3A.1: Linear Functions
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Skills to Develop
 Represent a linear function.
 Determine whether a linear function is increasing, decreasing, or constant.
 Interpret slope as a rate of change.
 Write and interpret an equation for a linear function.
 Graph linear functions.
 Determine whether lines are parallel or perpendicular.
 Write the equation of a line parallel or perpendicular to a given line.
There are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure \(\PageIndex{1}\)). It carries passengers comfortably for a 30kilometer trip from the airport to the subway station in only eight minutes.
Figure \(\PageIndex{1}\): Shanghai MagLev Train (credit: “kanegen”/Flickr)
Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate realworld situations such as the train’s distance from the station at a given point in time.
Representing Linear Functions
The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change; that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
Representing a Linear Function in Word Form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.
 The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.
The speed is the rate of change. The rate of change is a measure of how quickly the dependent variable changes with respect to the change of the independent variable. The rate of change for this example is constant. As the time (input) increases by 1 second, the corresponding distance (output) always increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.
Representing a Linear Function in Function Notation
Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slopeintercept form of a line, where \(x\) is the input value, \(m\) is the rate of change, and \(b\) is the initial value of the dependent variable. Keep in mind that \(m\) and \(b\) stand for constants, while \(x\) and \(y\) are variables.
\[\begin{align*} &\text{Equation form } &y=mx+b \\ &\text{Function notation } &f(x)=mx+b \end{align*}\]
In the example of the train, we might use the notation \(D(t)\) in which the total distance \(D\) is a function of the time \(t\). The rate, \(m\), is 83 meters per second. The initial value of the dependent variable \(b\) is the original distance from the station, 250 meters. We can write a function relationship to represent the motion of the train.
\[D(t)=83t+250 \nonumber\]
Representing a Linear Function in Tabular Form
A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure \(\PageIndex{2}\). From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.
Figure \(\PageIndex{2}\): Tabular representation of the function \(D\) showing selected input and output values
Can the input in the previous example be any real number?
No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of nonnegative real numbers.
Representing a Linear Function in Graphical Form
Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, \(D(t)=83t+250\), to draw a graph, represented in Figure \(\PageIndex{3}\). Notice the graph is a line. When we plot a linear function, the graph is always a line.
The rate of change, which is constant, determines the slant, or steepness, or slope of the line. The point at which the input value is zero is the vertical intercept, or yintercept, of the line. We can see from the graph in Figure \(\PageIndex{3}\) that the \(y\)intercept in the train example we just saw is \((0,250)\) and represents the distance of the train from the station when it began moving at a constant speed.
Figure \(\PageIndex{3}\): The graph of \(D(t)=83t+250\). Graphs of linear functions are lines because the rate of change is constant.
Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line \(f(x)=2x+1\). Ask yourself what numbers can be input to the function; that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.
definition: Linear Function
A linear function is a function whose graph is a line. Linear functions can be written in the slopeintercept form of a line
\[f(x)=mx+b \nonumber\]
where \(b\) is the initial or starting value of the function (when the input \(x=0\)), and \(m\) is the constant rate of change, or slope of the function. The yintercept is at \((0,b)\).
Example \(\PageIndex{1}\): Using a Linear Function to Find the Pressure on a Diver
The pressure, \(P\), in pounds per square inch (PSI) on the diver in Figure \(\PageIndex{4}\) depends upon her depth below the water surface, \(d\), in feet. This relationship may be modeled by the equation, \(P(d)=0.434d+14.696.\) Restate this function in words.
Figure \(\PageIndex{4}\): (credit: Ilse Reijs and JanNoud Hutten)
To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirtyfour thousandths times depth plus fourteen and six hundred ninetysix thousandths.
Analysis
The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.
Determining whether a Linear Function Is Increasing, Decreasing, or Constant
The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure \(\PageIndex{5}\)(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure \(\PageIndex{5}\)(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure \(\PageIndex{5}\)(c).
Figure \(\PageIndex{5}\): Three graphs depicting an increasing function, a decreasing function, and a constant function.
determine whether a linear function is Increasing, Decreasing, or constant
The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function.
