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Exercises 1 Let's pose two examples and then explain themFirst example Let's say your friend asks you if you want to go hiking on Sunday afternoon. You only want to go on that hike if it's going to be a sunny day, hiking in the rain would be miserable. So you tell your friend "it depends on the weather."Second example Let's say you ask your parent if you can go to the movies on Wednesday night. They say "it depends on whether or not you finished your homework." Variables work precisely in this manner. There are several possible outcomes but they depend on what happens. The independent variable is the thing that happens, and the dependent variable is the related outcome. Something happensIndependent variable ⇒ We have an outcome ⇒ Dependent variableExplaining the first example In the first example above, the weather is the independent variable and the hike is the dependent variable. Deciding to go on the hike won't change the weather, but the weather will change whether or not you go on the hike.Explaining the second example In the second example, finishing your homework is the independent variable and going to the movies is the dependent variable. Going to the movies won't mean that *poof* your homework is finished! But finishing your homework will mean that you can go to the movies. | |

Exercises 2 We are given four sentences and asked to find which is different. The easiest way to do this is by writing down the numeric values for each sentence. Remember that the domain is the set of all possible inputs and range is the set of all possible outputs.Range of the function represented by the table:7,5,-1Inputs of the function represented by the table:-1,0,1x-values of the function represented by (-1,7), (0,5), and, (1,-1):-1,0,1Domain of the function represented by (-1,7), (0,5), and, (1,-1): -1,0,1 The statement that is different is the one asking for the range of the function. Range is all possible outcomes or y-values while domain, x-values, and inputs are equivalent. | |

Exercises 3 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. (1,-2),(2,1),(3,6),(4,13),(5,22) Examining the ordered pairs, we can see that there is no x-value with multiple y-values. Therefore, the relation is a function. | |

Exercises 4 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. (7,4),(5,-1),(3,-8),(1,-5),(3,6) Examining the ordered pairs, we can see that there is at least one x-value with multiple y-values. Therefore, the relation is not a function. | |

Exercises 5 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a mapping, the x-values in the domain can only point to one y-value in the range.In the given mapping diagram, we can see that at least one of the x-values corresponds to multiple y-values. Since not all of the values in the domain only point to one value in the range, the relation is not a function. | |

Exercises 6 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a mapping, the x-values in the domain can only point to one y-value in the range.In the given mapping, we can see that none of the x-values correspond to multiple y-values. Since all of the values in the domain only point to one value in the range, the relation is a function. | |

Exercises 7 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values.Input, x1610116 Output, y-2-1012 If we examine the table, we can see that there is at least one x-value with multiple y-values. Therefore, the relation can't be a function. | |

Exercises 8 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values.Input, x-30369 Output, y115-1-7-13 If we examine the table, we can see that every x-value has exactly one y-value assigned to it. Therefore, the relation is a function. | |

Exercises 9 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a graph, we can check if each x-value has only one y-value by subjecting it to the vertical line test.On the given graph, there is no visible part where two points have the same y-value. Each vertical line passes through the graph at only one point, so the graph is a function. | |

Exercises 10 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a graph, we can check if each x-value has only one y-value by subjecting it to the vertical line test.Observing the graph, we can see that there is at least one x-value with more than one y-value. The graph fails the vertical line test, so the graph is not a function. | |

Exercises 11 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a graph, we can check if each x-value has only one y-value by subjecting it to the vertical line test.Observing the graph, we can see that there is at least one x-value with more than one y-value. The graph fails the vertical line test, so the graph is not a function. | |

Exercises 12 For a relation to be a function, each x-value can only be paired with one y-value, but one y-value can be paired with multiple x-values. In a graph, we can check if each x-value has only one y-value by subjecting it to the vertical line test.On the given graph, there is no visible part where two points have the same y-value. Each vertical line passes through the graph at only one point, so the graph is a function. | |

Exercises 13 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,-2),(-1,0),(0,2),(1,0),(2,-2) The domain of a function is found by listing the relation's x-values. The range is found by listing the relation's y-values. Domain: Range: {-2,-1,0,1,2}{-2,0,2} | |

Exercises 14 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. (-2,3),(0,3),(2,3),(4,3) The domain of a function is found by listing the relation's x-values. The range is found by listing the relation's y-values. Domain: Range: {-2,0,2,4}{3} | |

