4.2: Linear Functions (2024)

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Learning Objectives

In this section, you will:

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 agiven line.

Figure1ShanghaiMagLev Train (credit: "kanegen"/Flickr)

Just as with the growth of a bamboo plant, there are manysituations that involve constant change over time. Consider, forexample, the first commercial maglev train in the world, theShanghai MagLev Train (Figure1). It carries passengers comfortably for a 30-kilometertrip from the airport to the subway station in only eightminutes².

Suppose a maglev train travels a long distance, and maintains aconstant speed of 83 meters per second for a period of time once itis 250 meters from the station. How can we analyze the train’sdistance from the station as a function of time? In this section,we will investigate a kind of function that is useful for thispurpose, and use it to investigate real-world situations such asthe train’s distance from the station at a given point in time.

Representing LinearFunctions

The function describing the train’s motion is alinearfunction, which is defined as a function with a constant rate ofchange. This is a polynomial of degree 1. There are several ways torepresent a linear function, including word form, functionnotation, tabular form, and graphical form. We will describe thetrain’s motion as a function using each method.

Representing a Linear Function inWord Form

Let’s begin by describing the linear function in words. For thetrain problem we just considered, the following word sentence maybe used to describe the function relationship.

The train’s distance from the stationis a function of the time during which the train moves at aconstant speed plus its original distance from the station when itbegan moving at constant speed.

The speed is the rate of change. Recall that a rate of change isa measure of how quickly the dependent variable changes withrespect to the independent variable. The rate of change for thisexample is constant, which means that it is the same for each inputvalue. As the time (input) increases by 1 second, the correspondingdistance (output) increases by 83 meters. The train began moving atthis constant speed at a distance of 250 meters from thestation.

Representing a Linear Function inFunction Notation

Another approach to representing linear functions is by usingfunction notation. One example of function notation is an equationwritten in theslope-intercept formof a line,wherexxis the input value,mmis the rate ofchange, andbbis the initial value of the dependentvariable.

EquationformFunctionnotationy=mx+bf(x)=mx+bEquationformy=mx+bFunctionnotationf(x)=mx+b

In the example of the train, we might use thenotationD(t)D(t)where the totaldistanceDDis a function of the timet.t.Therate,m,m,is 83 meters per second. The initial value ofthe dependent variablebbis the original distance fromthe station, 250 meters. We can write a generalized equation torepresent the motion of the train.

D(t)=83t+250D(t)=83t+250

Representing a Linear Function inTabular Form

A third method of representing a linear function is through theuse of a table. The relationship between the distance from thestation and the time is represented inFigure2. From the table, we can see that the distance changes by83 meters for every 1 second increase in time.

Figure2Tabularrepresentation of the functionDDshowing selected inputand output values

Q&A

Can the input in the previous example be any realnumber?

No. The input represents time so whilenonnegative rational and irrational numbers are possible, negativereal numbers are not possible for this example. The input consistsof non-negative real numbers.

Representing a Linear Function inGraphical Form

Another way to represent linear functions is visually, using agraph. We can use the function relationship fromabove,D(t)=83t+250,D(t)=83t+250,to draw a graph asrepresented inFigure3. Notice the graph is a line.Whenwe plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant,orslopeof the line. The point at which the input valueis zero is the vertical intercept, ory-intercept, of the line. We can see from the graphthat they-intercept in thetrain example we just saw is(0,250)(0,250)andrepresents the distance of the train from the station when it beganmoving at a constant speed.

Figure3The graphofD(t)=83t+250D(t)=83t+250. Graphs of linear functionsare lines because the rate of change is constant.

Notice that the graph of the train example is restricted, butthis is not always the case. Consider the graph of thelinef(x)=2x+1.f(x)=2x+1.Ask yourself what numbers canbe input to the function. In other words, what is the domain of thefunction? The domain is comprised of all real numbers because anynumber may be doubled, and then have one added to the product.

LINEARFUNCTION

Alinear functionis a function whosegraph is a line. Linear functions can be written intheslope-intercept formof a line

f(x)=mx+bf(x)=mx+b

wherebbis the initial or starting value of thefunction (when input,x=0x=0), andmmis theconstant rate of change, or slope of the function.They-intercept isat(0,b).(0,b).

EXAMPLE1

Using a Linear Function to Find thePressure on a Diver

The pressure,P,P,in pounds per square inch (PSI) onthe diver inFigure4depends upon her depth below the watersurface,d,d,in feet. This relationship may be modeledby theequation,P(d)=0.434d+14.696.P(d)=0.434d+14.696.Restatethis function in words.

Figure4(credit:Ilse Reijs and Jan-Noud Hutten)

Answer

Analysis

The initial value, 14.696, is the pressure in PSI on the diverat a depth of 0 feet, which is the surface of the water. The rateof change, or slope, is 0.434 PSI per foot. This tells us that thepressure on the diver increases 0.434 PSI for each foot her depthincreases.

Determining Whether a Linear FunctionIs Increasing, Decreasing, or Constant

The linear functions we used in the two previous examplesincreased over time, but not every linear function does. A linearfunction may be increasing, decreasing, or constant. Foranincreasing function, as with the train example, the outputvalues increase as the input values increase. The graph of anincreasing function has a positive slope. A line with a positiveslope slants upward from left to right as inFigure5(a). For adecreasing function, theslope is negative. The output values decrease as the input valuesincrease. A line with a negative slope slants downward from left toright as inFigure5(b). If the function is constant, theoutput values are the same for all input values so the slope iszero. A line with a slope of zero is horizontal asinFigure5(c).

Figure5

INCREASING AND DECREASINGFUNCTIONS

The slope determines if the function isanincreasing linear function,adecreasing linear function, or a constantfunction.

f(x)=mx+bf(x)=mx+bis an increasing functionifm>0.m>0.
f(x)=mx+bf(x)=mx+bis a decreasing functionifm<0.m<0.
f(x)=mx+bf(x)=mx+bis a constant functionifm=0.m=0.

