrees every hour until it reached its maximum value at about 5 p.m. Complete the table of values for the temperature, \(T\text{,}\) at \(h\) hours after 6 a.m.
\(h\) \(0\) \(3\) \(6\) \(9\) \(10\)
\(T\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  1. Write an equation for the temperature, \(T\text{,}\) in terms of \(h\text{.}\)
  2. Graph the equation.
    grid
  3. How hot is it at noon? Illustrate the answer on your graph.
  4. When will the temperature be \(110\degree\)F? Illustrate the answer on your graph.

2.

The taxi out of Dulles Airport charges a traveler with one suitcase an initial fee of $\(2.00\text{,}\) plus $\(1.50\) for each mile traveled. Complete the table of values showing the charge, \(C\text{,}\) for a trip of \(n\) miles.
\(n\) \(0\) \(5\) \(10\) \(15\) \(20\) \(25\)
\(C\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  1. Write an equation for the charge, \(C\text{,}\) in terms of the number of miles traveled, \(n\text{.}\)
  2. Graph the equation.
    grid
  3. What is the charge for a trip to Mount Vernon, \(40\) miles from the airport? Illustrate the answer on your graph.
  4. If a ride to the National Institutes of Health (NIH) costs $\(39.50\text{,}\) how far is it from the airport to the NIH? Illustrate the answer on your graph.

3.

On October 31, Betty and Paul fill their \(250\)-gallon oil tank for their heater. Beginning in November, they use an average of \(15\) gallons of oil per week. Complete the table of values for the amount of oil, \(A\text{,}\) left in the tank after \(w\) weeks.
\(w\) \(0\) \(4\) \(8\) \(12\) \(16\)
\(A\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  1. Write an equation that expresses the amount of oil, \(A\text{,}\) in the tank in terms of the number of weeks, \(w\text{,}\) since October 31.
  2. Graph the equation.
    graph of gallons vs weeks
  3. How much did the amount of fuel oil in the tank decrease between the third week and the eighth week? Illustrate this amount on the graph.
  4. When will the tank contain more than \(175\) gallons of fuel oil? Illustrate on the graph.

4.

Leon’s camper has a \(20\)-gallon gas tank, and he gets \(12\) miles to the gallon. (That is, he uses \(\frac{1}{12}\) gallon per mile.) Complete the table of values for the amount of gas, \(g\text{,}\) left in Leon’s tank after driving \(m\) miles.
\(m\) \(0\) \(48\) \(96\) \(144\) \(192\)
\(g\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  1. Write an equation that expresses the amount of gas, \(g\text{,}\) in Leon’s fuel tank in terms of the number of miles, \(m\text{,}\) he has driven.
  2. Graph the equation.
    grid
  3. How much gas will Leon use between 8 a.m., when his odometer reads \(96\) miles, and 9 a.m., when the odometer reads \(144\) miles? Illustrate on the graph.
  4. If Leon has less than \(5\) gallons of gas left, how many miles has he driven? Illustrate on the graph.

5.

Phil and Ernie buy a used photocopier for $\(800\) and set up a copy service on their campus. For each hour that the copier runs, Phil and Ernie make $\(40\text{.}\)
  1. Write an equation that expresses Phil and Ernie’s profit (or loss), \(P\text{,}\) in terms of the number of hours, \(t\text{,}\) they run the copier.
  2. Find the intercepts and sketch the graph. (Suggestion: Scale the horizontal axis from \(0\) to \(40\) in increments of \(5\text{,}\) and scale the vertical axis from \(-1000\) to \(400\) in increments of \(100\text{.}\))
  3. What do the intercepts tell us about the profit?

6.

A deep-sea diver is taking some readings at a depth of \(400\) feet. He begins rising at \(20\) feet per minute.
  1. Write an equation that expresses the diver’s altitude, \(h\text{,}\) in terms of the number of minutes, \(m\text{,}\) elapsed. (Consider a depth of \(400\) feet as an altitude of \(-400\) feet.)
  2. Find the intercepts and sketch the graph. (Suggestion: Scale the horizontal axis from \(0\) to \(24\) in increments of \(2\text{,}\) and scale the vertical axis from \(-500\) to \(100\) in increments of \(50\text{.}\))
  3. What do the intercepts tell us about the diver’s depth?

