### Learning Objectives

In this section you will:

### 2.3.1 – Determining Whether a Linear Function Is Increasing, Decreasing, or Constant

A linear function may be increasing, decreasing, or constant. For an increasing function 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 1a. 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 1b. 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 1c.

### Increasing and Decreasing Functions

The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function.

• $f\left(x\right)=mx+b\,$ is an increasing function if $\,m>0.$
• $f\left(x\right)=mx+b\,$ is a decreasing function if $\,m<0.$
• $f\left(x\right)=mx+b\,$ is a constant function if $\,m=0.$

### Example 1 – Deciding Whether a Function Is Increasing, Decreasing, or Constant

Some recent studies suggest that a teenager sends an average of 60 texts per day[1] . 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.

1. 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.
2. 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.
3. 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. Analyze each function. 1. The function can be represented as $\,f\left(x\right)=60x\,$ 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. 2. The function can be represented as $\,f\left(x\right)=500-60x\,$ 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. 3. The cost function can be represented as $\,f\left(x\right)=50\,$ because the number of days does not affect the total cost. The slope is 0 so the function is constant. ### 2.3.2 – Interpreting Slope as a Rate of Change In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that 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 by a set of points, $\,\left({x}_{1}\text{, }{y}_{1}\right)\,$ and $\,\left({x}_{2}\text{, }{y}_{2}\right)$ —we can calculate the slope $m.$ $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ Note that in function notation we can obtain two corresponding values for the output $\,{y}_{1}\,$ and $\,{y}_{2}\,$ for the function $\,f,$ $\,{y}_{1}=f\left({x}_{1}\right)\,$ and $\,{y}_{2}=f\left({x}_{2}\right),$ so we could equivalently write $m=\frac{f\left({x}_{2}\right)–f\left({x}_{1}\right)}{{x}_{2}–{x}_{1}}$ Figure 2 indicates how the slope of the line between the points, $\,\left({x}_{1},{y}_{1}\right)\,$ and $\,\left({x}_{2},{y}_{2}\right),\,$ is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is. Are the units for slope always $\,\frac{\text{units for the output}}{\text{units for the input}}\,\text{?}$ 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. ### Calculate Slope The slope, or rate of change, of a function $\,m\,$ can be calculated according to the following: $m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ where $\,{x}_{1}\,$ and $\,{x}_{2}\,$ are input values, $\,{y}_{1}\,$ and $\,{y}_{2}\,$ are output values. ### How To Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. ### Example 2 – Finding the Slope of a Linear Function If $\,f\left(x\right)\,$ is a linear function, and $\,\left(3,-2\right)\,$ and $\,\left(8,1\right)\,$ are points on the line, find the slope. Is this function increasing or decreasing? The coordinate pairs are $\,\left(3,-2\right)\,$ and $\,\left(8,1\right).\,$ To find the rate of change, we divide the change in output by the change in input. $m=\frac{\text{change in output}}{\text{change in input}}=\frac{1-\left(-2\right)}{8-3}=\frac{3}{5}$ 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 y-coordinate, used corresponds with the first input value, or x-coordinate, used. Note that if we had reversed them, we would have obtained the same slope. $m=\frac{\left(-2\right)-\left(1\right)}{3-8}=\frac{-3}{-5}=\frac{3}{5}$ ### Try It If $\,f\left(x\right)\,$ is a linear function, and $\,\left(2,3\right)\,$ and $\,\left(0,4\right)\,$ are points on the line, find the slope. Is this function increasing or decreasing? Show answer $m=\frac{4-3}{0-2}=\frac{1}{-2}=-\frac{1}{2};$ decreasing because $m<0.$ ### Example 3 – 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. 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 four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. $\frac{\text{4,400 people}}{\text{4 years}}=\text{1,100 }\frac{\text{people}}{\text{year}}$ 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. ### Try It 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. Show answer $m=\frac{1,868-1,442}{2,012-2,009}=\frac{426}{3}=\text{142 people per year}$ ### 2.3.3 – Modeling Real-World 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 real-world problems. ### How To Given a linear function $\,f\,$ and the initial value and rate of change, evaluate $\,f\left(c\right).$ 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into $\,f\left(x\right)=mx+b.$ 3. Evaluate the function at $\,x=c.$ ### Example 4 – Using a Linear Function to Calculate Salary Based on 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 $\,I\left(n\right),$ and interpret the meaning of the components of the equation. The given information gives us two input-output pairs: $\,\left(3,760\right)\,$ and $\,\left(5,\text{92}0\right).\,$ We start by finding the rate of change. $\begin{array}{ccc}\hfill m& =& \frac{920-760}{5-3}\hfill \\ & =& \frac{160}{2 \text{policies}}\hfill \\ & =& \text{\}80\,\text{per}\,\text{policy}\hfill \end{array}$ 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.

