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Exercises 1 There are two common ways to write functions:standard notation function notation. Standard notation has x-values as the inputs and y-values as the outputs because those correspond to (x,y)-coordinates on a coordinate plane. Function notation also has x-values as the inputs, but it has f(x) to represent the outputs because it is read as: f(x)= "the value of the function at x". Therefore, f(x)=2x+10 is written in function notation. | |

Exercises 2 When x is the input and a function is written in function notation, f(x) represents the output of the function at x. Here we have a function h that represents your height. This means that h(x) represents your height at age x, where age is the input and height is the output. Now, h(14) represents your height at the age of 14 because 14 is the input, age, and h(14) represents your height dependent on age. | |

Exercises 3 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. f(x)=x+6x=-2f(-2)=-2+6Add termsf(-2)=4 When x=-2, the function's value is 4. We will evaluate this function for x=0 and x=5 in the same way using the table below.xx+6f(x) -2-2+64 00+66 55+611 | |

Exercises 4 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. g(x)=3xx=-2g(-2)=3(-2)a(-b)=-a⋅bg(-2)=-6 When x=-2, the function's value is -6. We will evaluate this function for x=0 and x=5 in the same way using the table below.x3xg(x) -23(-2)-6 03(0)0 53(5)15 | |

Exercises 5 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. h(x)=-2x+9x=-2h(-2)=-2(-2)+9-a(-b)=a⋅bh(-2)=4+9Add termsh(-2)=13 When x=-2, the function's value is 13. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-2x+9h(x) -2-2(-2)+913 0-2(0)+99 5-2(5)+9-1 | |

Exercises 6 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. r(x)=-x−7x=-2r(-2)=-(-2)−7-(-a)=ar(-2)=2−7Subtract termr(-2)=-5 When x=-2, the function's value is -5. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-x−7r(x) -2-(-2)−7-5 0-(0)−7-7 5-(5)−7-12 | |

Exercises 7 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. p(x)=-3+4xx=-2p(-2)=-3+4(-2)a(-b)=-a⋅bp(-2)=-3−8Subtract termp(-2)=-11 When x=-2, the function's value is -11. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-3+4xp(x) -2-3+4(-2)-11 0-3+4(0)-3 5-3+4(5)17 | |

Exercises 8 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. b(x)=18−0.5xx=-2b(-2)=18−0.5(-2)-a(-b)=a⋅bb(-2)=18+1Add termsb(-2)=19 When x=-2, the function's value is 19. We will evaluate this function for x=0 and x=5 in the same way using the table below.x18−0.5xb(x) -218−0.5(-2)19 018−0.5(0)18 518−0.5(5)15.5 | |

Exercises 9 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. v(x)=12−2x−5x=-2v(-2)=12−2(-2)−5-a(-b)=a⋅bv(-2)=12+4−5Add and subtract termsv(-2)=11 When x=-2, the function's value is 11. We will evaluate this function for x=0 and x=5 in the same way using the table below.x12−2x−5v(x) -212−2(-2)−511 012−2(0)−57 512−2(5)−5-3 | |

Exercises 10 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. n(x)=-1−x+4x=-2n(-2)=-1−(-2)+4-(-a)=an(-2)=-1+2+4Add termsn(-2)=5 When x=-2, the function's value is 5. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-1−x+4n(x) -2-1−(-2)+45 0-1−0+43 5-1−5+4-2 | |

Exercises 11 | |

Exercises 12 | |

Exercises 13 To find the value of x that will make the function equal to 63, we will substitute 63 for h(x) in the given function rule. Then we can solve for x. h(x)=-7xh(x)=6363=-7xLHS/-7=RHS/-7-9=xRearrange equationx=-9 When x=-9, the function equals 63. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-9 into the function and evaluating. h(x)=-7xx=-9h(-9)=-7(-9)-a(-b)=a⋅bh(-9)=63 When x=-9, h(x)=63, so our solution is correct. | |

Exercises 14 To find the value of x that will make the function equal to 24, we will substitute 24 for t(x) in the given function rule. Then we can solve for x. t(x)=3xt(x)=2424=3xLHS/3=RHS/38=xRearrange equationx=8 When x=8, the function equals 24. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=8 into the function and evaluating. t(x)=3xx=8t(8)=3(8)Multiplyt(8)=24 When x=8, t(x)=24, so our solution is correct. | |