7.

There are many formulas for estimating the annual cost of driving. The Automobile Club estimates that fixed costs for a small car—including insurance, registration, depreciation, and financing—total about $\(5000\) per year. The operating costs for gasoline, oil, maintenance, tires, and so forth are about \(12.5\) cents per mile. (Source: Automobile Association of America)
  1. Write an equation for the annual driving cost, \(C\text{,}\) in terms of \(d\text{,}\) the number of miles driven.
  2. Complete the table of values.
    Miles Driven \(4000\) \(8000\) \(12,000\) \(16,000\) \(20,000\)
    Cost ($) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  3. Choose scales for the axes and graph the equation.
  4. How much does the annual cost of driving increase when the mileage increases from \(8000\) to \(12,000\) miles? Illustrate this amount on the graph.
  5. How much mileage will cause the annual cost to exceed $\(7000\text{?}\) Illustrate on the graph.

8.

The boiling point of water changes with altitude. At sea level, water boils at \(212\degree\)F, and the boiling point diminishes by approximately \(0.002\degree\)F for each \(1\)-foot increase in altitude.
  1. Write an equation for the boiling point, \(B\text{,}\) in terms of \(a\text{,}\) the altitude in feet.
  2. Complete the table of values.
    Altitude (ft) \(-500\) \(0\) \(1000\) \(2000\) \(3000\) \(4000\) \(5000\)
    Boiling point (\(\degree\)F) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  3. Choose scales for the axes and graph the equation.
  4. How much does the boiling point decrease when the altitude increases from \(1000\) to \(3000\) feet? Illustrate this amount on the graph.
  5. At what altitudes is the boiling point less than \(204\degree\)F? Illustrate on the graph.

Exercise Group.

For each table, choose appropriate scales for the axes and plot the given points.
9.
\(x\) \(0\) \(80\) \(90\) \(120\)
\(y\) \(6\) \(2\) \(1.5\) \(1\)
10.
\(x\) \(300\) \(500\) \(800\) \(1100\)
\(y\) \(1.2\) \(1.3\) \(1.5\) \(1.9\)
11.
\(x\) \(0.01\) \(0.03\) \(0.06\) \(0.07\)
\(y\) \(-0.2\) \(-1\) \(-1.1\) \(-2\)
12.
\(x\) \(0.003\) \(0.005\) \(0.008\) \(0.011\)
\(y\) \(6\) \(2\) \(1.5\) \(1\)

Exercise Group.

For Problems 13-18,
  1. Find the intercepts of the graph.
  2. Graph the equation by the intercept method.
13.
\(x + 2y = 8\)
14.
\(2x - y = 6\)
15.
\(3x - 4y =12\)
16.
\(2x + 6y = 6\)
17.
\(\displaystyle{\frac{x}{9}- \frac{y}{4}= 1}\)
18.
\(\displaystyle{\frac{x}{5}+ \frac{y}{8}= 1}\)

Exercise Group.

For Problems 19-24,
  1. Find the intercepts of the graph.
  2. Use the intercepts to choose scales for the axes, and then graph the equation by the intercept method.
19.
\(20x = 30y - 45,000\)
20.
\(30x = 45y + 60,000\)
21.
\(0.4x + 1.2y = 4.8\)
22.
\(3.2x - 0.8y = 12.8\)
23.
\(\displaystyle{\frac{2x}{3}+ \frac{3y}{11}= 1}\)
24.
\(\displaystyle{\frac{8x}{7}- \frac{2y}{7}= 1}\)

25.

