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Modeling with Linear Functions |
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Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function{: data-type="term" .no-emphasis} models.
When modeling{: data-type="term" .no-emphasis} scenarios with linear functions and solving problems involving quantities with a constant rate of change{: data-type="term" .no-emphasis}, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:
- Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
- Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
- Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
- Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
- When needed, write a formula for the function.
- Solve or evaluate the function using the formula.
- Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
- Clearly convey your result using appropriate units, and answer in full sentences when necessary. {: type="1"}
Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.
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Output: M,
money remaining, in dollars
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Input: t,
time, in weeks {: data-bullet-style="none"}
So, the amount of money remaining depends on the number of weeks: M(t)
We can also identify the initial value and the rate of change.
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Initial Value: She saved $3,500, so $3,500 is the initial value for M.
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Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope. {: data-bullet-style="none"}
Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.
The rate of change{: data-type="term" .no-emphasis} is constant, so we can start with the linear model{: data-type="term" .no-emphasis} M( t )=mt+b.
Then we can substitute the intercept and slope provided.
intercept, we set the output to zero, and solve for the input.
The x-
intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.
When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-
intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0≤t≤8.75.
In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.
Some real-world problems provide the y-
intercept, which is the constant or initial value. Once the y-
intercept is known, the x-
intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-
intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.
Now we can set the function equal to 0, and solve for x
to find the x-
intercept.
The x-
intercept is the number of months it takes her to reach a balance of $0. The x-
intercept is 4 months, so it will take Hannah four months to pay off her loan.
Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.
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Identify the input and output values.
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Convert the data to two coordinate pairs.
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Find the slope.
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Write the linear model.
-
Use the model to make a prediction by evaluating the function at a given x-
value.
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Use the model to identify an x-
value that results in a given y-
value.
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Answer the question posed. {: type="1"}
- Predict the population in 2013.
- Identify the year in which the population will reach 15,000. {: type="a"}
intercept would correspond to the year 0, more than 2000 years ago!
To make computation a little nicer, we will define our input as the number of years since 2004:
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Input: t,
years since 2004
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Output: P(t),
the town’s population {: data-display="block"}
To predict the population in 2013 (t=9),
we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.
To determine the rate of change, we will use the change in output per change in input.
giving the point ( 0,6200 ).
Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t=5,
giving the point ( 5,8100 ).
The two coordinate pairs are ( 0,6200 )
and ( 5,8100 ).
Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.
To find when the population will reach 15,000, we can set P(t)=15000
and solve for t.
-
Write a linear model to represent the cost C
of the company as a function of x,
the number of doughnuts produced.
-
Find and interpret the y-intercept. {: type="a"}
- Predict the population in 2014.
- Identify the year in which the population will reach 54,000. {: type="a"}
It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.
“How long will it take them to be 2 miles apart?”
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In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.
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Input: t,
time in hours.
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Output: A(t),
distance in miles, and E(t),
distance in miles {: data-display="block" data-bullet-style="none"}
Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as [link].
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Initial Value: They both start at the same intersection so when t=0,
the distance traveled by each person should also be 0. Thus the initial value for each is 0.
Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A
is 4 and the slope for E
is 3.
Using those values, we can write formulas for the distance each person has walked.
which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E,
to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.
We can then define a third variable, D,
to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from [link].
Recall that we need to know how long it takes for D,
the distance between them, to equal 2 miles. Notice that for any given input t,
the outputs A( t ),E( t ),
and D( t )
represent distances.
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[link] shows us that we can use the Pythagorean Theorem because we have drawn a right angle.
Using the Pythagorean Theorem, we get:* * * {: data-type="newline" data-count="1"}
so our distance D( t )
will always be positive. We can simplify this answer to D(t)=5t.
This means that the distance between Anna and Emanuel is also a linear function. Because D
is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t)=2
and solve for t.
Yes. Sketch the figure and label the quantities and unknowns on the sketch.
and Eastborough at ( 20,0 ).
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Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:
Because the town of Eastborough is at the point (20, 0), we can find the equation:
Real-world situations including two or more linear functions may be modeled with a system of linear equations{: data-type="term" .no-emphasis}. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in [link].
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Identify the input and output of each linear model.
-
Identify the slope and y-intercept of each linear model.
-
Find the solution by setting the two linear functions equal to one another and solving for x,
or find the point of intersection on a graph. {: type="1"}
Input | d, distance driven in miles |
Outputs |
K(d):
cost, in dollars, for renting from Keep on Trucking M( d ) cost, in dollars, for renting from Move It Your Way |
Initial Value | Up-front fee: K( 0 )=20 and M( 0 )=16 |
Rate of Change | K(d)=$0.59 /mile and P(d)=$0.63 /mile |
Using the rates of change and initial charges, we can write the equations
The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K( d )
function is smaller.
These graphs are sketched in [link], with K( d )
in blue.
