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Exercises 1 Every continuous linear equation must have two things, a slope m and a y-intercept b. The slope tells you how many units to go up or down for each unit you move to the right. The y-intercept is the point at which the function crosses the y-axis. When we have these pieces of information, we can write an equation in slope-intercept form: y=mx+b. The m and b are constants when the equation is used to graph a line. Therefore, a linear equation in two variables is an equation that can be written in the form y=mx+b. | |

Exercises 2 To remember the difference between a linear function and a nonlinear function, just remember that the word linear is just the adjective to describe something that is like a line.Linear: Resembling a straight line. Nonlinear: Literally "not linear," not resembling a straight line. | |

Exercises 3 To remember the difference between a discrete domain and a continuous domain, it is best to remember what it means to be continuous. The word "continuous" in every day life means that something is constantly happening without interruption. If the radio is continuously playing, no one ever turns it off, it plays for 24 hours a day and 7 days a week.Continuous domain: the set of possible inputs is uninterrupted for the entire interval. Discrete domain: the possible inputs are specific values within an interval, there can be interruptions as many times as needed. | |

Exercises 4 To remember the difference between a discrete domain and a continuous domain, it is best to remember what it means to be continuous. The word "continuous" in every day life means that something is constantly happening without interruption. It's the same basic principle for the graph of a continuous function, a function with a continuous domain. Let's look at two different, but very similar, graphs.The graph above shows six distinct points that have a domain of {0,1,2,3,4,5}. The second graph below shows a function that is an uninterrupted line segment that happens to cross through those same points.However, the domain of this function is 0≤x≤5. We can tell which one has a continuous domain because it is a continuous line in the appropriate interval. The graph with the discrete domain is a set of specific values in the appropriate interval, it is not a continuous line. | |

Exercises 5 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that there are parts of the given graph that do not lie on the same straight line as other parts of the graph. Because of this, we can conclude that the function is nonlinear. | |

Exercises 6 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that all parts of the given graph lie on the same straight line. Because of this, we can conclude that the function is linear. | |

Exercises 7 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that all parts of the given graph lie on the same straight line. Because of this, we can conclude that the function is linear. | |

Exercises 8 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that all parts of the given graph lie on the same straight line. Because of this, we can conclude that the function is linear. | |

Exercises 9 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that there are parts of the given graph that do not lie on the same straight line as other parts of the graph. Because of this, we can conclude that the function is nonlinear. | |

Exercises 10 The graph of a linear function is portrayed as a single, straight line in a coordinate plane. One way to determine if the given graph is linear is by comparing it with a straight edge, such as a ruler.We can see that there are parts of the given graph that do not lie on the same straight line as other parts of the graph. Because of this, we can conclude that the function is nonlinear. | |

Exercises 11 The graph of a linear function can be portrayed as a single, straight line in a coordinate plane. To begin determining if the data given in the table represents a linear function, let's first plot the data as (x,y) coordinate pairs.If the function is linear, connecting these points will form a straight line. Otherwise, we will have shown that the function is nonlinear. Let's connect each of our points with a straight edge and observe the result.All of the given points lie on the same line in the coordinate plane. Therefore, the function is linear. | |

Exercises 12 The graph of a linear function can be portrayed as a single, straight line in a coordinate plane. To begin determining if the data given in the table represents a linear function, let's first plot the data as (x,y) coordinate pairs.If the function is linear, connecting these points will form a straight line. Otherwise, we will have shown that the function is nonlinear. Let's connect each of our points with a straight edge and observe the result.Now that all of our points are connected, we can see that they do not lie on the same line in the coordinate plane. Therefore, the function is nonlinear. | |

Exercises 13 The graph of a linear function can be portrayed as a single, straight line in a coordinate plane. To begin determining if the data given in the table represents a linear function, let's first plot the data as (x,y) coordinate pairs.If the function is linear, connecting these points will form a straight line. Otherwise, we will have shown that the function is nonlinear. Let's connect each of our points with a straight edge and observe the result.Now that all of our points are connected, we can see that they do not lie on the same line in the coordinate plane. Therefore, the function is nonlinear. | |