 \(f(x)=mx+b\) is an increasing function if \(m>0\).
 \(f(x)=mx+b\) is a decreasing function if \(m<0\).
 \(f(x)=mx+b\) is a constant function if \(m=0\).
Example \(\PageIndex{2}\): Deciding whether a Function Is Increasing, Decreasing, or Constant
Some recent studies suggest that a teenager sends an average of 60 texts per day. For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.
a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.
c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.
Solution
Analyze each function.
a. The function can be represented as \(f(x)=60x\) where \(x\) is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.
b. The function can be represented as \(f(x)=500−60x\) where \(x\) is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after \(x\) days.
c. The cost function can be represented as \(f(x)=0x+50=50\) because the number of days does not affect the total cost. The slope is 0 so the function is constant.
Calculating and Interpreting Slope
In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope, given input and output values.
Calculate Slope
Given two values for the input, \(x_1\) and \(x_2\), and two corresponding values for the output, \(y_1\) and \(y_2\) — which can be represented as points on the graph, \((x_1,y_1)\) and \((x_2,y_2)\) — we can calculate the slope \(m\), as follows:
\[m= \dfrac{\text{change in output (rise)}}{ \text{change in input (run)}} = \dfrac{{\Delta}y}{ {\Delta}x} = \dfrac{y_2−y_1}{x_2−x_1} \nonumber\]
where \({\Delta}y\) is the vertical displacement and \({\Delta}x\) is the horizontal displacement.
Note that in function notation two corresponding values for the output \(y_1\) and \(y_2\) for the function \(f\) are \(y_1=f(x_1)\) and \(y_2=f(x_2)\), so we could equivalently write
\[m=\dfrac{f(x_2)f(x_1)}{x_2x_1} \;\text{ (this form will be important in calculus).} \nonumber \]
Figure \(\PageIndex{6}\) indicates how the slope of the line between the points, \((x_1,y_1)\) and \((x_2,y_2)\), is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.
Figure \(\PageIndex{6}\): The slope of a function is calculated by the change in \(y\) divided by the change in \(x\). It does not matter which point is used as the \((x_2,y_2)\) and which is the \((x_1 ,y_1)\), as long as each calculation is started with the elements from the same coordinate pair.
Are the units for slope always \(\frac{\text{units for the output}}{ \text{units for the input}}\)?
Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.
Given two points from a linear function, calculate and interpret the slope.
 Determine the units for output values and input values.
 Calculate the change in output values (\(\Delta y\)) and change in input values (\(\Delta x\)).
 Interpret the slope as the change in output values per unit of the input value.
Example \(\PageIndex{3}\): Finding the Slope of a Linear Function
If \(f(x)\) is a linear function, and \((3,−2)\) and \((8,1)\) are points on the line, find the slope. Is this function increasing or decreasing?
Solution
The coordinate pairs are \((3,−2)\) and \((8,1)\). To find the rate of change, we divide the change in output by the change in input.
\[m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{1(2)}{83}=\dfrac{3}{5}\nonumber\]
We could also write the slope as \(m=0.6\). The function is increasing because \(m>0\).
Analysis
As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or ycoordinate, used corresponds with the first input value, or xcoordinate, used.
\(\PageIndex{1}\)
If \(f(x)\) is a linear function, and \((2, 3)\) and \((0,4)\) are points on the line, find the slope. Is this function increasing or decreasing?
 Answer

\(m=\frac{4−3}{0−2} =\frac{1}{2}=\frac{1}{2};\) decreasing because \(m<0\).
Example \(\PageIndex{4}\): Finding the Population Change from a Linear Function
The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.
Solution:
The rate of change relates the change in population to the change in time. The population increased by \(27,800−23,400=4400\) people over the fouryear time interval. To find the rate of change, divide the change in the number of people by the number of years.
\[\dfrac{4,400 \text{ people}}{4 \text{ years}} =1,100 \dfrac{\text{people}}{\text{year}}\nonumber\]
So the population increased by 1,100 people per year.
Analysis
Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.
\(\PageIndex{2}\)
The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.