Exercises 15 To begin, let's take a look at the given graph.The domain of a function is the set of all possible x-values. The range is the set of all possible y-values. Notice that, horizontally, our graph spans from x=-4 to x=2, inclusive. These values are the lower and upper boundaries of our domain. Domain: -4≤x≤2 Meanwhile, vertically, our graph spans from y=2 to y=6, inclusive. These values are the lower and upper boundaries of our range. Range: 2≤y≤6 | |

Exercises 16 To begin, let's take a look at the given graph.The domain of a function is the set of all possible x-values. The range is the set of all possible y-values. Notice that, horizontally, our graph spans from x=2 to x=7, exclusive. These values are the lower and upper boundaries of our domain. Domain: 2<x<7 Meanwhile, vertically, our graph spans from y=1 to y=6, exclusive. These values are the lower and upper boundaries of our range. Range: 1<y<6 | |

Exercises 17 aSince the monthly rent y depends on the number of days x the payment is late, we can say that y is the dependent variable and x is the independent variable.bTo determine the range of the function, we can substitute the given domain values into the equation and simplify.y25x+500x 025(0)+500500 125(1)+500525 225(2)+500550 325(3)+500575 425(4)+500600 525(5)+500625 From the table above, we can identify the y-values in the range that correspond to the x-values in the domain. Range: {500,525,550,575,600,625} | |

Exercises 18 aSince the cost y of a taxi ride depends on the number x of traveled miles, we can identify that y is the dependent variable and x is the independent variable.bThe minimum number of traveled miles is 0. To determine the cost that corresponds with this distance, we can substitute x=0 into the equation and solve for y. y=3.5x+2.8x=0y=3.5(0)+2.8Multiplyy=2.8 The minimum cost of a taxi ride is $2.80. It is given that there is enough money to travel most 20 miles. We can consider this the maximum distance. To determine the cost of such a ride, we will substitute x=20 into the equation and solve for y. y=3.5x+2.8x=20y=3.5(20)+2.8Multiplyy=70+2.8Add termsy=72.8 Since y=72.8 when x=20, the maximum cost of a ride is $72.80. From our work above, we have established that we can travel from 0 to 20 miles and the cost is from $2.80 to $72.80. These values represent the lower and upper boundaries of our domain and range. Domain: Range: 0≤x≤202.8≤y≤72.8 | |

Exercises 19 The statement says that this relation is not a function, because one output is paired with two inputs. However, a function can have one output paired with two inputs, the conclusion was incorrect. A function cannot have one input paired with more than one output. We can check this by analyzing the table.Input, x12345 Output, y67869 Therefore, this relation is a function and the correct statement would be: The relation is a function. Each input is paired with exactly one output. | |

Exercises 20 The statement says that the range is 1,2,3,4 and 5. This is wrong since these values are the input values, x. The range of a function is the set of all possible output values. In this case, the range is: 6,7,8, and 9. | |

Exercises 21 To identify which variable is independent and which is dependent, we need to ask ourselves which variable causes a change in the other variable.Independent variable: Causes change in the dependent variable. Dependent variable: Changes because of the independent variable.The number of quarters you put into a parking meter affects the amount of time you have on the meter. In other words, the time on the meter depends on how much is paid. Therefore, the number of quarters is the independent variable and the time on the parking meter is the dependent variable. | |

Exercises 22 To identify which variable is independent and which is dependent, we need to ask ourselves which variable causes a change in the other variable?Independent variable: Causes change in the dependent variable. Dependent variable: Changes because of the independent variable.The battery power remaining on the MP3 player depends on the length of time listening to it. Therefore, the listening time is the independent variable and the battery power remaining on the MP3 player is the dependent variable. | |

Exercises 23 aFrom the table, we can see that when x=0, the balance is $100. This means that at the beginning, we have $100 in our savings account. Each consecutive month, the balance increases by $25 for the course of 4 months.bIn an ordered pair (a,b), a represents the input value and b the output value. Since x-values are the inputs and y-values are the outputs, we can write five ordered pairs. (0,100),(1,125),(2,150),(3,175),(4,200)cLet's now plot the ordered pairs. Recall that the first coordinate corresponds to the horizontal axis, and the second coordinate to the vertical axis. | |