Exercises 14 The graph of a linear function can be portrayed as a single, straight line in a coordinate plane. To begin determining if the data given in the table represents a linear function, let's first plot the data as (x,y) coordinate pairs.If the function is linear, connecting these points will form a straight line. Otherwise, we will have shown that the function is nonlinear. Let's connect each of our points with a straight edge and observe the result.All of the given points lie on the same line in the coordinate plane. Therefore, the function is linear. | |

Exercises 15 For every step, x increases by 2. However, y increases by a factor of 4 which means the rate of change is not constant. Therefore, the function is not linear. By graphing these points, we can see that they do not form a straight line. | |

Exercises 16 A line is a function if it has only one y-value for each x-value. Although, as we can see from the graph, there are infinitely many different y-values which correspond to x=-3. So the vertical line is not a function, let alone a linear function. | |

Exercises 17 To determine if the function is linear, we will create a table of values.xx2+13y 1(1)2+1314 2(2)2+1317 3(3)2+1322 4(4)2+1329 Analyzing the x- and y-columns, we can see that as x increases by 1, the amount that y changes by is not constant. 14 ⟶3 17 ⟶5 22 ⟶7 29 Since the rate of change is not constant, the function is nonlinear. To verify, let's plot our points in a graph.We can see that, when graphed, the points do not all lie on one straight line. | |

Exercises 18 To determine if the function is linear, we will create a table of values.x7−3xy 17−3(1)4 27−3(2)1 37−3(3)-2 47−3(4)-5 Analyzing the x- and y-columns, we can see that as x increases by 1, y decreases by 3. 4 ⟶−3 1 ⟶−3 -2 ⟶−3 -5 Since the rate of change is constant, the function is linear. To verify, let's plot our points in a graph.We can see that, when graphed, the points all lie on one straight line. | |

Exercises 19 To determine if the function is linear, we will create a table of values. Notice that 38=2.x2−xy 12−11 22−20 32−3-1 42−4-2 Analyzing the x- and y-columns, we can see that as x increases by 1, y decreases by 1. 1 ⟶−1 0 ⟶−1 -1 ⟶−1 -2 Since the rate of change is constant, the function is linear. To verify, let's plot our points in a graph.We can see that, when graphed, the points all lie on one straight line. | |

Exercises 20 Let's start by rewriting the function. y=4x(8−x)Distribute 4xy=4x⋅8−4x⋅xMultiplyy=32x−4x2 A linear function can't have any variables that are raised to any exponent other than 1. From the rewritten version, we see that x is raised to the power of 2. Therefore, we can conclude that the function is not linear. | |

Exercises 21 Let's start by rewriting the function. 2+61y=3x+4LHS−2=RHS−261y=3x+2LHS⋅6=RHS⋅6y=18x+12 A linear function can't have any variables that are raised to an exponent other than 1. From the rewritten version, we see that the function fits this description. Therefore, we can conclude that the function is linear. | |

Exercises 22 To determine if the function is linear, we will transform it into slope-intercept form and create a table of values. y−x=2x−32yLHS+32y=RHS+32y35y−x=2xLHS+x=RHS+x35y=3xLHS⋅53=RHS⋅53y=59xCalculate quotienty=1.8x Let's substitute some arbitrary values into the function.x1.8xy 11.8(1)1.8 21.8(2)3.6 31.8(3)5.4 41.8(4)7.2 Analyzing the x- and y-columns, we can see that as x increases by 1, y increases by 1.8. 1.8 ⟶1.8 3.6 ⟶1.8 5.4 ⟶1.8 7.2 Since the rate of change is constant, the function is linear. To verify, let's plot our points in a graph.We can see that, when graphed, the points all lie on one straight line. | |

Exercises 23 To determine if the function is linear, we will transform it into slope-intercept form and create a table of values. 18x−2y=26LHS−18x=RHS−18x-2y=-18x+26LHS/-2=RHS/-2y=9x−13 Let's substitute some arbitrary values into the function.xy=9x−13y 1y=9(1)−13-4 2y=9(2)−135 3y=9(3)−1314 4y=9(4)−1323 Analyzing the x- and y-columns, we can see that as x increases by 1, y increases by 9. -4 ⟶9 5 ⟶9 14 ⟶9 23 Since the rate of change is constant, the function is linear. To verify, let's plot our points in a graph.We can see that, when graphed, the points all lie on one straight line. | |