EXAMPLE2

Deciding Whether a Function IsIncreasing, Decreasing, or Constant

Studies from the early 2010s indicated that teens sent about 60texts a day, while more recent data indicates much higher messagingrates among all users, particularly considering the various appswith which people can communicate.³. For each of thefollowing scenarios, find the linear function that describes therelationship between the input value and the output value. Then,determine whether the graph of the function is increasing,decreasing, or constant.

ⓐThe total number of texts a teen sends is considered afunction of time in days. The input is the number of days, andoutput is the total number of texts sent.
ⓑA person has a limit of 500 texts per month in their dataplan. The input is the number of days, and output is the totalnumber of texts remaining for the month.
ⓒA person has an unlimited number of texts in their data planfor a cost of $50 per month. The input is the number of days, andoutput is the total cost of texting each month.

Answer

Interpreting Slope as a Rate ofChange

In the examples we have seen so far, the slope was provided tous. However, we often need to calculate the slope given input andoutput values. Recall that given two values for theinput,x1x1andx2,x2,and two correspondingvalues for the output,y1y1andy2y2—which canbe represented by a set ofpoints,(x1,y1)(x1,y1)and(x2,y2)(x2,y2)—wecan calculate the slopem.m.

m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1

Note that in function notation we can obtain two correspondingvalues for the outputy1y1andy2y2for thefunctionf,f,y1=f(x1)y1=f(x1)andy2=f(x2),y2=f(x2),sowe could equivalently write

m=f(x2)–f(x1)x2–x1m=f(x2)–f(x1)x2–x1

Figure 6indicates how the slope of the line betweenthepoints,(x1,y1)(x1,y1)and(x2,y2),(x2,y2),iscalculated. Recall that the slope measures steepness, or slant. Thegreater the absolute value of the slope, the steeper the slantis.

Figure6The slopeof a function is calculated by the change inyydividedby the change inx.x.It does not matter which coordinateis used as the(x2,y2)(x2,y2)and which isthe(x1,y1),(x1,y1),as long as each calculation isstarted with the elements from the same coordinate pair.

Q&A

Are the units for slopealwaysunitsfortheoutputunitsfortheinput?unitsfortheoutputunitsfortheinput?

Yes. Think of the units as the changeof output value for each unit of change in input value. An exampleof slope could be miles per hour or dollars per day. Notice theunits appear as a ratio of units for the output per units for theinput.

CALCULATE SLOPE

The slope, or rate of change, of a functionmmcan becalculated according to the following:

m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1

wherex1x1andx2x2are inputvalues,y1y1andy2y2are output values.

HOWTO

Given two points from a linear function, calculate andinterpret the slope.

Determine the units for output and input values.
Calculate the change of output values and change of inputvalues.
Interpret the slope as the change in output values per unit ofthe input value.

EXAMPLE3

Finding the Slope of a LinearFunction

Iff(x)f(x)is a linear function,and(3,−2)(3,−2)and(8,1)(8,1)are points onthe line, find the slope. Is this function increasing ordecreasing?

Answer

Analysis

As noted earlier, the order in which we write the points doesnot matter when we compute the slope of the line as long as thefirst output value, ory-coordinate, used corresponds with the first inputvalue, orx-coordinate, used.Note that if we had reversed them, we would have obtained the sameslope.

m=(−2)−(1)3−8=−3−5=35m=(−2)−(1)3−8=−3−5=35

TRYIT#1

Iff(x)f(x)is a linear function,and(2,3)(2,3)and(0,4)(0,4)are points on theline, find the slope. Is this function increasing ordecreasing?

EXAMPLE4

Finding the Population Change from aLinear Function

The population of a city increased from 23,400 to 27,800 between2008 and 2012. Find the change of population per year if we assumethe change was constant from 2008 to 2012.

Answer

Analysis

Because we are told that the population increased, we wouldexpect the slope to be positive. This positive slope we calculatedis therefore reasonable.

TRYIT#2

The population of a small town increased from 1,442 to 1,868between 2009 and 2012. Find the change of population per year if weassume the change was constant from 2009 to 2012.

Writing and Interpreting an Equationfor a Linear Function

Recall fromEquationsand Inequalitiesthat we wrote equations in boththeslope-intercept formand thepoint-slope form.Now we can choose which method to use to write equations for linearfunctions based on the information we are given. That informationmay be provided in the form of a graph, a point and a slope, twopoints, and so on. Look at the graph of thefunctionffinFigure7.

Figure7

We are not given the slope of the line, but we can choose anytwo points on the line to find the slope. Let’schoose(0,7)(0,7)and(4,4).(4,4).

m===y2−y1x2−x14−74−0−34m=y2−y1x2−x1=4−74−0=−34

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

y−y1y−4==m(x−x1)−34(x−4)y−y1=m(x−x1)y−4=−34(x−4)

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

y−4y−4y===−34(x−4)−34x+3−34x+7y−4=−34(x−4)y−4=−34x+3y=−34x+7

If we want to find the slope-intercept form without firstwriting the point-slope form, we could have recognized that theline crosses they-axis whenthe output value is 7. Therefore,b=7.b=7.We now havethe initial valuebband the slopemmso we cansubstitutemmandbbinto the slope-interceptform of a line.

So the function isf(x)=−34x+7,f(x)=−34x+7,and thelinear equation would bey=−34x+7.y=−34x+7.

HOWTO

Given the graph of a linear function, write an equationto represent the function.

Identify two points on the line.
Use the two points to calculate the slope.
Determine where the line crosses they-axis to identify they-intercept by visual inspection.
Substitute the slope andy-intercept into the slope-intercept form of a lineequation.

EXAMPLE5

Writing an Equation for a LinearFunction

Write an equation for a linear function given a graphofffshown inFigure8.

Figure8

Answer

Analysis

This makes sense because we can see fromFigure9that the line crosses they-axis at the point(0,2),(0,2),which isthey-intercept,sob=2.b=2.

Figure9

EXAMPLE6

Writing an Equation for a Linear CostFunction

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 linearfunctionCCwhereC(x)C(x)is the costforxxitems produced in a given month.

