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Linear Functions

Represent a linear function.

Linear Functions

Find the slope of a line (IA 3.2.1)
Find an equation of the line given two points (IA 3.3.3)

Linear functions are a specific type of function that can be used to model many real-world applications, such as the growth of a plant, earned salary, the distance a train travels over time, or the costs to start a new business. In this section, we will explore linear functions, their graphs, and how to find them using data points.

A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

$f(x)=mx+b$

where $b$ is the initial or starting value of the function (when input, $x=0$), and $m$ is the constant rate of change, or slope of the function. The $y$ -intercept is at ( $0,b$ ).

$m=\text{slope}=\frac{\text{change in output}}{\text{change in input}}=\frac{∆y}{∆x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}=\frac{\text{rise}}{\text{run}}$

When interpreting slope, it will be important to consider the units of measurement. Make sure to always attach these units to both the numerator and denominator when they are provided to you.

Example

Try it.

Find the slope of the line shown.

Solution

Locate two points on the graph whose coordinates are integers. Label the coordinates of these points.
$(0,1)=({x}_{1},{y}_{1})$
$(5,-2)=({x}_{2},{y}_{2})\$
It may help to visualize this change as $m=\frac{rise}{run}$ Count the rise between the points. Since the line goes down, the slope is negative. Then count the run, or horizontal change. Using the slope formula: $m=\frac{rise}{run}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-2-1}{5-0}=\frac{-3}{5}$

Find the slope of the line.

Try it.

Find the slope of the line below:

Try it.

Find the slope of the following line: $y=3$

Try it.

Find the slope of the following line: $x=-2$

Try it.

Find the slope of the following line: $y=-5x-6$

Try it.

Find the slope of the following line: $2x-4y=5$

Try it.

Use the slope formula to find the slope of the line between the following pair of points. $(-2,4),(3,-1)$

Try it.

Use the slope formula to find the slope of the linear function satisfying the condition below. $f(-2)=-1,f(6)=5$

Condensed — the full section is in OpenStax College Algebra 2e.

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. This 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.

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. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept 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.

\[\begin{array}{ll}\text{Equation form}\ & y=mx+b \\ \text{Function notation}\ & f(x)=mx+b\end{array}\]

In the example of the train, we might use the notation $D(t)$ where 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 generalized equation to represent the motion of the train.

\[D(t)=83t+250\]

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 . From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Condensed — the full section is in OpenStax College Algebra 2e.

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 (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 (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 (c).

Condensed — the full section is in OpenStax College Algebra 2e.

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, $({x}_{1}\text{, }{y}_{1})$ and $({x}_{2}\text{, }{y}_{2})$ —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 $f,$ ${y}_{1}=f({x}_{1})$ and ${y}_{2}=f({x}_{2}),$ so we could equivalently write

\[m=\frac{f({x}_{2})-f({x}_{1})}{{x}_{2}-{x}_{1}}\]

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, or slant. The greater the absolute value of the slope, the steeper the slant is.

Example

Try it.

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.

\[m=\frac{\text{change in output}}{\text{change in input}}=\frac{1-(-2)}{8-3}=\frac{3}{5}\]

We could also write the slope as $m=0.6.$ The function is increasing because $m>0.$

Condensed — the full section is in OpenStax College Algebra 2e.

Writing and Interpreting an Equation for a Linear Function

Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions based on the information we are given. That 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 .

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 $(4,4).$

\[\begin{array}{lll}m & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{4-7}{4-0} \\ & = & -\frac{3}{4}\end{array}\]

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ \ y-4 & = & -\frac{3}{4}(x-4)\end{array}\]

If we want to rewrite the equation in the slope-intercept form, we would find

\[\begin{array}{lll}y-4 & = & -\frac{3}{4}(x-4) \\ y-4 & = & -\frac{3}{4}x+3 \\ y & = & -\frac{3}{4}x+7\end{array}\]

If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized 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 slope-intercept form of a line.

So the function is $f(x)=-\frac{3}{4}x+7,$ and the linear equation would be $y=-\frac{3}{4}x+7.$

Example

Try it.

Write an equation for a linear function given a graph of $f$ shown in .

Solution

Identify two points on the line, such as $(0,2)$ and $(-2,-4).$ Use the points to calculate the slope.

\[\begin{array}{lll}m & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{-4-2}{-2-0} \\ & = & \frac{-6}{-2} \\ & = & 3\end{array}\]

Substitute the slope and the coordinates of one of the points into the point-slope form.