Exercises 24 2x+3y=9xyLHS−3y=RHS−3y2x=9xy−3yFactor out y2x=y(9x−3)LHS/(9x−3)=RHS/(9x−3)9x−32x=yRearrange equationy=9x−32x Since the function doesn't match y=mx+b form, it's not linear. | |

Exercises 25 A linear equation can always be written in slope-intercept form y=mx+b. The x variable cannot have any type of exponent in a linear function. Let's look at each given equation and rewrite them with isolated y values to see if they can be written in slope-intercept form.OptionGiven EquationIsolated yLinear? A12=2x2+4y2y=3−21x2No By−x+3=xy=2x−3Yes Cx=8No y to isolateNo Dx=9−43yy=-34x+12Yes Ey=115xy=115xYes Fy=x+3y=x21+3No Options A,C, and F cannot be written in slope-intercept form. Options A and F have exponents on the x-variable and option C is a vertical line, which is undefined. These are, therefore, not linear functions. | |

Exercises 26 If the values in the table represent a linear function, there should be a constant rate of change. We can calculate this change using any two points that are on the line. Let's use the two points that are given to us. Slope=x2−x1y2−y1Substitute (25,11) & (5,-1)Slope=25−511−(-1) Simplify RHS a−(-b)=a+bSlope=25−511+1Add and subtract termsSlope=2012ba=b/4a/4 Slope=53 The slope between (5,-1) and (25,11) is 53. This means that for every 5 steps we take to the right, we take 3 steps up. This fits perfectly with the table as the x-values in the table are separated by 5. If we go 5 steps to the right and 3 steps up from (5,-1) we end up on the next point in the table, and so on.Now we can complete the table.x510152025 y-125811 | |

Exercises 27 To find the x-coordinate of a point, we move vertically until we hit the x-axis. Similarly, to find the y-coordinate, we move horizontally until we hit the y-axis.Now let's list the ordered pairs that we found. (2,6),(4,12),(6,18) The domain of a function is found by listing the relation's x-values. Domain: {2,4,6} Because the given function is a set of disconnected points, we can say that the function is discrete. | |

Exercises 28 The x-values corresponding to the given line segment fall into a specific interval. 0≤x≤7 This interval forms the domain of the given graph. Also, the line segment is an uninterrupted continuous line. Therefore, the domain is continuous over the interval. | |

Exercises 29 Before we look at the given table, let's recall the definitions of discrete and continuous.Discrete Domain: A set of input values that consist of only certain numbers in an interval. Continuous Domain: A set of input values that consist of all possible numbers in an interval.In the given situation, the input is the number of bags. It is only possible to have whole number values for the "number of bags." We can have 1 bag, 2 bags, or 8 bags, but not 3.14159 bags. Therefore, the domain is discrete. | |

Exercises 30 Before we look at the given table, let's recall the definitions of discrete and continuous.Discrete Domain: A set of input values that consist of only certain numbers in an interval. Continuous Domain: A set of input values that consist of all possible numbers in an interval.Here, the input is the number of years that a tree has been growing. This is a measure of time. Even though we are only given the height at specific times, the growth does not only happen at those times. The tree does not magically change height over night when a new year begins, it is gradual change over the interval. Therefore, the domain is continuous. | |

Exercises 31 The input determines the domain. Thus in order to determine if the domain is continuous or discrete, we look at the input. In this case, the input is the number of hours which is a measure of time. Time is continuous, therefore the domain is continuous. | |

Exercises 32 The input determines the domain. Thus in order to determine if the domain is continuous or discrete, we look at the input. In this case, the input is the number of relay teams. Since we are not interested in partial teams, this domain is discrete. | |