 Answer

\(m=\frac{1,868−1,442}{2,012−2,009} = \frac{426}{3} =\text{ 142 people per year}\)
Writing the PointSlope Form of a Linear Equation
Up until now, we have been using the slopeintercept form of a linear equation to describe linear functions. This is the only form that corresponds to the concept of a function; however, we can think of a linear function merely as a linear equation in two variables, and then there are two other forms of writing the equation that are commonly used.
The first form is called the standard form of a linear equation.
STANDARD FORM OF LINEAR EQUATION
A linear equation is in standard form when it is written \(Ax+By=C\).
Most people prefer to have A, B, and C be integers and \(A \geq 0\) when writing a linear equation in standard form, although it is not strictly necessary. This form is valuable because any straight line in the plane can be described using standard form. In this course, we will not have many occasions to use standard form
The second form, the pointslope form, is very valuable towards the beginning of any calculus course.
\[yy_1=m(xx_1)\]
The pointslope form is derived from the slope formula.
\[ \begin{align*} &m=\dfrac{yy_1}{xx_1} &\text{assuming }x{\neq}x_1 \\ &m(xx_1)=\dfrac{yy_1}{xx_1}(xx_1) &\text{Multiply each side of the equation by }(xx_1). \\ &m(xx_1)=yy_1 &\text{Simplify.} \\ &yy_1=m(xx_1) &\text{Reverse the sides of the equation.} \end{align*}\]
Both the slopeintercept form and the pointslope form are describing the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in pointslope form, \(y−4=− \frac{1}{2}(x−6).\) We can convert it to the slopeintercept form as shown:
\[\begin{align*} y4&=\dfrac{1}{2}(x6) \\ y4&=\dfrac{1}{2}x+3 &\text{Distribute the }\dfrac{1}{2}. \\ y&=\dfrac{1}{2}x+7 &\text{Add 4 to each side.}\end{align*}\]
Therefore, the same line can be described in slopeintercept form as \(y=\dfrac{1}{2}x+7\).
PointSlope Form of a Linear Equation
The pointslope form of a linear equation takes the form
\[yy_1=m(x−x_1) \nonumber\]
where \(m\) is the slope, and \(x_1\) and \(y_1\) are the \(x\) and \(y\) coordinates of a specific point through which the line passes.
Writing the Equation of a Line Using a Point and the Slope
The pointslope form is particularly useful if we know one point and the slope of a line, which will be very useful in calculus. Suppose, for example, we are told that a line has a slope of 2 and passes through the point \((4,1)\). We know that \(m=2\) and that \(x_1=4\) and \(y_1=1\). We can substitute these values into the general pointslope equation.
\[\begin{align*} y−y_1&=m(x−x_1) \\ y−1&=2(x−4) \end{align*}\]
If we wanted to then rewrite the equation in slopeintercept form, we apply algebraic techniques.
\[\begin{align*} y−1&=2(x−4) \\ y−1&=2x−8 &\text{Distribute the 2.} \\ y&=2x−7 &\text{Add 1 to each side.} \end{align*}\]
Both equations, \(y−1=2(x−4)\) and \(y=2x–7\), describe the same line. See Figure \(\PageIndex{7}\).
Figure \(\PageIndex{7}\)
Example \(\PageIndex{5}\): Writing Linear Equations Using a Point and the Slope
Write the pointslope form of an equation of a line with a slope of 3 that passes through the point \((6,–1)\). Then rewrite it in the slopeintercept form.
Solution
Let’s figure out what we know from the given information. The slope is 3, so \(m=3\). We also know one point, so we know \(x_1=6\) and \(y_1 =−1\). Now we can substitute these values into the general pointslope equation.
\[\begin{align*} yy_1&=m(xx_1) \\ y−(−1)&=3(x−6) &\text{Substitute known values.} \\ y+1&=3(x−6) &\text{Simplify the lefthand side.} \end{align*}\]
Then we use algebra to find the slopeintercept form.
\[\begin{align*} y+1&=3(x−6) \\ y+1&=3x−18 &\text{Distribute 3.} \\ y&=3x−19 &\text{Subtract 1 from both sides.} \end{align*}\]
\(\PageIndex{3}\)
Write the pointslope form of an equation of a line with a slope of –2 that passes through the point \((–2, 2)\). Then rewrite it in the slopeintercept form.