Exercises 24 aLet's isolate the variable y on the left side. 1.5x+0.5y=12LHS−1.5x=RHS−1.5x0.5y=12−1.5xLHS/0.5=RHS/0.5y=24−3xbLet's create a table and use a few values for the input x. Since x stands for the number of books it must be a whole number. Let's choose the first four positive integers 1,2,3, and 4.x24−3xy 124−3⋅121 224−3⋅218 324−3⋅315 424−3⋅412cLet's plot the ordered pairs we found as points in a coordinate plane. | |

Exercises 25 To find the input value corresponding to an output of 2, we have to determine the point on the function where the value of y is 2. Then we can identify its corresponding x value.The corresponding input value is -2. | |

Exercises 26 First, we want t to be the independent variable. This means that it will represent the input values of the function. For the relation to be a function each input value has to pair with exactly one output value. Let's fill the table with 4 different values of t and 4 corresponding v-values. It doesn't matter what values v takes on.t1234 v5678 To fulfill the second condition, when t is the dependent variable, we want this relation to no longer be a function. This will happen if one input value v pairs with two different output values t. Since all v-values are different in the table above, let's change one of them to match another. For example, we can replace the 6 with a 5.t1234 v5578 Now, the relation above satisfies both conditions. When t is the input, there are no repeats and it is a function. When v is the input, 5 gives two different outputs, and it is not a function. | |

Exercises 27 aAnalyzing the vending machine, we can see that each letter-number combination is paired with exactly one food or drink item. If you use the vending machine and select a letter-number combination, you will get the exact one item that you wanted, and only that one item. This is exactly how a function works!bTo buy an item from the vending machine, we first have to select a letter-number combination. Then we get the corresponding food or drink item. Therefore, it must be that: Dependent variable: Independent variable: Item bought. Combination selected.cThe domain of a function is the set of all possible input values. Domain{A1, A2, A3,B1, B2, B3, B4,C1, C2, C3, C4} The range is the set of all possible output values. In this case, it is the items from the vending machine. Range{popcorn, nuts, pretzels, protein bar,granola bar, cereal, energy bar,orange juice, water, milk} | |

Exercises 28 aWe can see that h is a function of t by performing the vertical line test.The line only crosses the graph once. In fact, no matter where we would draw the vertical line, it will only intersect the graph once.bTo approximate the height of the projectile at the given times, we have to state the h-values when t=0.5 and t=1.25.The corresponding h-values for t=0.5 and t=1.25 are t=20 and t≈28, respectively.cThe domain of this function is all possible t-values. From the graph, we can see that it starts from t=0 and ends somewhere after t=2.5. Since we are asked for an approximate domain, we can approximate this value at t≈2.7. Domain: 0≤x≤2.7dIf t is a function of h, then the horizontal axis shows the input values. In a function, each input value pairs with exactly one output value. This is not the case of the given graph. For example, for h=20 there are 2 corresponding t-values, t=0.5 and t=2.Therefore, t is not a function of h. | |

Exercises 29 A line is a function if for every x-value there is only one y-value. For example, a horizontal line is a function because each x-value corresponds to only one y-value.For a vertical line however there exists an infinite number of y-values for a single x-value. We have an example of this below.This, a vertical line is not a function. Your friend wasn't correct. | |

Exercises 30 Can we make a function out of people? How can we pair each person with exactly one output? Well, something which surely connects everyone is the fact that we were all born from one woman. We can say: Domain: Range: People of the world.Women who gave birth to them. This represents a function because we were all born and it is impossible for two women to have given birth to the same child. Therefore, the independent variable is people of the world, and the dependent variable are women who have given birth. | |

Exercises 31 An item can be sold for any price the store determines is best. Therefore, two items that cost the same to make could be sold for different prices. This means that the selling price of an item is not a function of the cost of making it. | |

Exercises 32 The sales tax in a given state is always the same, so an item of a certain price will always have one corresponding sales tax. Therefore, the sales tax on a purchased item in a given state is a function of the selling price. | |