Exercises 33 The domain of a function is the complete set of possible values the independent variable x can take. In this particular graph, we have a discrete function that can take on all counting numbers from 1 to 6. Our domain is then: Domain:{1,2,3,4,5,6} As we can see 2.5 is not in the domain. 2.5 is, however, in the range. | |

Exercises 34 A discrete domain is a set of input values that consist of only certain numbers within an interval. On the other hand, a continuous domain describes an interval that consist of all real numbers within that interval. However, the function doesn't have to continue forever! As long as the function is uninterrupted within its interval, it is continuous. | |

Exercises 35 aThe domain of a function is the values the independent variable can take. In our case, it's the number of books we buy. Since we can only buy a whole number of books, the domain is the counting numbers: 0,1,2,3 and so on. This means the domain is discrete. To determine the domain, let's substitute b with counting numbers starting from 1 and stop when m is less than or equal to 0. This is when we have no money left.b55−8.5bm 155−8.5⋅146.5 255−8.5⋅238 355−8.5⋅329.5 455−8.5⋅421 555−8.5⋅512.5 655−8.5⋅64 755−8.5⋅7-4.5 We can only buy 6 books with our money as 7 books will put us in the hole for $4.50. Hence, the domain is: {1,2,3,4,5,6}.bTo graph the function, we will only mark the points from the table in part A: (1,46.5),(2,38),(3,29.5)(4,21),(5,12.5),(6,4). Do not draw a line through these points as that would suggest that we can buy fractions of books. | |

Exercises 36 aYou can't rock climb for a negative number of hours because that just doesn't make sense. However, theoretically, you can do it for as long as you want. Therefore, the duration of your rock climbing from x=0 and onward forms the domain: x≥0 We do not count but rather measure time because it makes sense to talk about fractional time. So our domain is continuous.bLet's graph the function. Remember that we have to limit the domain to non negative numbers. The graph is a straight line emanating from the origin and traveling to infinity and beyond! | |

Exercises 37 aTo check if the function is linear, we can compare the rates of change between each consecutive set of ordered pairs. If they are the same, then the function is linear. t: 21221→2121421→2216121→2218211→22110d: 0.434→0.4340.868→0.4341.302→0.4341.736→0.4342.170 As we can see, for every 2 steps to the right, we take 0.434 steps up. This is a constant rate of change and, therefore, the function is linear. The slope of this linear function is: m=runrise⇒20.434=0.217.bThe distance d that sound has traveled is dependent on the amount of time t that has passed. This means that time is our independent variable and the possible values for t represent our domain. Time cannot be negative so we must have: t≥0. Note, this domain is continuous because time does not happen in perfect, whole number increments. Time always continues to pass.cTo graph this function, we can begin by plotting the given points.Now we can connect those points with a straight line. But, remember that our domain only allows values of t greater than or equal to 0.The function of this line can be written as: d=0.217t. | |

Exercises 38 aWhen a function is linear, it can be written in the slope-intercept form, y=mx+b. We are given that the function representing the cost of grooming services is: y=30+5x. This is already in slope-intercept form and, therefore, the function is linear.bGrooming services can be purchased selectively, customers pay $30 for the basic grooming only and can select from any number of the five extra services for an additional $5 per service. The variable x represents the number of extra services selected. The domain is then the set: D:{0,1,2,3,4,5}, where the values show whether 0, 1, 2, 3, 4, or 5 extra services were chosen for the customer's dog. This domain is discrete because the customers cannot choose to have only a fraction of a paw treatment or only a few of the dog's teeth cleaned, they either purchase the whole treatment or they don't.cTo graph this function, we can begin by finding the (x,y)-coordinates for each value in the domain.x30+5xy 030+5⋅030 130+5⋅135 230+5⋅240 330+5⋅345 430+5⋅450 530+5⋅555 Now we can plot the points on a coordinate plane to have a graph of the function. | |

Exercises 39 Imagine you are at the supermarket buying lemons to make lemonade. Each pack of 4 lemons costs $1. This means that if you want to buy two packs of 4 lemons, or 8 lemons, you would spend $2. This situation could be represented by the given graph. The plotted points, (4,1) and (8,2), would show how much money you would spend depending on the number of lemons you chose to purchase. This domain is discrete because you can only purchase lemons in packs of 4, you cannot buy a fraction of a lemon. | |