Exercises 15 To find the value of x that will make the function equal to 7, we will substitute 7 for m(x) in the given function rule. Then we can solve for x. m(x)=4x+15m(x)=77=4x+15LHS−15=RHS−15-8=4xLHS/4=RHS/4-2=xRearrange equationx=-2 When x=-2, the function equals 7. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-2 into the function and evaluating. m(x)=4x+15x=-2m(-2)=4(-2)+15a(-b)=-a⋅bm(-2)=-8+15Add termsm(-2)=7 When x=-2, m(x)=7, so our solution is correct. | |

Exercises 16 To find the value of x that will make the function equal to 18, we will substitute 18 for k(x) in the given function rule. Then we can solve for x. k(x)=6x−12k(x)=1818=6x−12LHS+12=RHS+1230=6xLHS/6=RHS/65=xRearrange equationx=5 When x=5, the function equals 18. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=5 into the function and evaluating. k(x)=6x−12x=5k(5)=6(5)−12Multiplyk(5)=30−12Subtract termk(5)=18 When x=5, k(x)=18, so our solution is correct. | |

Exercises 17 To find the value of x that will make the function equal to -4, we will substitute -4 for q(x) in the given function rule. Then we can solve for x. q(x)=21x−3q(x)=-4-4=21x−3LHS+3=RHS+3-1=21xLHS⋅2=RHS⋅2-2=xRearrange equationx=-2 When x=-2, the function equals -4. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-2 into the function and evaluating. q(x)=21x−3x=-2q(-2)=21(-2)−3a(-b)=-a⋅bq(-2)=-1−3Subtract termq(-2)=-4 When x=-2, q(x)=-4, so our solution is correct. | |

Exercises 18 To find the value of x that will make the function equal to -5, we will substitute -5 for j(x) in the given function rule. Then we can solve for x. j(x)=-54x+7j(x)=-5-5=-54x+7LHS−7=RHS−7-12=-54xLHS⋅5=RHS⋅5-60=-4xLHS/-4=RHS/-415=xRearrange equationx=15 When x=15, the function equals -5. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=15 into the equation and evaluating. j(x)=-54x+7x=15j(15)=-54(15)+7(-a)b=-abj(15)=-12+7Add termsj(15)=-5 When x=15, j(x)=-5, so our solution is correct. | |

Exercises 19 Usually when we think about points on coordinate planes, we think of them as x-coordinates and y-coordinates. However, when we use function notation, we use f(x) instead of y. (x,y)⇔(x,f(x)) We want to know the value of x when f(x)=7. To do this, we need to find the corresponding x-value of the function when then y-value is 7.We can see that the function crosses through the point (5,7). Therefore, when f(x)=7, x=5. | |

Exercises 20 Usually when we think about points on coordinate planes, we think of them as x-coordinates and y-coordinates. However, when we use function notation, we use f(x) instead of y. (x,y)⇔(x,f(x)) We want to know the value of x when f(x)=7. To do this, we need to find the corresponding x-value of the function when the y-value is 7.We can see that the function crosses through the point (-3,7). Therefore, when f(x)=7, x=-3. | |

Exercises 21 | |

Exercises 22 | |

Exercises 23 We can graph the linear function using a table. First, let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for p(x) will give us the corresponding values from the range.xp(x)=4xSimplify 0p(0)=4(0)p(0)=0 1p(1)=4(1)p(1)=4 2p(2)=4(2)p(2)=8 Using these x and p(x) values as (x,p(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 24 The first thing to notice about this equation is that it is missing a variable term. Let's try to rewrite it in standard form, Ax+By=C, using 0 as the coefficient for our missing variable." h(x)=-5⇔0x+1h(x)=-5 We can graph the linear equation using a table. First let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for h(x) will give us the corresponding values from the range.x0x+h(x)=-5Simplify 00(0)+h(0)=-5h(0)=-5 20(2)+h(2)=-5h(2)=-5 40(4)+h(4)=-5h(4)=-5 Using these x and h(x) values as (x,h(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 25 We can graph the linear function using a table. First, let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for d(x) will give us the corresponding values from the range.xd(x)=-21x−3Simplify 0d(0)=-21(0)−3d(0)=-3 2d(2)=-21(2)−3d(2)=-4 4w(4)=-21(4)−3w(4)=-5 Using these x and d(x) values as (x,d(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 26 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for w(x) will give us the corresponding values from the range.xw(x)=53x+2Simplify 0w(0)=53(0)+2w(0)=2 5w(5)=53(5)+2w(5)=5 10w(10)=53(10)+2w(10)=8 Using these x and w(x) values as (x,w(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 27 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for g(x) will give us the corresponding values from the range.xg(x)=-4+7xSimplify 0g(0)=-4+7(0)g(0)=-4 1g(0)=-4+7(1)g(1)=3 2g(2)=-4+7(2)g(2)=10 Using these x and g(x) values as (x,g(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 28 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for f(x) will give us the corresponding values from the range.xf(x)=3−6xSimplify 0f(0)=3−6(0)f(0)=3 1f(1)=3−6(1)f(1)=-3 2f(2)=3−6(2)f(2)=-9 Using these x and f(x) values as (x,f(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation. | |