 Answer

\(y−2=−2(x+2)\) ; \(y=−2x−2\)
Writing the Equation of a Line Using Two Points
The pointslope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points \((0, 1)\) and \((3, 2)\). We can use the coordinates of the two points to find the slope.
\[\begin{align*} m&=\dfrac{y_2y_1}{x_2x_1} \\ &=\dfrac{21}{30} \\ &=\dfrac{1}{3} \end{align*}\]
Now we can use the slope we found and the coordinates of either one of the points to find the equation for the line. Let's use \((0,1)\) for our point.
\[\begin{align*} yy_1&=m(xx_1) \\ y1&=\dfrac{1}{3}(x0) \end{align*}\]
As before, we can use algebra to rewrite the equation in slopeintercept form.
\[\begin{align*} y1&=\dfrac{1}{3}(x0) \\ y1&=\dfrac{1}{3}x &\text{Distribute the }\dfrac{1}{3}. \\ y&=\dfrac{1}{3}x+1 &\text{Add 1 to each side.} \end{align*}\]
Both equations describe the line shown in Figure \(\PageIndex{8}\).
Figure \(\PageIndex{8}\)
Example \(\PageIndex{6}\): Writing Linear Equations Using Two Points
Write the pointslope form of an equation of a line that passes through the points \((5,1)\) and \((8, 7)\). Then rewrite it in slopeintercept form.
Solution:
Let’s begin by finding the slope.
\[\begin{align*} m &= \dfrac{y_2y_1}{x_2x_1} \\ &=\dfrac{71}{85} \\ &=\dfrac{6}{3} \\ &= 2 \end{align*}\]
So \(m=2\). Next, we substitute the slope and the coordinates for one of the points into the general pointslope equation. We can choose either point. We will use \((5,1)\).
\[\begin{align*} yy_1&= m(xx_1) \\ y1&=2(x5) \end{align*}\]
The pointslope equation of the line is \(y–1=2(x–5)\). To rewrite the equation in slopeintercept form, we use algebra.
\[\begin{align*} y1&=2(x5) \\ y1&=2x10 \\ y&=2x9 \end{align*}\]
The slopeintercept equation of the line is \(y=2x–9\).
\(\PageIndex{4}\): Write the pointslope form of an equation of a line that passes through the points \((–1,3)\) and \((0,0)\). Then rewrite it in the slopeintercept form.
Solution
\(y−0=−3(x−0)\); \(y=−3x\)
Writing and Interpreting an Equation for a Linear Function
Now that we have written equations for linear functions in both the slopeintercept form and the pointslope form, we can choose which method to use based on the information we are given. Information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function \(f\) in Figure \(\PageIndex{9}\).
Figure \(\PageIndex{9}\): Graph depicting how to calculate the slope of a line.
We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose \((0,7)\) and \((4, 4)\). We can use these points to calculate the slope.
\[\begin{align*} m&=\dfrac{y_2y_1}{x_2x_1} \\ &=\dfrac{47}{40} \\&=\dfrac{3}{4}\end{align*}\]
Now we can substitute the slope and the coordinates of one of the points into pointslope form.
\[\begin{align*} yy_1&=m(xx_1) \\ y4&=\dfrac{3}{4}(x4) \end{align*}\]
If we want to rewrite the equation in slopeintercept form, we would find
\[\begin{align*} y4&=\dfrac{3}{4}(x4) \\ y4 &=\dfrac{3}{4}x+3 \\ y&=\dfrac{3}{4}x+7\end{align*}\]
If we wanted to find the slopeintercept form without first writing the pointslope form, we could have observed that the line crosses the \(y\)axis when the output value is 7. Therefore, \(b=7\). We now have the initial value \(b\) and the slope \(m\) so we can substitute \(m\) and \(b\) into the slopeintercept form of a line. If we want to use function notation we can substitute \(f(x)\) in place of \(y\).