Answer

Analysis

If Ben produces 100 items in a month, his monthly cost is foundby substituting 100 forx.x.

C(100)==1250+37.5(100)5000C(100)=1250+37.5(100)=5000

So his monthly cost would be $5,000.

EXAMPLE7

Writing an Equation for a LinearFunction Given Two Points

Ifffis a linear function,withf(3)=−2,f(3)=−2,andf(8)=1,f(8)=1,findan equation for the function in slope-intercept form.

Answer

TRYIT#3

Iff(x)f(x)is a linear function,withf(2)=–11,f(2)=–11,andf(4)=−25,f(4)=−25,writean equation for the function in slope-intercept form.

Modeling Real-World Problems withLinear Functions

In the real world, problems are not always explicitly stated interms of a function or represented with a graph. Fortunately, wecan analyze the problem by first representing it as a linearfunction and then interpreting the components of the function. Aslong as we know, or can figure out, the initial value and the rateof change of a linear function, we can solve many different kindsof real-world problems.

HOWTO

Given a linear functionffand the initialvalue and rate of change, evaluatef(c).f(c).

Determine the initial value and the rate of change(slope).
Substitute the values intof(x)=mx+b.f(x)=mx+b.
Evaluate the function atx=c.x=c.

EXAMPLE8

Using a Linear Function to Determinethe Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Everymonth, he adds 15 new songs. Write a formula for the number ofsongs,N,N,in his collection as a function oftime,t,t,the number of months. How many songs will heown at the end of one year?

Answer

Analysis

Notice thatNis anincreasing linear function. As the input (the number of months)increases, the output (number of songs) increases as well.

EXAMPLE9

Using a Linear Function to CalculateSalary Based on Commission

Working as an insurance salesperson, Ilya earns a base salaryplus a commission on each new policy. Therefore, Ilya’s weeklyincomeI,I,depends on the number of newpolicies,n,n,he sells during the week. Last week hesold 3 new policies, and earned $760 for the week. The week before,he sold 5 new policies and earned $920. Find an equationforI(n),I(n),and interpret the meaning of thecomponents of the equation.

Answer

EXAMPLE10

Using Tabular Form to Write anEquation for a Linear Function

Table 1relates the number of rats in a population totime, in weeks. Use the table to write a linear equation.

number ofweeks,w	0	2	4	6
number ofrats,P(w)	1000	1080	1160	1240

Table1

Q&A

Is the initial value always provided in a table ofvalues likeTable1?

No. Sometimes the initial value isprovided in a table of values, but sometimes it is not. If you seean input of 0, then the initial value would be the correspondingoutput. If the initial value is not provided because there is novalue of input on the table equal to 0, find the slope, substituteone coordinate pair and the slopeintof(x)=mx+b,f(x)=mx+b,and solveforb.b.

TRYIT#4

A new plant food was introduced to a young tree to test itseffect on the height of the tree.Table2shows the height of the tree, infeet,xxmonths since the measurements began. Write alinear function,H(x),H(x),wherexxis thenumber 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

Table2

Graphing LinearFunctions

Now that we’ve seen and interpreted graphs of linear functions,let’s take a look at how to create the graphs. There are threebasic methods of graphing linear functions. The first is byplotting points and then drawing a line through the points. Thesecond is by using they-intercept and slope. And the third method is byusing transformations of the identityfunctionf(x)=x.f(x)=x.

Graphing a Function by PlottingPoints

To find points of a function, we can choose input values,evaluate the function at these input values, and calculate outputvalues. The input values and corresponding output values formcoordinate pairs. We then plot the coordinate pairs on a grid. Ingeneral, we should evaluate the function at a minimum of two inputsin order to find at least two points on the graph. For example,given the function,f(x)=2x,f(x)=2x,we might use theinput values 1 and 2. Evaluating the function for an input value of1 yields an output value of 2, which is represented by thepoint(1,2).(1,2).Evaluating the function for an inputvalue of 2 yields an output value of 4, which is represented by thepoint(2,4).(2,4).Choosing three points is oftenadvisable because if all three points do not fall on the same line,we know we made an error.

HOWTO

Given a linear function, graph by plottingpoints.

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

EXAMPLE11

Graphing by PlottingPoints

Graphf(x)=−23x+5f(x)=−23x+5by plotting points.

Answer

Analysis

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

TRYIT#5

Graphf(x)=−34x+6f(x)=−34x+6by plotting points.

Graphing a FunctionUsingy-intercept andSlope

Another way to graph linear functions is by using specificcharacteristics of the function rather than plotting points. Thefirst characteristic is itsy-intercept, which is the point at which the inputvalue is zero. To find they-intercept, we can setx=0x=0in theequation.

The other characteristic of the linear function is itsslope.

Let’s consider the following function.

f(x)=12x+1f(x)=12x+1

The slope is12.12.Because the slope is positive, weknow the graph will slant upward from left to right.They-intercept is the point onthe graph whenx=0.x=0.The graph crossesthey-axisat(0,1).(0,1).Now we know the slope andthey-intercept. We can begingraphing by plotting the point(0,1).(0,1).We know thatthe slope is the change in they-coordinate over the change inthex-coordinate. This iscommonly referred to as rise overrun,m=riserun.m=riserun.From our example, wehavem=12,m=12,which means that the rise is 1 and therun is 2. So starting from oury-intercept(0,1),(0,1),we can rise 1 andthen run 2, or run 2 and then rise 1. We repeat until we have a fewpoints, and then we draw a line through the points as showninFigure12.

Figure12

GRAPHICAL INTERPRETATION OF A LINEARFUNCTION

In the equationf(x)=mx+bf(x)=mx+b

bbis they-interceptof the graph and indicates the point(0,b)(0,b)at whichthe graph crosses they-axis.
mmis the slope of the line and indicates the verticaldisplacement (rise) and horizontal displacement (run) between eachsuccessive pair of points. Recall the formula for the slope:

m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1m=changeinoutput(rise)changeininput(run)=ΔyΔx=y2−y1x2−x1

Q&A

Do all linear functions havey-intercepts?