Exercises 40 Imagine there is a big snow storm coming and the weather report expected drops in temperature overnight. You are interested in tracking the temperature and how it is affected by the snow. When you went to bed, the temperature was 2∘C and when you woke up, 6 hours later, the temperature was -1∘C. If time is shown on the x-axis and the temperature is shown the y-axis, the average change overnight can be shown by the given graph. It is continuous because time is always continually continuing. | |

Exercises 41 Imagine you want to make some extra spending money for the holidays by selling homemade Puppy Chow (the dessert, not dog food). You spend $300 to buy all the necessary ingredients, packaging materials, and a kitchen scale. The scale was purchased because you want customers to fill their baggies with however much they want and pay you based on the weight. Your base charge being $10 per kilogram of Puppy Chow. This situation can be shown on the drawn graph. You are starting your business $300 in debt and will reach the break-even point, start making profits, after you have sold 30 kg of Puppy Chow. This function is continuous because customers have no whole number limits on the amount of Puppy Chow they can buy. | |

Exercises 42 Imagine that the local gym charges $10 for every 2 hours of classes, but you have a new member's coupon that allows you to join your first 2 hours of classes for free! This situation can shown with the given graph if the x-axis is the number of hours you spent in classes and the y-axis shows the amount of money you spent on classes. This function is discrete because you can only join classes in 2 hour increments. | |

Exercises 43 aSince we want the table to represent a linear function, we need to find the rate of change between the values that we already have and then use that to solve for the missing y-value. We can use the points (4,40.80) and (6,61.20) to solve for the slope: m=x2−x1y2−y1⇒6−461.20−40.80=10.20 Now that we have our slope, let's solve for y using the point (4,40.80) and the fact that the x=5 for the missing y. m=x2−x1y2−y1Substitute values10.20=5−4y2−40.80 Solve for y2 Subtract term10.20=1y2−40.801a=a10.20=y2−40.80LHS+40.80=RHS+40.8051.00=y2Rearrange equation y2=51.00 The missing y-value from the table is 51.00.bBecause the table shows time in hours and earnings in dollars, the rate of change is going to be your hourly wage. In this case, your hourly pay rate is $10.20. | |

Exercises 44 aTo decide if the table shows a linear function, we need to look at the rate of change between each consecutive ordered pair. t: d: 111→2113 1→211560 →120 180 →130 310 We can see that the rate of change between hours 1 and 3 is 120÷2=60 while the rate of change between hours 3 and 5 is 130÷2=65. Because this rate of change is not constant, the function is not linear.bAccording to the function given, the rate for Car A is 50 miles per hour. We found in the previous exercise that the rate for Car B was 60 miles per hour at first and then 65 miles per hour. This means that your friend is correct, Car B is moving faster. | |

Exercises 45 The formula for volume of a rectangular prism is: V=l⋅w⋅h, where l is the length, w is the width, and h is the height. In this exercise, we are given that both the length and width are s and the height is 9 m. If we plug these values into the formula and simplify, we can determine whether our equation is linear or not. V=l⋅w⋅hSubstitute valuesV=s⋅s⋅9MultiplyV=9s2 In this circumstance, with volume as a function of s, the function is not linear. A linear function cannot have an exponent. | |

Exercises 46 The formula for volume of a triangular prism is: V=21⋅b⋅h⋅d, where b is the base, h is the height, and d is the depth. In this exercise, we are given that the base is b, the height is 3 in, and the depth is 4 in. If we plug these values into the formula and simplify, we can determine whether our equation is linear or not. V=21⋅b⋅h⋅dSubstitute valuesV=21⋅b⋅3⋅4MultiplyV=6b In this circumstance, where volume is now a function of the base b, our function is linear. A linear function can be written in slope-intercept form, y=mx+b. This function can, it has a slope of 6 and a y-intercept of 0. | |