Figure \(\PageIndex{10}\)
The function is \(f(x)=−\frac{3}{4}x+7,\) and the corresponding linear equation would be \(y=−\frac{3}{4}x+7.\)
Given the graph of a linear function, write an equation to represent the function
 Identify two points on the line.
 Use the two points to calculate the slope.
 Determine where the line crosses the \(y\)axis to identify the \(y\)intercept by visual inspection.
 Substitute the slope and \(y\)intercept into the slopeintercept form of a line equation.
Example \(\PageIndex{7}\): Writing an Equation for a Linear Function
Write an equation for a linear function given the graph of \(f\) shown in Figure \(\PageIndex{11}\).
Figure \(\PageIndex{11}\)
Solution
Identify two points on the line, such as \((0, 2)\) and \((−2,−4)\). Use the points to calculate the slope.
\[\begin{align*} m&=\dfrac{y_2y_1}{x_2x_1} \\ &=\dfrac{42}{20} \\ &=\dfrac{6}{2} \\ &=3 \end{align*}\]
Substitute the slope and the coordinates of one of the points into pointslope form.
\[\begin{align*} yy_1&=m(xx_1) \\ y(4)&=3(x(2)) \\ y+4 &= 3(x+2)\end{align*}\]
We can use algebra to rewrite the equation in slopeintercept form.
\[\begin{align*} y+4&= 3(x+2) \\ y+4&= 3x+6 \\ y & = 3x + 2 \end{align*}\]
Analysis
This makes sense because we can see from Figure \(\PageIndex{12}\) that the line crosses the \(y\)axis at the point \((0, 2)\), which is the \(y\)intercept, so \(b=2\). Furthermore, the positive slope corresponds to the fact that the line is increasing from left to right.
Figure \(\PageIndex{12}\): Graph of an increasing line with points at \((0, 2)\) and \((2, 4)\).
Example \(\PageIndex{8}\): Writing an Equation for a Linear Cost Function
Suppose Brenda starts a company to produce designer phone cases, in which she incurs a fixed cost of $1,250 per month for the overhead. Her production costs are $37.50 per phone case. Write a linear function \(C\) where \(C(x)\) is the cost for \(x\) phone cases produced in a given month.
Solution
The fixed cost is present every month, $1,250. The cost that varies is the cost to produce each case, which is dependent on how many cases are produced. The variable cost per case, called the marginal cost, is represented by $37.50. The cost Brenda incurs is the sum of these two costs, represented by \(C(x)=1250+37.5x\).
Analysis
If Brenda produces 100 cases in a month, her monthly cost is represented by
\[\begin{align*} C(100)&=1250+37.5(100) \\ &=5000 \end{align*}\]
So her monthly cost that month would be $5,000.
Example \(\PageIndex{9}\): Writing an Equation for a Linear Function Given Two Points
If \(f\) is a linear function, with \(f(3)=−2\), and \(f(8)=1\), find an equation for the function in slopeintercept form.
Solution
We can write the given points using coordinates.
\[\begin{align*} f(3)&= 2 \rightarrow (3,2) \\ f(8)&=1 \rightarrow (8,1) \end{align*}\]
We can then use the points to calculate the slope.
\[\begin{align*} m&=\dfrac{y_2y_1}{x_2x_1} \\[2pt] &=\dfrac{1(2)}{83} \\[2pt] &=\dfrac{3}{5} \end{align*}\]
Substitute the slope and the coordinates of one of the points into the pointslope form.
\[\begin{align*} yy_1&=m(xx_1) \\ y(2)&=\dfrac{3}{5}(x3) \end{align*}\]
We can use algebra to rewrite the equation in the slopeintercept form.
\[\begin{align*} y+2&=\dfrac{3}{5}(x3) \\[2pt] y+2&=\dfrac{3}{5}x\dfrac{9}{5} \\[2pt] y&=\dfrac{3}{5}x\dfrac{19}{5} \end{align*}\]
\(\PageIndex{5}\)
If \(f(x)\) is a linear function, with \(f(2)= 11\), and \(f(4)= 25\), find an equation for the function in slopeintercept form.
 Answer

\(y=−7x+3\)
Modeling RealWorld Problems with Linear Functions
In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of realworld problems.