$\begin{array}{cccc}\hfill I\left(n\right)& =& 80n+b\hfill & \\ \hfill 760& =& 80\left(3\right)+b\hfill & \phantom{\rule{1em}{0ex}}\text{ }\text{when}\,n=3,I\left(3\right)=760\hfill \\ \hfill 760& -& 80\left(3\right)=b\hfill & \\ \hfill 520& =& b\hfill & \end{array}$

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.

$I\left(n\right)=80n+520$

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional$80 commission for each policy sold.

### Try It

A new plant food was introduced to a young tree to test its effect on the height of the tree. The table below shows the height of the tree, in feet, $\,x\,$ months since the measurements began. Write a linear function, $\,H\left(x\right),$ where $\,x\,$ is the number of months since the start of the experiment.

 x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5

$H\left(x\right)=0.5x+12.5$

### 2.3.4 – Graphing Linear Functions

Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function $\,f\left(x\right)=x.$

#### Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, $\,f\left(x\right)=2x,$ we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point $\,\left(1,2\right).\,$ Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point $\,\left(2,4\right).\,$ Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

### How To

Given a linear function, graph by plotting points.

1. Choose a minimum of two input values.
2. Evaluate the function at each input value.
3. Use the resulting output values to identify coordinate pairs.
4. Plot the coordinate pairs on a grid.
5. Draw a line through the points.

### Example 5 – Graphing by Plotting Points

Graph $\,f\left(x\right)=-\frac{2}{3}x+5\,$ by plotting points.

Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

Evaluate the function at each input value, and use the output value to identify coordinate pairs.

$\begin{array}{cc}\hfill x=0& \phantom{\rule{2em}{0ex}}f\left(0\right)=-\frac{2}{3}\left(0\right)+5=5⇒\left(0,5\right)\hfill \\ \hfill x=3& \phantom{\rule{2em}{0ex}}f\left(3\right)=-\frac{2}{3}\left(3\right)+5=3⇒\left(3,3\right)\hfill \\ \hfill x=6& \phantom{\rule{2em}{0ex}}f\left(6\right)=-\frac{2}{3}\left(6\right)+5=1⇒\left(6,1\right)\hfill \end{array}$

Plot the coordinate pairs and draw a line through the points. Figure 3 represents the graph of the function $\,f\left(x\right)=-\frac{2}{3}x+5.$

#### Analysis

The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function.

### Try It

Graph $\,f\left(x\right)=-\frac{3}{4}x+6\,$ by plotting points.

#### Graphing a Function Using y-intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set $\,x=0\,$ in the equation.

The other characteristic of the linear function is its slope.

Let’s consider the following function.

$f\left(x\right)=\frac{1}{2}x+1$

The slope is $\,\frac{1}{2}.\,$ Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when $\,x=0.\,$ The graph crosses the y-axis at $\,\left(0,1\right).\,$ Now we know the slope and the y-intercept. We can begin graphing by plotting the point $\,\left(0,1\right).\,$ We know that the slope is the change in the y-coordinate over the change in the x-coordinate. This is commonly referred to as rise over run, $\,m=\frac{\text{rise}}{\text{run}}.\,$ From our example, we have $\,m=\frac{1}{2},$ which means that the rise is 1 and the run is 2. So starting from our y-intercept $\,\left(0,1\right),$ we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 4.

### Graphical Interpretation of a Linear Function

In the equation $\,f\left(x\right)=mx+b$

• $b\,$ is the y-intercept of the graph and indicates the point $\,\left(0,b\right)\,$ at which the graph crosses the y-axis.
• $m\,$ is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:
$m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

Do all linear functions have y-intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line is parallel to the y-axis does not have a y-intercept, but it is not a function.)

### How To

Given the equation for a linear function, graph the function using the y-intercept and slope.

1. Evaluate the function at an input value of zero to find the y-intercept.
2. Identify the slope as the rate of change of the input value.
3. Plot the point represented by the y-intercept.
4. Use $\,\frac{\text{rise}}{\text{run}}\,$ to determine at least two more points on the line.
5. Sketch the line that passes through the points.

### Example 6 – Graphing by Using the y-intercept and Slope

Graph $\,f\left(x\right)=-\frac{2}{3}x+5\,$ using the y-intercept and slope.