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

HOWTO

Given the equation for a linear function, graph thefunction using they-interceptand slope.

Evaluate the function at an input value of zero to findthey-intercept.
Identify the slope as the rate of change of the inputvalue.
Plot the point represented by they-intercept.
Useriserunriserunto determine at least two morepoints on the line.
Sketch the line that passes through the points.

EXAMPLE12

Graphing by Usingthey-intercept andSlope

Graphf(x)=−23x+5f(x)=−23x+5usingthey-intercept and slope.

Answer

Analysis

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

TRYIT#6

Find a point on the graph we drew inExample12that has a negativex-value.

Graphing a Function UsingTransformations

Another option for graphing is to useatransformationof the identityfunctionf(x)=x.f(x)=x.A function may be transformed bya shift up, down, left, or right. A function may also betransformed using a reflection, stretch, or compression.

Vertical Stretch orCompression

In the equationf(x)=mx,f(x)=mx,themmisacting as theverticalstretchorcompressionof the identity function.Whenmmis negative, there is also a vertical reflectionof the graph. Notice inFigure14that multiplying the equationoff(x)=xf(x)=xbymmstretches the graphofffby a factor ofmmunitsifm>1m>1and compresses the graphofffby a factor ofmmunitsif0<m<1.0<m<1.This means the larger theabsolute value ofm,m,the steeper the slope.

Figure14Verticalstretches and compressions and reflections on thefunctionf(x)=xf(x)=x

Vertical Shift

Inf(x)=mx+b,f(x)=mx+b,thebbacts asthevertical shift, moving the graph up and down withoutaffecting the slope of the line. Notice inFigure15that adding a value ofbbto the equationoff(x)=xf(x)=xshifts the graph offfa totalofbbunits up ifbbis positiveand|b||b|units down ifbbis negative.

Figure15This graphillustrates vertical shifts of the functionf(x)=x.f(x)=x.

Using vertical stretches or compressions along with verticalshifts is another way to look at identifying different types oflinear functions. Although this may not be the easiest way to graphthis type of function, it is still important to practice eachmethod.

HOWTO

Given the equation of a linear function, usetransformations to graph the linear function in theformf(x)=mx+b.f(x)=mx+b.

Graphf(x)=x.f(x)=x.
Vertically stretch or compress the graph by afactorm.m.
Shift the graph up or downbbunits.

EXAMPLE13

Graphing by UsingTransformations

Graphf(x)=12x−3f(x)=12x−3using transformations.

Answer

TRYIT#7

Graphf(x)=4+2xf(x)=4+2xusing transformations.

Q&A

InExample15, could we have sketched the graph by reversing the orderof the transformations?

No. The order of the transformationsfollows the order of operations. When the function is evaluated ata given input, the corresponding output is calculated by followingthe order of operations. This is why we performed the compressionfirst. For example, following the order: Let the input be2.

f(2)===12(2)−31−3−2f(2)=12(2)−3=1−3=−2

Writing the Equation for a Functionfrom the Graph of a Line

Earlier, we wrote the equation for a linear function from agraph. Now we can extend what we know about graphing linearfunctions to analyze graphs a little more closely. Begin by takinga look atFigure18. We can see right away that the graph crossesthey-axis at thepoint(0,4)(0,4)so this is they-intercept.

Figure18

Then we can calculate the slope by finding the rise and run. Wecan choose any two points, but let’s look at thepoint(–2,0).(–2,0).To get from this point tothey-intercept, we must moveup 4 units (rise) and to the right 2 units (run). So the slope mustbe

m=riserun=42=2m=riserun=42=2

Substituting the slope andy-intercept into the slope-intercept form of a linegives

y=2x+4y=2x+4

HOWTO

Given a graph of linear function, find the equation todescribe the function.

Identify they-intercept ofan equation.
Choose two points to determine the slope.
Substitute they-interceptand slope into the slope-intercept form of a line.

EXAMPLE14

Matching Linear Functions to TheirGraphs

Match each equation of the linear functions with one of thelines inFigure19.

ⓐf(x)=2x+3f(x)=2x+3
ⓑg(x)=2x−3g(x)=2x−3
ⓒh(x)=−2x+3h(x)=−2x+3
ⓓj(x)=12x+3j(x)=12x+3

Figure19

Answer

Finding thex-intercept of a Line

So far we have been finding they-intercepts of a function: the point at which thegraph of the function crosses they-axis. Recall that a function may also haveanx-intercept, which isthex-coordinate of the pointwhere the graph of the function crosses thex-axis. In other words, it is the input value whenthe output value is zero.

To find thex-intercept, seta functionf(x)f(x)equal to zero and solve for the valueofx.x.For example, consider the function shown.

f(x)=3x−6f(x)=3x−6

Set the function equal to 0 and solve forx.x.

062x====3x−63xx20=3x−66=3x2=xx=2

The graph of the function crosses thex-axis at the point(2,0).(2,0).

Q&A

Do all linear functions havex-intercepts?

No. However, linear functions of theformy=c,y=c,whereccis a nonzero real numberare the only examples of linear functions with no x-intercept. Forexample,y=5y=5is a horizontal line 5 units above thex-axis. This function has no x-intercepts, as showninFigure21.

Figure21

X-INTERCEPT

Thex-intercept of thefunction is value ofxxwhenf(x)=0.f(x)=0.Itcan be solved by the equation0=mx+b.0=mx+b.

EXAMPLE15

Finding anx-intercept

Find thex-interceptoff(x)=12x−3.f(x)=12x−3.

Answer

Analysis

A graph of the function is shown inFigure22. We can see that thex-intercept is(6,0)(6,0)as weexpected.

Figure22

TRYIT#8

Find thex-interceptoff(x)=14x−4.f(x)=14x−4.