Given a linear function \(f\) and the initial value and rate of change, evaluate \(f(c)\).
 Determine the initial value and the rate of change (slope).
 Substitute the values into \(f(x)=mx+b\).
 Evaluate the function at \(x=c\).
Example \(\PageIndex{10}\): Using a Linear Function to Determine the Number of Songs in a Collection
Marcos currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, \(N\), in his collection as a function of time, \(t\), the number of months. How many songs will he own in a year?
Solution
The initial value for this function is 200 because he currently owns 200 songs, so \(N(0)=200\), which means that \(b=200\).
The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that \(m=15\). We can substitute the initial value and the rate of change into the slopeintercept form of a line.
Figure \(\PageIndex{13}\)
We can write the formula \(N(t)=15t+200.\)
With this formula, we can then predict how many songs Marcos will have in 1 year (12 months). In other words, we can evaluate the function at \(t=12\).
\[\begin{align*} N(12)&=15(12)+200 \\ &=180+200 \\ &= 380 \end{align*}\]
Marcos will have 380 songs in 12 months.
Analysis
Notice that \(N\) is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.
Example \(\PageIndex{11}\): Using a Linear Function to Calculate Salary Plus Commission
Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I,depends on the number of new policies, \(n\), he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for \(I(n)\), and interpret the meaning of the components of the equation.
Solution
The given information gives us two inputoutput pairs: \((3,760)\) and \((5,920)\). We start by finding the rate of change.
\[\begin{align*} m&=\dfrac{920760}{53} \\ &=\dfrac{$160}{2 \text{ policies}} \\ &=$80 \text{ per policy} \end{align*}\]
Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.
We can then solve for the initial value.
\[\begin{align*} I(n)&=80n+b \\ 760&=80(3)+b \quad \text{ when } n=3, I(3)=760 \\ 76080(3)&=b \\ 520 & =b \end{align*}\]
The value of \(b\) is the starting value for the function and represents Ilya’s income when \(n=0\), or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.
We can now write the final equation.
\[I(n)=80n+520 \nonumber]
Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.
Key Equations
 slopeintercept form of a line: \(f(x)=mx+b\)
 slope: \(m=\dfrac{\text{change in output (rise)}}{\text{change in input (run)}}=\dfrac{{\Delta}y}{{\Delta}x}=\dfrac{y_2y_1}{x_2x_1}\)
 pointslope form of a line: \(y−y_1 =m(xx_1)\)
Key Concepts
 The ordered pairs given by a linear function represent points on a line.
 Linear functions can be represented in words, function notation, tabular form, and graphical form.
 The rate of change of a linear function is also known as the slope.
 An equation in the slopeintercept form of a line includes the slope and the initial value of the function.
 The initial value, or yintercept, is the output value when the input of a linear function is zero. It is the yvalue of the point at which the line crosses the yaxis.
 An increasing linear function results in a graph that slants upward from left to right and has a positive slope.
 A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.
 A constant linear function results in a graph that is a horizontal line.
 Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant.
 The slope of a linear function can be calculated by dividing the difference between yvalues by the difference in corresponding xvalues of any two points on the line.
 The slope and initial value can be determined given a graph or any two points on the line.
 One type of function notation is the slopeintercept form of an equation.
 The pointslope form is useful for finding a linear equation when given the slope of a line and one point.
 The pointslope form is also convenient for finding a linear equation when given two points through which a line passes.
 The equation for a linear function can be written if the slope \(m\) and initial value \(b\) are known.
 A linear function can be used to solve realworld problems.
 A linear function can be written from tabular form.
Section Exercises
Verbal
Terry is skiing down a steep hill. Terry's elevation, \(E(t)\), in feet after \(t\) seconds is given by \(E(t)=3000−70t\). Write a complete sentence describing Terry’s starting elevation and how it is changing over time.
Terry starts at an elevation of 3000 feet and descends 70 feet per second.
Maria is climbing a mountain. Maria's elevation, \(E(t)\), in feet after \(t\) minutes is given by \(E(t)=1200+40t\). Write a complete sentence describing Maria’s starting elevation and how it is changing over time.
Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?
3 miles per hour
Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home \(t\) hours from now.