Exercises 33 In a school, each class has one homeroom teacher. Therefore, each student will be paired with exactly one homeroom teacher. We can thus conclude that a relation which pairs each student with a homeroom teacher is a function. | |

Exercises 34 A proper function pairs each input value with exactly one output value. In this case, the chaperone is the input value and they are paired with 10 students. Hence, we can conclude that it isn't a function. | |

Exercises 35 A relation is any set of points on a coordinate plane, a collection of ordered pairs. A function, on the other hand, is a specific type of relation in which no two different ordered pairs have the same x-value. In other words, a function is relation such that each element x of the first set is related to one and only one element y of the second set. | |

Exercises 36 A relation is any set of ordered pairs. A function, however, is a relation such that each element of the domain of the function is associated with exactly one element of the range of the function. Let's look at an example: (1,1),(1,2),(2,3),(2,4). The list above represents a relation, it is a set of ordered pairs. However, the input x=1 gives two different outputs y=1 and y=2. The input x=2 also gives two different outputs. This relation is not a function. Therefore, a relation is not always a function but a function is always a relation. | |

Exercises 37 A function is a relation in which each element of the domain is paired with exactly one element of the range. It is not possible to obtain two output values from the same input. Consider a function represented by a mapping diagram.Although we see that there are two inputs paired with the same output, every input is paired with exactly one output. Therefore, the above relation is a function. Let's see what happens if we switch the inputs and outputs.We see that there is an input, 4, that is mapped onto two outputs. Therefore, in this case, switching the inputs and outputs does not result in a function. The statement is false. | |

Exercises 38 When the domain of a function has an infinite number of values, the range does not always have an infinite number of values. For example, the function y=3. Let's graph it.Here, the domain is all real numbers which is an infinite number of values. But the range is only one value, 3. Therefore, this statement is false. | |

Exercises 39 aThe perimeter P of a triangle is the sum of its three sides. Therefore, we can write the function as: P=h+13+10⇔P=h+23.bThe perimeter depends on the value of the third side h. Therefore, P is the dependent variable and h is the independent variable.cWe will find the domain first, and then determine the corresponding range.Domain Since the sum of the lengths of any two sides of the triangle is greater than the length of the remaining side, we can write the following inequalities: 13+10h+10h+13>h>13>10. Let's isolate h in all of these inequalities and analyze the results.InequalityIsolate h 13+10>hh<23 h+10>13h>3 h+13>10h>-3 Right off the bat we can ignore the last inequality as it doesn't make sense for a side to be greater than -3. From the first inequality, we determine that h<23 and from the second one we know that h>3. Combining these we can form the following compound inequality: 3<h<23, which is the domain of the function.Range The range is the corresponding perimeter P that the function can take on given the domain. Because our function is a continuous linear function, we can find the lower and upper endpoints of the range by substituting the lower and upper endpoints of the domain into the function. P=h+23h=3P=3+23Add termsP=26 The lower boundary of the range is 26. Let's find the upper boundary. P=h+23h=23P=23+23Add termsP=46 The upper boundary is 46. Combining the boundaries gives us the following range: 26<P<46. | |

Exercises 40 To find the domain and range of the function we can graph it in a coordinate plane. y=∣x∣ Notice that the function is an absolute value function, so the output value will never be negative. Let's calculate the corresponding y-values for a few x-values.x∣x∣y -3∣-3∣3 -2∣-2∣2 -1∣-1∣1 0∣0∣0 1∣1∣1 2∣2∣2 3∣3∣3 We can plot these coordinates and connect them with a line to draw our function.Now we can identify the domain and range. Remember, domain is the set of all possible x-values and the range is the set of all possible y-values. Looking horizontally, we can see that there are no restrictions to the possible x-values. Domain: All real numbers Looking vertically, we can see that the range will never be negative. Range: y≥0 | |

Exercises 41 To find the domain and range of the function we can graph it in a coordinate plane. y=-∣x∣ Notice that the function is a negative absolute value function, so the output value will never be positive. Let's calculate the corresponding y-values for a few x-values.x-∣x∣y -3-∣-3∣-3 -2-∣-2∣-2 -1-∣-1∣-1 0-∣0∣0 1-∣1∣-1 2-∣2∣-2 3-∣3∣-3 We can plot these coordinates and connect them with a line to draw our function.Now we can identify the domain and range. Remember, domain is the set of all possible x-values and the range is the set of all possible y-values. Looking horizontally, we can see that there are no restrictions to the possible x-values. Domain: All real numbers Looking vertically, we can see that the range will never be positive. Range: y≤0 | |