The owner of a gas station has $\(19,200\) to spend on unleaded gas this month. Regular unleaded costs him $\(2.40\) per gallon, and premium unleaded costs $\(3.20\) per gallon.
  1. How much do \(x\) gallons of regular cost? How much do \(y\) gallons of premium cost?
  2. Write an equation in general form that relates the amount of regular unleaded gasoline, \(x\text{,}\) the owner can buy and the amount of premium unleaded, \(y\text{.}\)
  3. Find the intercepts and sketch the graph.
  4. What do the intercepts tell us about the amount of gasoline the owner can purchase?

26.

Five pounds of body fat is equivalent to \(16,000\) calories. Carol can burn \(600\) calories per hour bicycling and \(400\) calories per hour swimming.
  1. How many calories will Carol burn in \(x\) hours of cycling? How many calories will she burn in \(y\) hours of swimming?
  2. Write an equation in general form that relates the number of hours, \(x\text{,}\) of cycling and the number of hours, \(y\text{,}\) of swimming Carol needs to perform in order to lose \(5\) pounds.
  3. Find the intercepts and sketch the graph.
  4. What do the intercepts tell us about Carol’s exercise program?

27.

Delbert must increase his daily potassium intake by \(1800\) mg. He decides to eat a combination of figs and bananas, which are both low in sodium. There are \(9\) mg potassium per gram of fig, and \(4\) mg potassium per gram of banana.
  1. How much potassium is in \(x\) grams of fig? How much potassium is in \(y\) grams of banana?
  2. Write an equation in general form that relates the number of grams, \(x\text{,}\) of fig and the number of grams, \(y\text{,}\) of banana Delbert needs to get \(1800\) mg of potassium.
  3. Find the intercepts and sketch the graph.
  4. What do the intercepts tell us about Delbert’s diet?

28.

Leslie plans to invest some money in two CD accounts. The first account pays \(3.6\%\) interest per year, and the second account pays \(2.8\%\) interest per year. Leslie would like to earn $\(500\) per year on her investment.
  1. If Leslie invests \(x\) dollars in the first account, how much interest will she earn? How much interest will she earn if she invests \(y\) dollars in the second account?
  2. Write an equation in general form that relates \(x\) and \(y\) if Leslie earns \(\$500\) interest.
  3. Find the intercepts and sketch the graph.
  4. What do the intercepts tell us about Leslie’s investments?

29.

Find the intercepts of the graph for each equation.
  1. \(\displaystyle \displaystyle{\frac{x}{3}+\frac{y}{5}=1} \)
  2. \(\displaystyle \displaystyle{2x - 4y = 1} \)
  3. \(\displaystyle \displaystyle{\frac{2x}{5}-\frac{2y}{3}=1} \)
  4. \(\displaystyle \displaystyle{\frac{x}{p}+\frac{y}{q}=1} \)
\(\hphantom{00}\) e. Why is the equation \(\displaystyle{\frac{x}{a}+\frac{y}{b}=1} \) called the intercept form for a line?

30.

Write an equation in intercept form (see Problem 29) for the line with the given intercepts. Then write the equation in general form.
  1. \(\displaystyle (6, 0), (0, 2) \)
  2. \(\displaystyle (-3, 0), (0, 8) \)
  3. \(\displaystyle \left(\dfrac{3}{4}, 0\right), \left(0, \dfrac{-1}{4}\right) \)
  4. \(\displaystyle (v, 0), (0, -w) \)
  5. \(\displaystyle \left(\dfrac{1}{H}, 0\right), \left(0, \dfrac{1}{T}\right) \)

31.

  1. Find the \(y\)-intercept of the line \(y = mx + b\text{.}\)
  2. Find the \(x\)-intercept of the line \(y = mx + b\text{.}\)

32.

  1. Find the \(y\)-intercept of the line \(Ax + By = C\text{.}\)
  2. Find the \(x\)-intercept of the line \(Ax + By = C\text{.}\)

Exercise Group.

Write an equation in general form for each line.
33.
graph of line
34.
graph of line
35.
graph of line
36.
graph of line

Exercise Group.