\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ y-(-4) & = & 3(x-(-2)) \\ y+4 & = & 3(x+2)\end{array}\]

We can use algebra to rewrite the equation in the slope-intercept form.

\[\begin{array}{lll}y+4 & = & 3(x+2) \\ y+4 & = & 3x+6 \\ y & = & 3x+2\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

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.

Example

Try it.

Marcus 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 at the end of one 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 slope-intercept form of a line.

We can write the formula $N(t)=15t+200.$

With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at $t=12.$

\[\begin{array}{lll}N(12) & = & 15(12)+200 \\ & = & 180+200 \\ & = & 380\end{array}\]

Marcus will have 380 songs in 12 months.

Example

Try it.

relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

number of weeks, w	0	2	4	6
number of rats, P(w)	1000	1080	1160	1240

Solution

We can see from the table that the initial value for the number of rats is 1000, so $b=1000.$

Rather than solving for $m,$ we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

\[P(w)=40w+1000\]

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using $(2,1080)$ and $(6,1240)$

\[\begin{array}{lll}m & = & \frac{1240-1080}{6-2} \\ & = & \frac{160}{4} \\ & = & 40\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

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 $f(x)=x.$

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, $f(x)=2x,$ 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 $(1,2).$ Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point $(2,4).$ Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

Example

Try it.

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

Solution

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.

\[\begin{array}{ll}x=0 & \ f(0)=-\frac{2}{3}(0)+5=5⇒(0,5) \\ x=3 & \ f(3)=-\frac{2}{3}(3)+5=3⇒(3,3) \\ x=6 & \ f(6)=-\frac{2}{3}(6)+5=1⇒(6,1)\end{array}\]

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

Condensed — the full section is in OpenStax College Algebra 2e.

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 . We can see right away that the graph crosses the y-axis at the point $(0,\text{4})$ 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 $(-2,0).$ 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

\[y=2x+4\]

Example

Try it.

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

ⓐ $f(x)=2x+3$
ⓑ $g(x)=2x-3$
ⓒ $h(x)=-2x+3$
ⓓ $j(x)=\frac{1}{2}x+3$

Solution

Analyze the information for each function.

ⓐ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 $(0,\text{3})$ so $f$ must be represented by line I.
ⓑThis function also has a slope of 2, but a y-intercept of $-3.$ It must pass through the point $(0,-3)$ and slant upward from left to right. It must be represented by line III.
ⓒ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.
ⓓ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 $(0,\text{3}),$ 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 .

Condensed — the full section is in OpenStax College Algebra 2e.

Determining Whether Lines are Parallel or Perpendicular

The two lines in are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the other, they would become coincident.

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

\[\begin{array}{ll}\begin{array}{l}f(x)=-2x+6 \\ f(x)=-2x-4\end{array}\}\ \text{parallel}\ & \begin{array}{l}f(x)=3x+2 \\ f(x)=2x+2\end{array}\}\ \text{not parallel}\end{array}\]

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in are perpendicular.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is $1.$ So, if ${m}_{1}\ \text{and }{m}_{2}$ are negative reciprocals of one another, they can be multiplied together to yield $-1.$

\[{m}_{1}{m}_{2}=-1\]

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is $\frac{1}{8},$ and the reciprocal of $\frac{1}{8}$ is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

\[\begin{array}{llll}f(x) & = & \frac{1}{4}x+2 & \ \text{negative reciprocal of}\ \frac{1}{4}\ \text{is }-4 \\ f(x) & = & -4x+3 & \ \text{negative reciprocal of}\ -4\ \text{is }\frac{1}{4}\end{array}\]

The product of the slopes is –1.

\[-4(\frac{1}{4})=-1\]

Condensed — the full section is in OpenStax College Algebra 2e.

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Suppose for example, we are given the equation shown.

\[f(x)=3x+1\]

We know that the slope of the line formed by the function is 3. We also know that the y-intercept is $(0,1).$ Any other line with a slope of 3 will be parallel to $f(x).$ So the lines formed by all of the following functions will be parallel to $f(x).$

\[\begin{array}{lll}g(x) & = & 3x+6 \\ h(x) & = & 3x+1 \\ p(x) & = & 3x+\frac{2}{3}\end{array}\]

Suppose then we want to write the equation of a line that is parallel to $f$ and passes through the point $(1,\text{7}).$ This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of $b$ will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ y-7 & = & 3(x-1) \\ y-7 & = & 3x-3 \\ y & = & 3x+4\end{array}\]

So $g(x)=3x+4$ is parallel to $f(x)=3x+1$ and passes through the point $(1,\text{7}).$

Example

Try it.