Describing Horizontal and VerticalLines

There are two special cases of lines on a graph—horizontal andvertical lines. Ahorizontal lineindicates a constantoutput, ory-value.InFigure23, we see that the output has a value of 2 for every inputvalue. The change in outputs between any two points, therefore, is0. In the slope formula, the numerator is 0, so the slope is 0. Ifwe usem=0m=0in theequationf(x)=mx+b,f(x)=mx+b,the equation simplifiestof(x)=b.f(x)=b.In other words, the value of thefunction is a constant. This graph represents thefunctionf(x)=2.f(x)=2.

Figure23Ahorizontal line representing the functionf(x)=2f(x)=2

Avertical lineindicates a constant input,orx-value. We can see that theinput value for every point on the line is 2, but the output valuevaries. Because this input value is mapped to more than one outputvalue, a vertical line does not represent a function. Notice thatbetween any two points, the change in the input values is zero. Inthe slope formula, the denominator will be zero, so the slope of avertical line is undefined.

Figure24Example ofhow a line has a vertical slope. 0 in the denominator of theslope.

A vertical line, such as the one inFigure25,has anx-intercept, but noy-intercept unless it’s thelinex=0.x=0.This graph represents thelinex=2.x=2.

Figure25Thevertical line,x=2,x=2,which does not represent afunction

HORIZONTAL AND VERTICALLINES

Lines can be horizontal or vertical.

Ahorizontal lineis a line definedby an equation in the formf(x)=b.f(x)=b.

Avertical lineis a line defined byan equation in the formx=a.x=a.

EXAMPLE16

Writing the Equation of a HorizontalLine

Write the equation of the line graphed inFigure26.

Figure26

Answer

EXAMPLE17

Writing the Equation of a VerticalLine

Write the equation of the line graphed inFigure27.

Figure27

Answer

Determining Whether Lines areParallel or Perpendicular

The two lines inFigure28areparallel lines: they will never intersect.They have exactly the same steepness, which means their slopes areidentical. The only difference between the two lines isthey-intercept. If we shiftedone line vertically toward the other, they would becomecoincident.

Figure28Parallellines

We can determine from their equations whether two lines areparallel by comparing their slopes. If the slopes are the same andthey-intercepts are different,the lines are parallel. If the slopes are different, the lines arenot parallel.

f(x)=−2x+6f(x)=−2x−4}parallelf(x)=3x+2f(x)=2x+2}notparallelf(x)=−2x+6f(x)=−2x−4}parallelf(x)=3x+2f(x)=2x+2}notparallel

Unlike parallel lines,perpendicular linesdointersect. Their intersection forms a right, or 90-degree, angle.The two lines inFigure29are perpendicular.

Figure29Perpendicularlines

Perpendicular lines do not have the same slope. The slopes ofperpendicular lines are different from one another in a specificway. The slope of one line is the negative reciprocal of the slopeof the other line. The product of a number and its reciprocalis1.1.So, ifm1andm2m1andm2arenegative reciprocals of one another, they can be multipliedtogether to yield–1.–1.

m1m2=−1m1m2=−1

To find the reciprocal of a number, divide 1 by the number. Sothe reciprocal of 8 is18,18,and the reciprocalof1818is 8. To find the negative reciprocal, first findthe reciprocal and then change the sign.

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

f(x)f(x)==14x+2−4x+3negativereciprocalof14is−4negativereciprocalof−4is14f(x)=14x+2negativereciprocalof14is−4f(x)=−4x+3negativereciprocalof−4is14

The product of the slopes is –1.

−4(14)=−1−4(14)=−1

PARALLEL AND PERPENDICULARLINES

Two lines areparallel linesif theydo not intersect. The slopes of the lines are the same.

f(x)=m1x+b1andg(x)=m2x+b2areparallelifandonlyifm1=m2f(x)=m1x+b1andg(x)=m2x+b2areparallelifandonlyifm1=m2

If and onlyifb1=b2b1=b2andm1=m2,m1=m2,we say the linescoincide. Coincident lines are the same line.

Two lines areperpendicular linesifthey intersect to form a right angle.

f(x)=m1x+b1andg(x)=m2x+b2areperpendicularifandonlyiff(x)=m1x+b1andg(x)=m2x+b2areperpendicularifandonlyif

m1m2=−1,som2=−1m1m1m2=−1,som2=−1m1

EXAMPLE18

Identifying Parallel andPerpendicular Lines

Given the functions below, identify the functions whose graphsare a pair of parallel lines and a pair of perpendicular lines.

f(x)g(x)==2x+312x−4h(x)j(x)==−2x+22x−6f(x)=2x+3h(x)=−2x+2g(x)=12x−4j(x)=2x−6

Answer

Analysis

A graph of the lines is shown inFigure30.

Figure30

The graph shows that thelinesf(x)=2x+3f(x)=2x+3andj(x)=2x–6j(x)=2x–6areparallel, and thelinesg(x)=12x–4g(x)=12x–4andh(x)=−2x+2h(x)=−2x+2areperpendicular.

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

If we know the equation of a line, we can use what we know aboutslope to write the equation of a line that is either parallel orperpendicular to the given line.

Writing Equations of ParallelLines

Suppose for example, we are given the equation shown.

f(x)=3x+1f(x)=3x+1

We know that the slope of the line formed by the function is 3.We also know that they-intercept is(0,1).(0,1).Any other linewith a slope of 3 will be parallel tof(x).f(x).So thelines formed by all of the following functions will be paralleltof(x).f(x).

g(x)h(x)p(x)===3x+63x+13x+23g(x)=3x+6h(x)=3x+1p(x)=3x+23

Suppose then we want to write the equation of a line that isparallel toffand passes through thepoint(1,7).(1,7).This type of problem is oftendescribed as a point-slope problem because we have a point and aslope. In our example, we know that the slope is 3. We need todetermine which value ofbbwill 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.

y−y1y−7y−7y====m(x−x1)3(x−1)3x−33x+4y−y1=m(x−x1)y−7=3(x−1)y−7=3x−3y=3x+4

Sog(x)=3x+4g(x)=3x+4is paralleltof(x)=3x+1f(x)=3x+1and passes through thepoint(1,7).(1,7).