Exercises 42 To find the domain and range of the function: y=∣x∣−6 we can graph it in a coordinate plane. To do this, let's calculate the corresponding y-values for a few x-values.x∣x∣−6y -3∣-3∣−6-3 -2∣-2∣−6-4 -1∣-1∣−6-5 0∣0∣−6-6 1∣1∣−6-5 2∣2∣−6-4 3∣3∣−6-3 We can plot these coordinates and connect them with a line to draw our function.Now we can identify the domain and range. Remember, domain is the set of all possible x-values and the range is the set of all possible y-values. Looking horizontally, we can see that there are no restrictions to the possible x-values. Domain: All real numbers Looking vertically, we can see that the lower bound of the range is -6 and there are no other restrictions. Range: y≥-6 | |

Exercises 43 To find the domain and range of the function: y=4−∣x∣ we can graph it in a coordinate plane. To do this let's calculate the corresponding y-values for a few x-values.x4−∣x∣y -34−∣-3∣1 -24−∣-2∣2 -14−∣-1∣3 04−∣0∣4 14−∣1∣3 24−∣2∣2 34−∣3∣1 We can plot these coordinates and connect them with a line to draw our function.Now we can identify the domain and range. Remember, domain is the set of all possible x-values and the range is the set of all possible y-values. Looking horizontally, we can see that there are no restrictions to the possible x-values. Domain: All real numbers Looking vertically, we can see that the upper bound of the range is 4 and there are no other restrictions. Range: y≤4 | |

Exercises 44 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "less than," which is represented by the symbol <. …<… On the left-hand side of the inequality we have "a number y". y<… On the right-hand side of the inequality, we have "16." We can now form our final answer. a number y is less than 16y<16 | |

Exercises 45 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "no less than," which is represented by the symbol ≥. …≥… On the left-hand side of the inequality we have "three." 3≥… On the right-hand side of the inequality, we have "a number x." We can now form our final answer. Three is no less than a number n3≥x | |

Exercises 46 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "at most," which is represented by the symbol ≤. …≤… On the left-hand side of the inequality we have "seven." 7≤… On the left-hand side of the inequality, we have "the quotient of a number d and -5," which we will express as a fraction. We can now form our final answer. Seven is at most the quotient of a number d and -57≤-5d | |

Exercises 47 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "more than," which is represented by the symbol >. …>… On the left-hand side of the inequality we have "the sum of a number w and 4." w+4>… On the right-hand side of the inequality, we have "-12." We can now form our final answer. The sum of a number w and 4 is more than -12w + 4>-12 | |

Exercises 48 The expression 112, or "11 squared," means that we multiply 11 by itself. 112=11⋅11 To calculate this, we can type either 112 or 11⋅11 into the calculator. Both expressions give us the same answer. | |

Exercises 49 The expression (-3)4, or -3 to the fourth power, can be written as the product of four -3 terms. Note that when multiplying two negative numbers, we obtain a positive number. (-a)(-a)=a⋅a This means that any real negative number raised to an even power will result in a positive result. (-3)4Rewrite (-3)4 as (-3)⋅(-3)⋅(-3)⋅(-3)(-3)⋅(-3)⋅(-3)⋅(-3)(-a)(-b)=a⋅b3⋅3⋅3⋅3Multiply81 Using a calculator, we can either enter (-3)4 or 3⋅3⋅3⋅3. Both expressions give us the same answer. | |

Exercises 50 The power -52 is not the same as (-5)2. Note that in this case, there are no parentheses surrounding the negative. This means that we should evaluate as though parentheses surround the exponent, and then multiply the result by -1. -52⇔-(52) The term 52, or "5 squared," means that we need to multiply 5 by itself. 52=5⋅5 To calculate the final result, we can either enter -52 or -(5⋅5) into a calculator. Both expressions give us the same answer. | |

Exercises 51 |

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##### Other subchapters in Graphing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Linear 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