Find a line parallel to the graph of $f(x)=3x+6$ that passes through the point $(3,\text{0}).$

Solution

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute $m=3,x=3,$ and $f(x)=0$ into the slope-intercept form to find the y-intercept.

\[\begin{array}{lll}g(x) & = & 3x+b \\ 0 & = & 3(3)+b \\ b & = & -9\end{array}\]

The line parallel to $f(x)$ that passes through $(3,\text{0})$ is $g(x)=3x-9.$

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

Linear functions can be represented in words, function notation, tabular form, and graphical form. See .
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 .
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 and .
An equation for a linear function can be written from a graph. See .
The equation for a linear function can be written if the slope $m$ and initial value $b$ are known. See and .
A linear function can be used to solve real-world problems given information in different forms. See , , and .
Linear functions can be graphed by plotting points or by using the y-intercept and slope. See and .
Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See .
The equation for a linear function can be written by interpreting the graph. See .
The x-intercept is the point at which the graph of a linear function crosses the x-axis. See .
Horizontal lines are written in the form, $f(x)=b.$ See .
Vertical lines are written in the form, $x=b.$ See .
Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See .
A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, $f(x)=mx+b,$ and using the $b$ that results. Similarly, the point-slope form of an equation can also be used. See .
A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the slope of the line shown.

Odkrij odgovor

Locate two points on the graph whose coordinates are integers. Label the coordinates of these points.
$(0,1)=({x}_{1},{y}_{1})$
$(5,-2)=({x}_{2},{y}_{2})\$
It may help to visualize this change as $m=\frac{rise}{run}$ Count the rise between the points. Since the line goes down, the slope is negative. Then count the run, or horizontal change. Using the slope formula: $m=\frac{rise}{run}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{-2-1}{5-0}=\frac{-3}{5}$
Find the slope of the line below:
Find the slope of the following line: $y=3$
Find the slope of the following line: $x=-2$
Find the slope of the following line: $y=-5x-6$
Find the slope of the following line: $2x-4y=5$
Use the slope formula to find the slope of the line between the following pair of points. $(-2,4),(3,-1)$
Use the slope formula to find the slope of the linear function satisfying the condition below. $f(-2)=-1,f(6)=5$
Find the equation of a line containing the points (−4, −3) and (1, −5)
Odkrij odgovor
1. Find the slope of the line using the given points.
  $m=\frac{-5-(-3)}{1-(-4)}=\frac{-5+3}{1+4}=\frac{-2}{5}$
2. Choose one point and label these coordinates $(-4,-3)={x}_{1},{y}_{1}$
3. Plug $m,{x}_{2}$ and ${y}_{2}$ into point-slope for, $y-{y}_{1}=m(x-{x}_{1})$ .
  $y-(-3)=\frac{-2}{5}(x-(-4))$
  $y+3=\frac{-2}{5}(x+4)$
4. Write the equation in slope-intercept form, $y=mx+b$ .
  $y+3=\frac{-2}{5}x-\frac{8}{5}$
  $y=\frac{-2}{5}x-\frac{8}{5}-3$
  $y=\frac{-2}{5}x-\frac{8}{5}-\frac{15}{5}$
  $y=\frac{-2}{5}x-\frac{23}{5}$
Find the equation of a line containing the given points. Write the equation in slope-intercept form. $(4,3)$ and $(8,1)$
Find the equation of a line containing the given points. Write the equation in slope-intercept form. $(-5,-3)$ and $(4,-6)$
Find the equation of a line containing the given points. Write the equation in slope-intercept form. $f(-2)=8,f(4)=6$
Derek notices the amount he receives in tips each night is a linear function of the number of tables he waits on. On Friday evening he waits on 22 tables and receives $87 in tips, and on Tuesday evening he waits on 18 tables and receives $73 in tips.
1. ⓐ Record the information given above as two data points.
2. ⓑ Find the slope of this linear function in terms of $\frac{\$}{table}$ .
3. ⓒ Write a linear function S(t) that will let him estimate the amount in $ in tips he will receive as a function of tables, t, waited on.
Complete the following:
Verbal Description Formula
Slope between two points
Point-slope form of a line
Slope-intercept form of a line
The pressure, $P,$ in pounds per square inch (PSI) on the diver in 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.