A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after \(t\) hours.
\(d(t) =100−10t\)
Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding \(n\) rides?
Algebraic
For the following exercises, determine whether the equation of the curve can be written as a linear function.
\(y= \frac{1}{4}x+6\)
Yes.
\(y=3x−5\)
\(y=3 x^2 −2\)
No.
\(3x+5y=15\)
\(3 x^2 +5y=15\)
No.
\(3x+5 y^ 2 =15\)
\(−2 x^2 +3y^2 =6\)
No.
\(−\frac{x−3}{5} =2y\)
For the following exercises, determine whether each function is increasing or decreasing.
\(f(x)=4x+3\)
Increasing.
\(g(x)=5x+6\)
\(a(x)=5−2x\)
Decreasing.
\(b(x)=8−3x\)
\(h(x)=−2x+4\)
Decreasing.
\(k(x)=−4x+1\)
\(j(x)=\frac{1}{2}x−3\)
Increasing.
\(p(x)=\frac{1}{4}x−5\)
\(n(x)=−\frac{1}{3}x−2\)
Decreasing.
\(m(x)=−\frac{3}{8}x+3\)
For the following exercises, find the slope of the line that passes through the two given points.
\((2, 4)\) and \((4, 10)\)
3
\((1, 5)\) and \((4, 11)\)
\((−1,4)\) and \((5,2)\)
\(–\frac{1}{3}\)
\((8,−2)\) and \((4,6)\)
(6, 11) and (−4, 3)
\(\frac{4}{5}\)
For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
\(f(−5)=−4\), and \(f(5)=2\)
\(f(−1)=4\) and \(f(5)=1\)
\(f(x)=−\frac{1}{2}x+\frac{7}{2}\)
\((2,4)\) and \((4,10)\)
Passes through \((1,5)\) and \((4,11)\)
\(y=2x+3\)
Passes through \((−1, 4)\) and \((5, 2)\)
Passes through \((−2, 8)\) and \((4, 6)\)
\(y=−\frac{1}{3}x+\frac{22}{3}\)
\(x\) intercept at (−2, 0) and \(y\) intercept at (0,−3)
\(x\) intercept at (−5, 0) and \(y\) intercept at (0, 4)
\(y=\frac{4}{5}x+4\)
Graphical
For the following exercises, find the slope of the lines graphed.
\(−\frac{5}{4}\)
For the following exercises, write an equation for the lines graphed.
\(y= \frac{2}{3} x+1\)
\(y=−2x+3\)
\(y=3\)
Numeric
For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.
\(x\) 
0  5  10  15 
\(g(x)\) 
5  –10  –25  –40 
Linear, \(g(x)=−3x+5\)
\(x\)  0  5  10  15 
\(h(x)\)  5  30  105  230 
\(x\)  0  5  10  15 
\(f(x)\)  –5  20  45  70 
Linear, \(f(x)=5x−5\)
\(x\) 
5  10  20  25 
\(k(x)\) 
28  13  58  73 
\(x\) 
0  2  4  6 
\(g(x)\) 
6  –19  –44  –69 
Linear, \(g(x)=−\frac{25}{2}x+6\)
\(x\) 
2  4  6  8 
\(f(x)\) 
–4  16  36  56 
\(x\) 
2  4  6  8 
\(f(x)\) 
–4  16  36  56 
Linear, \(f(x)=10x−24\)
\(x\) 
0  2  6  8 
\(k(x)\) 
6  31  106  231 
Technology
If \(f\) is a linear function,\(f(0.1)=11.5\), and \(f(0.4)=–5.9\), find an equation for the function.
\(f(x)=−58x+17.3\)
Graph the function \(f\) on a domain of \([ –10, 10 ]: f(x)=0.02x−0.01\). Enter the function in a graphing utility. For the viewing window, set the minimum value of \(x\) to be −10 and the maximum value of \(x\) to be 10.
Graph the function \(f\) on a domain of \([ –10, 10 ]:fx)=2,500x+4,000\)
[link] shows the input, \(w\), and output, \(k\), for a linear function \(k\). a. Fill in the missing values of the table. \(b\). Write the linear function \(k\), round to 3 decimal places.