Exercises 47 The formula for volume of a cylinder is: V=π⋅r2⋅h, where r is the radius and h is the height. In this exercise, we are given that the radius is 2 cm and the height is h. If we plug these values into the formula and simplify, we can determine whether our equation is linear or not. Remember, π is a constant. V=π⋅r2⋅hSubstitute valuesV=π⋅22⋅hCalculate powerV=π⋅4⋅hMultiplyV=4πh In this circumstance, where volume is now a function of the height h, our function is linear. A linear function can be written in slope-intercept form, y=mx+b. This function can, it has a slope of 4π and a y-intercept of 0. | |

Exercises 48 The formula for volume of a circular cone is: V=31⋅π⋅r2⋅h, where r is the radius and h is the height. In this exercise, we are given that the radius is r and the height is 15 ft. If we plug these values into the formula and simplify, we can determine whether our equation is linear or not. Remember, π is a constant. V=31⋅π⋅r2⋅hSubstitute valuesV=31⋅π⋅r2⋅15MultiplyV=5πr2 In this circumstance, with volume as a function of r, the function is not linear. A linear function cannot have an exponent. | |

Exercises 49 Before we can decipher what each expression tells us, let's identify the key pieces of information given in the problem. x=gallons of water in the first type of jugy=gallons of water in the second type of jugA=number of the first type of jugB=number of the second type of jug Now let's look at each expression.x+y The first expression, x+y, can be expressed in words as "the number of gallons in the first type of jug plus the number of gallons in the second type of jug." Therefore, we can say that it represents the total gallons of water if we have one jug of each type. Because water does not need to be measured in perfect integer amounts, this expression results in a continuous set of values.A+B The second expression, A+B, can be expressed in words as "the number of the first type of jug plus the number of the second type of jug." Therefore, we can say that it represents the total number of jugs. Because we cannot have a fraction of a jug, this expression results in a discrete set of values.Ax The third expression, Ax, can be expressed in words as "the number of the first type of jug multiplied by the number of gallons of water held by each of the first type of jug." Therefore, we can say that it represents the total gallons of water in all of the jugs of the first type. Because water does not need to be measured in perfect integer amounts, this expression results in a continuous set of values.Ax+By The third expression, Ax+By, can be expressed in words as "the number of the first type of jug multiplied by the number of gallons of water held by each of the first type of jug PLUS the number of the second type of jug multiplied by the number of gallons of water held by each of the second type of jug." Therefore, we can say that it represents the total gallons of water in all of the jugs of both types. Because water does not need to be measured in perfect integer amounts, this expression results in a continuous set of values. | |

Exercises 50 Imagine we are at a farmer's market and want to buy some tomatoes. What does it cost to buy 1 tomato? 2 tomatoes? What does it cost to buy a pound of tomatoes? In the exercise, we are not told if the tomatoes are sold individually or by weight. These two graphs would look very different from one another. Let's compare!Tomatoes sold individually If tomatoes are sold individually, then we would have one set price per tomato. The domain of this function would be discrete because you could not purchase fractions of tomatoes. The graph of the cost would then look like this.Notice that we have left the values of the y-axis blank as we don't know the cost of a single tomato from this market. However, the cost will increase at a steady rate because the cost of two tomatoes will be twice as much as the cost of one tomato, the cost of three will be three times as much as one, and so on.Tomatoes sold by weight If the tomatoes are priced and sold based on weight, the function will be continuous. You could buy one small tomato and it would cost less than one very large tomato. You could buy any combination of sizes and numbers of tomatoes. The function would then still follow the same pattern as the discrete function, but it would be a continuous line to represent all possible weights.Notice that the domain is restricted to x≥0 because we cannot buy a negative amount of tomatoes. Unless we went to the market and then tried to sell the tomato guy our own, home-grown tomatoes. But, that'd be weird. center | |

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##### Other subchapters in Graphing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Functions
- Function Notation
- Quiz
- Graphing Linear Equations in Standard Form
- Graphing Linear Equations in Slope-Intercept Form
- Transformations of Graphs of Linear Functions
- Graphing Absolute Value Functions
- Chapter Review
- Chapter Test
- Cumulative Assessment