Odkrij odgovor

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 thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.
Studies from the early 2010s indicated that teens sent about 60 texts a day, while more recent data indicates much higher messaging rates among all users, particularly considering the various apps with which people can communicate.http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/. 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 person has a limit of 500 texts per month in their data plan. The input is the number of days, and output is the total number of texts remaining for the month.
3. ⓒA person has an unlimited number of texts in their 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.
Odkrij odgovor
Analyze each function.
1. ⓐ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.
2. ⓑ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.
3. ⓒThe cost function can be represented as $f(x)=50$ because the number of days does not affect the total cost. The slope is 0 so the function is constant.
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?

Odkrij odgovor

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.
\[m=\frac{\text{change in output}}{\text{change in input}}=\frac{1-(-2)}{8-3}=\frac{3}{5}\]
We could also write the slope as $m=0.6.$ The function is increasing because $m>0.$
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?

Odkrij odgovor

$m=\frac{4-3}{0-2}=\frac{1}{-2}=-\frac{1}{2};$ decreasing because $m<0.$
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.

Odkrij odgovor

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.
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.

Odkrij odgovor

\[m=\frac{1,868-1,442}{2,012-2,009}=\frac{426}{3}=\text{142 people per year}\]
Write an equation for a linear function given a graph of $f$ shown in .

Odkrij odgovor

Identify two points on the line, such as $(0,2)$ and $(-2,-4).$ Use the points to calculate the slope.
\[\begin{array}{lll}m & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{-4-2}{-2-0} \\ & = & \frac{-6}{-2} \\ & = & 3\end{array}\]
Substitute the slope and the coordinates of one of the points into the point-slope form.
\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ y-(-4) & = & 3(x-(-2)) \\ y+4 & = & 3(x+2)\end{array}\]
We can use algebra to rewrite the equation in the slope-intercept form.
\[\begin{array}{lll}y+4 & = & 3(x+2) \\ y+4 & = & 3x+6 \\ y & = & 3x+2\end{array}\]
Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function $C$ where $C(x)$ is the cost for $x$ items produced in a given month.

Odkrij odgovor

The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50. The variable cost, called the marginal cost, is represented by $37.5.$ The cost Ben incurs is the sum of these two costs, represented by $C(x)=1250+37.5x.$
If $f$ is a linear function, with $f(3)=-2,$ and $f(8)=1,$ find an equation for the function in slope-intercept form.

Odkrij odgovor

We can write the given points using coordinates.
\[\begin{array}{lll}f(3) & = & -2\to (3,-2) \\ f(8) & = & 1\to (8,1)\end{array}\]
We can then use the points to calculate the slope.
\[\begin{array}{lll}m & = & \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\ & = & \frac{1-(-2)}{8-3} \\ & = & \frac{3}{5}\end{array}\]
Substitute the slope and the coordinates of one of the points into the point-slope form.
\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ y-(-2) & = & \frac{3}{5}(x-3)\end{array}\]
We can use algebra to rewrite the equation in the slope-intercept form.
\[\begin{array}{lll}y+2 & = & \frac{3}{5}(x-3) \\ y+2 & = & \frac{3}{5}x-\frac{9}{5} \\ y & = & \frac{3}{5}x-\frac{19}{5}\end{array}\]
If $f(x)$ is a linear function, with $f(2)=-11,$ and $f(4)=-25,$ write an equation for the function in slope-intercept form.

Odkrij odgovor

$y=-7x+3$
Marcus 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 at the end of one year?

Odkrij odgovor

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 slope-intercept form of a line.
We can write the formula $N(t)=15t+200.$
With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at $t=12.$
\[\begin{array}{lll}N(12) & = & 15(12)+200 \\ & = & 180+200 \\ & = & 380\end{array}\]
Marcus will have 380 songs in 12 months.
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.