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To find the intersection, we set the equations equal and solve:
is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d>100.
which is setting the expression mx+b
equal to 0.
- The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See [link].
- A graph of the system may be used to identify the points where one line falls below (or above) the other line.
the line f(x)=1+2x,
and the line parallel to f(x)
passing through ( 2, 7 ).
and the line perpendicular to f(x)
that passes through the origin.
and the line perpendicular to f(x)
that passes through the origin.
the line f(x)=3x,
and the line parallel to f(x)
passing through (6,1).
as a function of the year, t,
where t
is the number of years since the model began.
is graphed, find and interpret the x- and y-intercepts.
is graphed, find and interpret the slope of the function.
as a function of the age of the baby, in months, t.
.
is graphed, find and interpret the x- and y-intercepts.
: The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. ( 0, 7.5 )
: The baby weighed 7.5 pounds at birth.
as a function of the year, t.
is graphed, find and interpret the x- and y-intercepts.
: In roughly 59 years, the number of people inflicted with the common cold would be 0. (0,12,025)
: Initially there were 12,025 people afflicted by the common cold.
is graphed, find and interpret the slope of the function.
For the following exercises, use the graph in [link], which shows the profit, y,
in thousands of dollars, of a company in a given year, t,
where t
represents the number of years since 1980.
{: #CNX_Precalc_Figure_02_03_201}
where y
depends on t,
the number of years since 1980.
in thousands of dollars, of a company in a given year, t,
where t
represents the number of years since 1980.
{: #CNX_Precalc_Figure_02_03_202}
where y
depends on t,
the number of years since 1980.
For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in [link]. Assume that the house values are changing linearly.
| Year | Mississippi | Hawaii | |---------- | 1950 | $25,200 | $74,400 | | 2000 | $71,400 | $272,700 | {: #Table_02_03_03 summary=".."}
| Year | Indiana | Alabama | |---------- | 1950 | $37,700 | $27,100 | | 2000 | $94,300 | $85,100 | {: #Table_02_03_04 summary=".."}
-
How much did the population grow between the year 2004 and 2008?
-
How long did it take the population to grow from 1,001 students to 1,697 students?
-
What is the average population growth per year?
-
What was the population in the year 2000?
-
Find an equation for the population, P,
of the school t years after 2000.
-
Using your equation, predict the population of the school in 2011. {: type="a"}
-
How much did the population grow between the year 2003 and 2007?
-
How long did it take the population to grow from 1,431 people to 2,134 people?
-
What is the average population growth per year?
-
What was the population in the year 2000?
-
Find an equation for the population, P
of the town t
years after 2000.
-
Using your equation, predict the population of the town in 2014. {: type="a"}
- Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used.
- Interpret the slope and y-intercept of the equation.
- Use your equation to find the total monthly cost if 687 minutes are used. {: type="a"}
-
Find a linear equation for the monthly cost of the data plan as a function of x
, the number of MB used.
-
Interpret the slope and y-intercept of the equation.
-
Use your equation to find the total monthly cost if 250 MB are used. {: type="a"}
- Find a formula for the moose population, P since 1990.
- What does your model predict the moose population to be in 2003? {: type="a"}
-
Find a formula for the owl population, P.
Let the input be years since 2003.
-
What does your model predict the owl population to be in 2012? {: type="a"}
-
Give a linear equation for the remaining federal helium reserves, R,
in terms of t,
the number of years since 2010.
-
In 2015, what will the helium reserves be?
-
If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted? {: type="a"}
-
Give a linear equation for the remaining oil reserves, R,
in terms of t,
the number of years since now.
-
Seven years from now, what will the oil reserves be?
-
If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted? {: type="a"}
- Option A: Base salary of $17,000 a year with a commission of 12% of your sales
- Option B: Base salary of $20,000 a year with a commission of 5% of your sales {: data-bullet-style="none"}
How much jewelry would you need to sell for option A to produce a larger income?
- Option A: Base salary of $14,000 a year with a commission of 10% of your sales
- Option B: Base salary of $19,000 a year with a commission of 4% of your sales {: data-bullet-style="none"}
How much electronics would you need to sell for option A to produce a larger income?
- Option A: Base salary of $20,000 a year with a commission of 12% of your sales
- Option B: Base salary of $26,000 a year with a commission of 3% of your sales {: data-bullet-style="none"}
How much electronics would you need to sell for option A to produce a larger income?
- Option A: Base salary of $10,000 a year with a commission of 9% of your sales
- Option B: Base salary of $20,000 a year with a commission of 4% of your sales {: data-bullet-style="none"}
How much electronics would you need to sell for option A to produce a larger income?
- {: data-type="footnote-ref" #footnote1} 1{: data-type="footnote-ref-link"} Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ {: data-list-type="bulleted" data-bullet-style="none"}