HOWTO

Given the equation of a function and a point throughwhich its graph passes, write the equation of a line parallel tothe given line that passes through the given point.

Find the slope of the function.
Substitute the given values into either the general point-slopeequation or the slope-intercept equation for a line.
Simplify.

EXAMPLE19

Finding a Line Parallel to a GivenLine

Find a line parallel to the graphoff(x)=3x+6f(x)=3x+6that passes through thepoint(3,0).(3,0).

Answer

Analysis

We can confirm that the two lines are parallel by graphingthem.Figure31shows that the two lines will never intersect.

Figure31

Writing Equations of PerpendicularLines

We can use a very similar process to write the equation for aline perpendicular to a given line. Instead of using the sameslope, however, we use the negative reciprocal of the given slope.Suppose we are given the function shown.

f(x)=2x+4f(x)=2x+4

The slope of the line is 2, and its negative reciprocalis−12.−12.Any function with a slopeof−12−12will be perpendiculartof(x).f(x).So the lines formed by all of the followingfunctions will be perpendicular tof(x).f(x).

g(x)h(x)p(x)===−12x+4−12x+2−12x−12g(x)=−12x+4h(x)=−12x+2p(x)=−12x−12

As before, we can narrow down our choices for a particularperpendicular line if we know that it passes through a given point.Suppose then we want to write the equation of a line that isperpendicular tof(x)f(x)and passes through thepoint(4,0).(4,0).We already know that the slopeis−12.−12.Now we can use the point to findthey-intercept by substitutingthe given values into the slope-intercept form of a line andsolving forb.b.

g(x)002b=====mx+b−12(4)+b−2+bb2g(x)=mx+b0=−12(4)+b0=−2+b2=bb=2

The equation for the function with a slopeof−12−12and ay-intercept of 2 is

g(x)=−12x+2g(x)=−12x+2

Sog(x)=−12x+2g(x)=−12x+2is perpendiculartof(x)=2x+4f(x)=2x+4and passes through thepoint(4,0).(4,0).Be aware that perpendicular lines maynot look obviously perpendicular on a graphing calculator unless weuse the square zoom feature.

Q&A

A horizontal line has a slope of zero and a verticalline has an undefined slope. These two lines are perpendicular, butthe product of their slopes is not –1. Doesn’t this fact contradictthe definition of perpendicular lines?

No. For two perpendicular linearfunctions, the product of their slopes is –1. However, a verticalline is not a function so the definition is notcontradicted.

HOWTO

Given the equation of a function and a point throughwhich its graph passes, write the equation of a line perpendicularto the given line.

Find the slope of the function.
Determine the negative reciprocal of the slope.
Substitute the new slope and the valuesforxxandyyfrom the coordinate pair providedintog(x)=mx+b.g(x)=mx+b.
Solve forb.b.
Write the equation of the line.

EXAMPLE20

Finding the Equation of aPerpendicular Line

Find the equation of a line perpendiculartof(x)=3x+3f(x)=3x+3that passes through thepoint(3,0).(3,0).

Answer

Analysis

A graph of the two lines is shown inFigure32.

Figure32

Note that that if we graph perpendicular lines on a graphingcalculator using standard zoom, the lines may not appear to beperpendicular. Adjusting the window will make it possible to zoomin further to see the intersection more closely.

TRYIT#9

Given the functionh(x)=2x−4,h(x)=2x−4,write anequation for the line passing through(0,0)(0,0)thatis

ⓐparallel toh(x)h(x)
ⓑperpendicular toh(x)h(x)

HOWTO

Given two points on a line and a third point, write theequation of the perpendicular line that passes through thepoint.

Determine the slope of the line passing through thepoints.
Find the negative reciprocal of the slope.
Use the slope-intercept form or point-slope form to write theequation by substituting the known values.
Simplify.

EXAMPLE21

Finding the Equation of a LinePerpendicular to a Given Line Passing through a Point

A line passes through thepoints(−2,6)(−2,6)and(4,5).(4,5).Find theequation of a perpendicular line that passes through thepoint(4,5).(4,5).

Answer

TRYIT#10

A line passes through thepoints,(−2,−15)(−2,−15)and(2,−3).(2,−3).Findthe equation of a perpendicular line that passes through thepoint,(6,4).(6,4).

MEDIA

Access this online resource for additional instruction andpractice with linear functions.

4.1 SectionExercises

Verbal

Terry is skiing down a steep hill. Terry'selevation,E(t),E(t),in feet afterttsecondsis given byE(t)=3000−70t.E(t)=3000−70t.Write a completesentence describing Terry’s starting elevation and how it ischanging over time.

Jessica is walking home from a friend’s house. After 2 minutesshe is 1.4 miles from home. Twelve minutes after leaving, she is0.9 miles from home. What is her rate in miles per hour?

A boat is 100 miles away from the marina, sailing directlytoward it at 10 miles per hour. Write an equation for the distanceof the boat from the marina afterthours.

If the graphs of two linear functions are perpendicular,describe the relationship between the slopes andthey-intercepts.

If a horizontal line has the equationf(x)=af(x)=aanda vertical line has the equationx=a,x=a,what is thepoint of intersection? Explain why what you found is the point ofintersection.

Algebraic

For the following exercises, determine whether the equation ofthe curve can be written as a linear function.

y=14x+6y=14x+6

y=3x−5y=3x−5

y=3x2−2y=3x2−2

3x+5y=153x+5y=15

10.

3x2+5y=153x2+5y=15

11.

3x+5y2=153x+5y2=15

12.

−2x2+3y2=6−2x2+3y2=6

13.

−x−35=2y−x−35=2y

For the following exercises, determine whether each function isincreasing or decreasing.

14.

f(x)=4x+3f(x)=4x+3

15.

g(x)=5x+6g(x)=5x+6

16.

a(x)=5−2xa(x)=5−2x

17.

b(x)=8−3xb(x)=8−3x

18.

h(x)=−2x+4h(x)=−2x+4

19.

k(x)=−4x+1k(x)=−4x+1

20.

j(x)=12x−3j(x)=12x−3

21.

p(x)=14x−5p(x)=14x−5

22.

n(x)=−13x−2n(x)=−13x−2

23.

m(x)=−38x+3m(x)=−38x+3

For the following exercises, find the slope of the line thatpasses through the two given points.