Odkrij odgovor

The given information gives us two input-output pairs: $(3,760)$ and $(5,\text{92}0).$ We start by finding the rate of change.
\[\begin{array}{lll}m & = & \frac{920-760}{5-3} \\ & = & \frac{\$160}{2\text{policies}} \\ & = & \text{\$}80\ \text{per}\ \text{policy}\end{array}\]
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.
\[\begin{array}{llll}I(n) & = & 80n+b & \\ 760 & = & 80(3)+b & \ \text{when}\ n=3,I(3)=760 \\ 760 & - & 80(3)=b & \\ 520 & = & b & \end{array}\]
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.
\[I(n)=80n+520\]
Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.
relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.
number of weeks, w 0 2 4 6
number of rats, P(w) 1000 1080 1160 1240

Odkrij odgovor

We can see from the table that the initial value for the number of rats is 1000, so $b=1000.$
Rather than solving for $m,$ we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.
\[P(w)=40w+1000\]
If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using $(2,1080)$ and $(6,1240)$
\[\begin{array}{lll}m & = & \frac{1240-1080}{6-2} \\ & = & \frac{160}{4} \\ & = & 40\end{array}\]
A new plant food was introduced to a young tree to test its effect on the height of the tree. shows the height of the tree, in feet, $x$ months since the measurements began. Write a linear function, $H(x),$ 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

Odkrij odgovor

$H(x)=0.5x+12.5$
Graph $f(x)=-\frac{2}{3}x+5$ by plotting points.

Odkrij odgovor

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.
\[\begin{array}{ll}x=0 & \ f(0)=-\frac{2}{3}(0)+5=5⇒(0,5) \\ x=3 & \ f(3)=-\frac{2}{3}(3)+5=3⇒(3,3) \\ x=6 & \ f(6)=-\frac{2}{3}(6)+5=1⇒(6,1)\end{array}\]
Plot the coordinate pairs and draw a line through the points. represents the graph of the function $f(x)=-\frac{2}{3}x+5.$
Graph $f(x)=-\frac{3}{4}x+6$ by plotting points.
Graph $f(x)=-\frac{2}{3}x+5$ using the y-intercept and slope.

Odkrij odgovor

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 $(0,5).$
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 . From the initial value $(0,5)$ 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.
Find a point on the graph we drew in that has a negative x-value.

Odkrij odgovor

Possible answers include $(-3,7),$ $(-6,9),$ or $(-9,11).$
Graph $f(x)=\frac{1}{2}x-3$ using transformations.

Odkrij odgovor

The equation for the function shows that $m=\frac{1}{2}$ so the identity function is vertically compressed by $\frac{1}{2}.$ The equation for the function also shows that $b=-3$ so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in .
Then show the vertical shift as in .
Graph $f(x)=4+2x$ using transformations.
Match each equation of the linear functions with one of the lines in .
1. ⓐ $f(x)=2x+3$
2. ⓑ $g(x)=2x-3$
3. ⓒ $h(x)=-2x+3$
4. ⓓ $j(x)=\frac{1}{2}x+3$
Odkrij odgovor
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 $(0,\text{3})$ 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 $(0,-3)$ 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 $(0,\text{3}),$ 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 .
Find the x-intercept of $f(x)=\frac{1}{2}x-3.$

Odkrij odgovor

Set the function equal to zero to solve for $x.$
\[\begin{array}{lll}0 & = & \frac{1}{2}x-3 \\ 3 & = & \frac{1}{2}x \\ 6 & = & x \\ x & = & 6\end{array}\]
The graph crosses the x-axis at the point $(6,\text{0}).$
Find the x-intercept of $f(x)=\frac{1}{4}x-4.$

Odkrij odgovor

$(16,\ \text{0})$
Write the equation of the line graphed in .

Odkrij odgovor

For any x-value, the y-value is $-4,$ so the equation is $y=-4.$
Write the equation of the line graphed in .

Odkrij odgovor

The constant x-value is $7,$ so the equation is $x=7.$
Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.
\[\begin{array}{llllll}f(x) & = & 2x+3 & \ h(x) & = & -2x+2 \\ g(x) & = & \frac{1}{2}x-4 & \ j(x) & = & 2x-6\end{array}\]

Odkrij odgovor

Parallel lines have the same slope. Because the functions $f(x)=2x+3$ and $j(x)=2x-6$ each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and $\frac{1}{2}$ are negative reciprocals, the functions $g(x)=\frac{1}{2}x-4$ and $h(x)=-2x+2$ represent perpendicular lines.

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Linear Functions

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.
Find the slope of a line (IA 3.2.1)

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Poskusi sam.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Več v Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Verbal Description	Formula
Slope between two points
Point-slope form of a line
Slope-intercept form of a line

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