Evaluate the function at $\,x=0\,$ to find the y-intercept. The output value when $\,x=0\,$ is 5, so the graph will cross the y-axis at $\,\left(0,5\right).$

According to the equation for the function, the slope of the line is $\,-\frac{2}{3}.\,$ This tells us that for each vertical decrease in the “rise” of $\,–2\,$ units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 5. From the initial value $\,\left(0,5\right)\,$ we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then drawing a line through the points.

#### Analysis

The graph slants downward from left to right, which means it has a negative slope as expected.

### Try It

Find a point on the graph we drew in Example 8 that has a negative x-value.

Possible answers include $\,\left(-3,7\right),\,$
$\left(-6,9\right),\,$ or $\,\left(-9,11\right).$

### 2.3.5 – Writing the Equation for a Function from the Graph of a Line

Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 6. We can see right away that the graph crosses the y-axis at the point $\,\left(0,\text{4}\right)\,$ so this is the y-intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point $\,\left(–2,0\right).\,$ To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

$m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2$

Substituting the slope and y-intercept into the slope-intercept form of a line gives

$y=2x+4$

### How To

Given a graph of linear function, find the equation to describe the function.

1. Identify the y-intercept of an equation.
2. Choose two points to determine the slope.
3. Substitute the y-intercept and slope into the slope-intercept form of a line.

### Example 7 – Matching Linear Functions to Their Graphs

Match each equation of the linear functions with one of the lines in Figure 7.

$\begin{array}{cccc}a\text{.}\hfill & \hfill \text{ }f\left(x\right)& =& 2x+3\hfill \\ b\text{.}\hfill & \hfill g\left(x\right)& =& 2x-3\hfill \\ c\text{.}\hfill & \hfill h\left(x\right)& =& -2x+3\hfill \\ d\text{. }\hfill & \hfill j\left(x\right)& =& \frac{1}{2}x+3\hfill \end{array}$

Analyze the information for each function.

1. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function $\,g\,$ has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through $\,\left(0,\text{3}\right)\,$ so $\,f\,$ must be represented by line I.
2. This function also has a slope of 2, but a y-intercept of $\,-3.\,$ It must pass through the point $\,\left(0,-3\right)\,$ and slant upward from left to right. It must be represented by line III.
3. This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
4. This function has a slope of $\,\frac{1}{2}\,$ and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through $\,\left(0,\text{3}\right),$ but the slope of $\,j\,$ is less than the slope of $\,f\,$ so the line for $\,j\,$ must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 8.

#### Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 11, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use $\,m=0\,$ in the equation $\,f\left(x\right)=mx+b,$ the equation simplifies to $\,f\left(x\right)=b.\,$ In other words, the value of the function is a constant. This graph represents the function $\,f\left(x\right)=2.$

A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

A vertical line, such as the one in Figure 13, has an x-intercept, but no y-intercept unless it’s the line $\,x=0.\,$ This graph represents the line $\,x=2.$

### Horizontal and Vertical Lines

Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation in the form $\,f\left(x\right)=b.$

A vertical line is a line defined by an equation in the form $\,x=a.$

### Example 8 – Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 14.

For any x-value, the y-value is $\,-4,\,$ so the equation is $\,y=-4.$

### Example 9 – Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 15.

The constant x-value is $\,7,$ so the equation is $\,x=7.$

Access this online resource for additional instruction and practice with linear functions.

### Key Concepts

• 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. See Example 1.
• Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 2 and Example 3.
• A linear function can be used to solve real-world problems given information in different forms. See Example 4.
• Linear functions can be graphed by plotting points or by using the y-intercept and slope. See Example 5 and Example 6.
• The equation for a linear function can be written by interpreting the graph. See Example 7.
• Horizontal lines are written in the form, $\,f\left(x\right)=b.\,$ See Example 8.
• Vertical lines are written in the form, $\,x=b.\,$ See Example 9.

### Glossary

decreasing linear function
a function with a negative slope: If $\,f\left(x\right)=mx+b, \text{then} m<0.$
horizontal line
a line defined by $\,f\left(x\right)=b,$ where $\,b\,$ is a real number. The slope of a horizontal line is 0.
increasing linear function
a function with a positive slope: If $\,f\left(x\right)=mx+b, \text{then} m>0.$
linear function
a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line
point-slope form
the equation for a line that represents a linear function of the form $\,y-{y}_{1}=m\left(x-{x}_{1}\right)$
slope
the ratio of the change in output values to the change in input values; a measure of the steepness of a line
slope-intercept form
the equation for a line that represents a linear function in the form $\,f\left(x\right)=mx+b$
vertical line
a line defined by $\,x=a,$ where $\,a\,$ is a real number. The slope of a vertical line is undefined.