For Problems 37–44,
  1. Solve each equation for \(y\) in terms of \(x\text{.}\) (See the Algebra Skills Refresher Section A.2 to review this skill.)
  2. Graph the equation with your graphing utility in the specified window.
  3. Make a pencil and paper sketch of the graph. Label the scales on your axes, and the coordinates of the intercepts.
37.
\(2+y=6\)
\begin{align*} {\text{Xmin}} \amp = -10 \amp\amp {\text{Ymin}} = -10\\ {\text{Xmax}} \amp = 10 \amp\amp {\text{Ymax}} = 10\\ {\text{Xscl}} \amp = 1 \amp\amp {\text{Yscl}} = 1 \end{align*}
38.
\(8 - y + 3x = 0\)
\begin{align*} {\text{Xmin}} \amp = -10 \amp\amp {\text{Ymin}} = -10\\ {\text{Xmax}} \amp = 10 \amp\amp {\text{Ymax}} = 10\\ {\text{Xscl}} \amp = 1 \amp\amp {\text{Yscl}} = 1 \end{align*}
39.
\(3x - 4y = 1200\)
\begin{align*} {\text{Xmin}} \amp = -1000 \amp\amp {\text{Ymin}} = -1000\\ {\text{Xmax}} \amp = 1000 \amp\amp {\text{Ymax}} = 1000\\ {\text{Xscl}} \amp = 100 \amp\amp {\text{Yscl}} = 100 \end{align*}
40.
\(x + 2y = 500\)
\begin{align*} {\text{Xmin}} \amp = -1000 \amp\amp {\text{Ymin}} = -1000\\ {\text{Xmax}} \amp = 1000 \amp\amp {\text{Ymax}} = 1000\\ {\text{Xscl}} \amp = 100 \amp\amp {\text{Yscl}} = 100 \end{align*}
41.
\(0.2x + 5y = 0.1\)
\begin{align*} {\text{Xmin}} \amp = -1 \amp\amp {\text{Ymin}} = -0.1\\ {\text{Xmax}} \amp = 1 \amp\amp {\text{Ymax}} = 0.1\\ {\text{Xscl}} \amp = 0.1 \amp\amp {\text{Yscl}} = 0.01 \end{align*}
42.
\(1.2x - 4.2y = 3.6\)
\begin{align*} {\text{Xmin}} \amp = -1 \amp\amp {\text{Ymin}} = -1\\ {\text{Xmax}} \amp = 4 \amp\amp {\text{Ymax}} = 1\\ {\text{Xscl}} \amp = 0.2 \amp\amp {\text{Yscl}} = 0.1 \end{align*}
43.
\(70x + 3y = y + 420\)
\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Ymin}} = 0\\ {\text{Xmax}} \amp = 10 \amp\amp {\text{Ymax}} = 250\\ {\text{Xscl}} \amp = 1 \amp\amp {\text{Yscl}} = 25 \end{align*}
44.
\(40y - 5x = 780 - 20y\)
\begin{align*} {\text{Xmin}} \amp = -200 \amp\amp {\text{Ymin}} = 0\\ {\text{Xmax}} \amp = 0 \amp\amp {\text{Ymax}} = 20\\ {\text{Xscl}} \amp = 20 \amp\amp {\text{Yscl}} = 2 \end{align*}

Exercise Group.

For Problems 45–52,
  1. Find the \(x\)- and \(y\)-intercepts.
  2. Solve the equation for \(y\text{.}\)
  3. Choose a graphing window in which both intercepts are visible, and graph the equation with your graphing utility.
45.
\(x + 4y = 100\)
46.
\(2x - 3y = -72\)
47.
\(25x - 20y = 1\)
48.
\(4x + 75y = 60,000\)
49.
\(\dfrac{y}{12} - \dfrac{x}{60}= 1\)
50.
\(\dfrac{x}{80} + \dfrac{y}{400}= 1\)
51.
\(-2x = 3y + 84\)
52.
\(7x = 91 - 13y\)
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