24.

(2,4)(2,4)and(4,10)(4,10)

25.

(1,5)(1,5)and(4,11)(4,11)

26.

(–1,4)(–1,4)and(5,2)(5,2)

27.

(8,–2)(8,–2)and(4,6)(4,6)

28.

(6,11)(6,11)and(–4,3)(–4,3)

For the following exercises, given each set of information, finda linear equation satisfying the conditions, if possible.

29.

f(−5)=−4,f(−5)=−4,andf(5)=2f(5)=2

30.

f(−1)=4,f(−1)=4,andf(5)=1f(5)=1

31.

Passes through(2,4)(2,4)and(4,10)(4,10)

32.

Passes through(1,5)(1,5)and(4,11)(4,11)

33.

Passes through(−1,4)(−1,4)and(5,2)(5,2)

34.

Passes through(−2,8)(−2,8)and(4,6)(4,6)

35.

xinterceptat(−2,0)(−2,0)andyintercept at(0,−3)(0,−3)

36.

xinterceptat(−5,0)(−5,0)andyintercept at(0,4)(0,4)

For the following exercises, determine whether the lines givenby the equations below are parallel, perpendicular, or neither.

37.

4x−7y=107x+4y=14x−7y=107x+4y=1

38.

3y+x=12−y=8x+13y+x=12−y=8x+1

39.

3y+4x=12−6y=8x+13y+4x=12−6y=8x+1

40.

6x−9y=103x+2y=16x−9y=103x+2y=1

For the following exercises, find thex- andy-intercepts of each equation.

41.

f(x)=−x+2f(x)=−x+2

42.

g(x)=2x+4g(x)=2x+4

43.

h(x)=3x−5h(x)=3x−5

44.

k(x)=−5x+1k(x)=−5x+1

45.

−2x+5y=20−2x+5y=20

46.

7x+2y=567x+2y=56

For the following exercises, use the descriptions of each pairof lines given below to find the slopes of Line 1 and Line 2. Iseach pair of lines parallel, perpendicular, or neither?

47.

Line 1: Passesthrough(0,6)(0,6)and(3,−24)(3,−24)

Line 2: Passesthrough(−1,19)(−1,19)and(8,−71)(8,−71)

48.

Line 1: Passesthrough(−8,−55)(−8,−55)and(10,89)(10,89)

Line 2: Passesthrough(9,−44)(9,−44)and(4,−14)(4,−14)

49.

Line 1: Passesthrough(2,3)(2,3)and(4,−1)(4,−1)

Line 2: Passesthrough(6,3)(6,3)and(8,5)(8,5)

50.

Line 1: Passesthrough(1,7)(1,7)and(5,5)(5,5)

Line 2: Passesthrough(−1,−3)(−1,−3)and(1,1)(1,1)

51.

Line 1: Passesthrough(2,5)(2,5)and(5,−1)(5,−1)

Line 2: Passesthrough(−3,7)(−3,7)and(3,−5)(3,−5)

For the following exercises, write an equation for the linedescribed.

52.

Write an equation for a line paralleltof(x)=−5x−3f(x)=−5x−3and passing through thepoint(2,–12).(2,–12).

53.

Write an equation for a line paralleltog(x)=3x−1g(x)=3x−1and passing through thepoint(4,9).(4,9).

54.

Write an equation for a line perpendiculartoh(t)=−2t+4h(t)=−2t+4and passing through thepoint(−4,–1).(−4,–1).

55.

Write an equation for a line perpendiculartop(t)=3t+4p(t)=3t+4and passing through thepoint(3,1).(3,1).

Graphical

For the following exercises, find the slope of thelinegraphed.

56.

57.

For the following exercises, write an equation for the linegraphed.

58.

59.

60.

61.

62.

63.

For the following exercises, match the given linear equationwith its graph inFigure33.

Figure33

64.

f(x)=−x−1f(x)=−x−1

65.

f(x)=−3x−1f(x)=−3x−1

66.

f(x)=−12x−1f(x)=−12x−1

67.

f(x)=2f(x)=2

68.

f(x)=2+xf(x)=2+x

69.

f(x)=3x+2f(x)=3x+2

For the following exercises, sketch a line with the givenfeatures.

70.

Anx-interceptof(–4,0)(–4,0)andy-intercept of(0,–2)(0,–2)

71.

Anx-intercept(–2,0)(–2,0)andy-interceptof(0,4)(0,4)

72.

Ay-interceptof(0,7)(0,7)and slope−32−32

73.

Ay-interceptof(0,3)(0,3)and slope2525

74.

Passing through thepoints(–6,–2)(–6,–2)and(6,–6)(6,–6)

75.

Passing through thepoints(–3,–4)(–3,–4)and(3,0)(3,0)

For the following exercises, sketch the graph of eachequation.

76.

f(x)=−2x−1f(x)=−2x−1

77.

f(x)=−3x+2f(x)=−3x+2

78.

f(x)=13x+2f(x)=13x+2

79.

f(x)=23x−3f(x)=23x−3

80.

f(t)=3+2tf(t)=3+2t

81.

p(t)=−2+3tp(t)=−2+3t

82.

x=3x=3

83.

x=−2x=−2

84.

r(x)=4r(x)=4

For the following exercises, write the equation of the lineshown in the graph.

85.

86.

87.

88.

Numeric

For the following exercises, which of the tables could representa linear function? For each that could be linear, find a linearequation that models the data.

89.

xx	0	5	10	15
g(x)g(x)	5	–10	–25	–40

90.

xx	0	5	10	15
h(x)h(x)	5	30	105	230

91.

xx	0	5	10	15
f(x)f(x)	–5	20	45	70

92.

xx	5	10	20	25
k(x)k(x)	13	28	58	73

93.

xx	0	2	4	6
g(x)g(x)	6	–19	–44	–69

94.

xx	2	4	8	10
h(x)h(x)	13	23	43	53

95.

xx	2	4	6	8
f(x)f(x)	–4	16	36	56

96.

xx	0	2	6	8
k(x)k(x)	6	31	106	231

Technology

For the following exercises, use a calculator or graphingtechnology to complete the task.