\(w\)  –10  5.5  67.5  b 
\(k\)  30  –26  a  –44 
[link] shows the input,p,and output,q,for a linear function \(q\). a. Fill in the missing values of the table. \(b\). Write the linear function \(k\).
\(p\)  0.5  0.8  12  b 
\(q\)  400  700  a  1,000,000 
a. \(a=11\),900; \(b=1001\).1 \(b\). \(q(p)=1000p−100\)
Graph the linear function \(f\) on a domain of \([ −10,10 ]\) for the function whose slope is \(\frac{1}{8}\) and yintercept is \(\frac{31}{16}\). Label the points for the input values of \(−10\) and \(10\).
Graph the linear function \(f\) on a domain of \([ −0.1,0.1 ]\) for the function whose slope is 75 and yintercept is −22.5. Label the points for the input values of −0.1 and 0.1.
Graph the linear function \(f\) where \(f(x)=ax+b\) on the same set of axes on a domain of \([ −4,4 ]\) for the following values of \(a\) and \(b\).
\(a=2;\) \(b=3\)
\(a=2; \) \(b=4\)
\(a=2;\) \(b=–4\)
\(a=2;\) \(b=–5\)
Extensions
Find the value of \(x\) if a linear function goes through the following points and has the following slope: \((x,2),(−4,6),\) \(m=3\)
\(x=−\frac{16}{3}\)
Find the value of \(y\) if a linear function goes through the following points and has the following slope: \((10,y),(25,100),\) \(m=−5\)
Find the equation of the line that passes through the following points: \((a, b)\) and \(( a, b+1 )\)
\(x=a\)
Find the equation of the line that passes through the following points: \((2a, b)\) and (a, b+1)
Find the equation of the line that passes through the following points: \((a, 0)\) and \((c, d)\)
\(y=\frac{d}{c−a}x−\frac{ad}{c−a}\)
RealWorld Applications
At noon, a barista notices that she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves \(n\) more customers during her shift?
A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?
$45 per training session.
A clothing business finds there is a linear relationship between the number of shirts, \(n\), it can sell and the price, \(p\), it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form \(p(n)=mn+b\) that gives the price \(p\) they can charge for \(n\) shirts.
A phone company charges for service according to the formula: \(C(n)=24+0\).1n, where \(n\) is the number of minutes talked, and \(C(n)\) is the monthly charge, in dollars. Find and interpret the rate of change and initial value.
The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.
A farmer finds there is a linear relationship between the number of bean stalks, \(n\), she plants and the yield, \(y\), each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form \(y=mn+b\) that gives the yield when \(n\) stalks are planted.
A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.
The slope is −400. This means for every year between 1960 and 1989, the population dropped by 400 per year in the city.
A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, \(P(t)\), for the population \(t\) years after 2003.
Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: \(I(x)=1054x+23\),\(286,\) where xis the number of years after 1990. Which of the following interprets the slope in the context of the problem?
As of 1990, average annual income was $23,286.
In the tenyear period from 1990–1999, average annual income increased by a total of $1,054.
Each year in the decade of the 1990s, average annual income increased by $1,054.
Average annual income rose to a level of $23,286 by the end of 1999.
\(c\).
When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of \(C\), the Celsius temperature, \(F(C)\).
Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
Find and interpret \(F(28)\).
Find and interpret \(F(–40)\).
Footnotes
1 http://www.chinahighlights.com/shang...glevtrain.htm
2 http://www.cbsnews.com/8301501465_1...aystudysays/
Glossary
decreasing linear function
a function with a negative slope: If \(f(x)=mx+b\), then \(m<0\).
increasing linear function
a function with a positive slope: If \(f(x)=mx+b\), then \(m>0\).
linear function
a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line
pointslope form
the equation for a line that represents a linear function of the form \(y−y_1=m(x−x_1)
slope
the ratio of the change in output values to the change in input values; a measure of the steepness of a line
slopeintercept form
the equation for a line that represents a linear function in the form \(f(x)=mx+b\)
yintercept
the value of a function when the input value is zero; also known as initial value
Contributors
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.