97.

Ifffis a linearfunction,f(0.1)=11.5f(0.1)=11.5,andf(0.4)=–5.9f(0.4)=–5.9, find an equation for thefunction.

98.

Graph the functionffon a domainof[–10,10]:f(x)=0.02x−0.01.[–10,10]:f(x)=0.02x−0.01.Enterthe function in a graphing utility. For the viewing window, set theminimum value ofxxto be−10−10and themaximum value ofxxto be1010.

99.

Graph the functionffon a domainof[–10,10]:fx)=2,500x+4,000[–10,10]:fx)=2,500x+4,000

100.

Table 3shows the input,w,w,andoutput,k,k,for a linear functionk.k.

ⓐFill in the missing values of the table.
ⓑWrite the linear function

k,k,round to 3 decimal places.

w	–10	5.5	67.5	b
k	30	–26	a	–44

Table3

101.

Table 4shows the input,p,p,andoutput,q,q,for a linear functionq.q.

ⓐFill in the missing values of the table.
ⓑWrite the linear function

k.k.

p	0.5	0.8	12	b
q	400	700	a	1,000,000

Table4

102.

Graph the linear functionffon a domainof[−10,10][−10,10]for the function whoseslope is1818andy-intercept is3116.3116.Label the pointsfor the input values of−10−10and10.10.

103.

Graph the linear functionffon a domainof[−0.1,0.1][−0.1,0.1]for the functionwhose slope is 75 andy-intercept is−22.5.−22.5.Label thepoints for the input valuesof−0.1−0.1and0.1.0.1.

104.

Graph the linearfunctionffwheref(x)=ax+bf(x)=ax+bon thesame set of axes on a domainof[−4,4][−4,4]for the following valuesofaaandb.b.

ⓐa=2;b=3a=2;b=3
ⓑa=2;b=4a=2;b=4
ⓓa=2;b=–5a=2;b=–5

Extensions

105.

Find the value ofxxif a linear function goes throughthe following points and has the followingslope:(x,2),(−4,6),m=3(x,2),(−4,6),m=3

106.

Find the value ofyifa linear function goes through the following points and has thefollowing slope:(10,y),(25,100),m=−5(10,y),(25,100),m=−5

107.

Find the equation of the line that passes through the followingpoints:

(a,b)(a,b)and(a,b+1)(a,b+1)

108.

Find the equation of the line that passes through the followingpoints:

(2a,b)(2a,b)and(a,b+1)(a,b+1)

109.

Find the equation of the line that passes through the followingpoints:

(a,0)(a,0)and(c,d)(c,d)

110.

Find the equation of the line parallel to thelineg(x)=−0.01x+2.01g(x)=−0.01x+2.01through thepoint(1,2).(1,2).

111.

Find the equation of the line perpendicular to thelineg(x)=−0.01x+2.01g(x)=−0.01x+2.01through thepoint(1,2).(1,2).

For the following exercises, use thefunctionsf(x)=−0.1x+200andg(x)=20x+0.1.f(x)=−0.1x+200andg(x)=20x+0.1.

112.

Find the point of intersection of thelinesffandg.g.

113.

Where isf(x)f(x)greaterthang(x)?g(x)?Where isg(x)g(x)greaterthanf(x)?f(x)?

Real-World Applications

114.

At noon, a barista notices that they have $20 in their tip jar.If the barista makes an average of $0.50 from each customer, howmuch will they have in the tip jar if they servennmorecustomers during the shift?

115.

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?

116.

A clothing business finds there is a linear relationship betweenthe number of shirts,n,n,it can sell and theprice,p,p,it can charge per shirt. In particular,historical data shows that 1,000 shirts can be sold at a priceof$30,$30,while 3,000 shirts can be sold at a price of$22. Find a linear equation in theformp(n)=mn+bp(n)=mn+bthat gives thepriceppthey can charge fornnshirts.

117.

A phone company charges for service according to theformula:C(n)=24+0.1n,C(n)=24+0.1n,wherennisthe number of minutes talked, andC(n)C(n)is the monthlycharge, in dollars. Find and interpret the rate of change andinitial value.

118.

A farmer finds there is a linear relationship between the numberof bean stalks,n,n,she plants and theyield,y,y,each plant produces. When she plants 30stalks, each plant yields 30 oz of beans. When she plants 34stalks, each plant produces 28 oz of beans. Find a linearrelationships in the formy=mn+by=mn+bthat gives theyield whennnstalks are planted.

119.

A city’s population in the year 1960 was 287,500. In 1989 thepopulation was 275,900. Compute the rate of growth of thepopulation and make a statement about the population rate of changein people per year.

120.

A town’s population has been growing linearly. In 2003, thepopulation was 45,000, and the population has been growing by 1,700people each year. Write an equation,P(t),P(t),for thepopulationttyears after 2003.

121.

Suppose that average annual income (in dollars) for the years1990 through 1999 is given by the linearfunction:I(x)=1054x+23,286I(x)=1054x+23,286,wherexxis the number of years after 1990. Which of thefollowing interprets the slope in the context of the problem?

ⓐAs of 1990, average annual income was $23,286.
ⓑIn the ten-year period from 1990–1999, average annual incomeincreased by a total of $1,054.
ⓓAverage annual income rose to a level of $23,286 by the end of1999.

122.

When temperature is 0 degrees Celsius, the Fahrenheittemperature is 32. When the Celsius temperature is 100, thecorresponding Fahrenheit temperature is 212. Express the Fahrenheittemperature as a linear function ofC,C,the Celsiustemperature,F(C).F(C).

ⓐFind the rate of change of Fahrenheit temperature for eachunit change temperature of Celsius.
ⓑFind and interpretF(28).F(28).

Footnotes

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