Exercises 29 To determine which battery lasts longer, we can compare the number of hours it takes for each battery to have a 0% charge. We will begin by finding when the laptop's battery dies from the given graph.Notice on the graph that the laptop's battery charge reaches p=0 at t=5. This means that the laptop's battery lasts for 5 hours.The tablet Next we will determine how long the tablet's battery lasts. It is given that the tablet's initial charge is 75% and that the battery loses 12.5% each hour. We can write each of these percents in decimal form as 0.75 and 0.125, respectively. Now we can model the remaining charge of the tablet at any hour with the following function: p=0.75−0.125t As was the case with the graph, the tablet's battery will die when p=0. So we can substitute this into the equation above and solve for t. p=0.75−0.125tp=00=0.75−0.125tLHS+0.125t=RHS+0.125t0.125t=0.75LHS/0.125=RHS/0.125t=6 The solution t=6 means that the tablet's batter will last for 6 hours. Since the laptop's battery only lasted 5 hours, we can conclude that the tablet's battery lasts longer. | |

Exercises 30 This exercise asks us to compare the total cost of labor for 8 hours of work from two different companies. In this situation, our variables are x, the number of hours worked, and C(x), the cost. Because we want to know the cost for 8 hours of labor, we will focus on x=8. Let's start by determining the cost for Certified Remodeling.Certified Remodeling We will use the given equation for Certified Remodeling, C(x)=25x+50, to determine the cost they will charge for 8 hours of labor. To do this, we will substitute x=8 into the equation and solve. C(x)=25x+50x=8C(8)=25⋅8+50MultiplyC(8)=200+50Add termsC(8)=250 Certified Remodeling will charge $200 for 8 hours of labor.Master Remodeling We are given a table showing the rates offered by the main competitor, Master Remodeling.HoursCosts 2$130 4$160 6$190 Notice that in the table, the hours increase by 2 while the cost increases by $30. We can add another row onto the bottom of the table by following this rate of change. Increasing 6 hours by 2 gives 8 hours. Increasing $190 by 30 gives $220.HoursCosts 2$130 4$160 6$190 8$220 The main competitor, Master Remodeling, will charge $220 for 8 hours of labor.Conclusion In the work above, we found that Certified Remodeling will charge $250 while the main competitor, Master Remodeling, will charge $220. Thus, we should choose the main competitor. | |

Exercises 31 In function notation, x represents the input while f(x) represents the output. Then, P(x+1) represents the output of the function when the input is x+1, and P(x) represents the output when the input is x. There are three cases for how P(x+1) and P(x) relate.P(x+1)>P(x). It is possible (likely even) that the number of people with cell phones in year x+1 is greater than the number of people with cell phones in year x. Hence, we would see an increase. P(x+1)=P(x). It is also possible that the number of people with cell phones in year x+1 is equal to the number of people with cell phones in year x. Hence, we would not see a change. P(x+1)<P(x). But it is also possible that the number of people with cell phones in year x+1 is less than the number of people with cell phones in year x. Hence, we would see a decrease. In other words, just because x increases, it doesn't mean the functions value also does. Thus, we cannot conclude with certainty that P(x+1)>P(x). | |

Exercises 32 This exercise asks us to compare the output values of a function for different input values. Here, t represents the number of days and B(t) represents the balance in a bank account on day t. It is given that B(0)<B(4)<B(2). Notice that the expression is given in function notation, f(x)=y. In function notation, x represents the input while f(x) represents the output. Then,B(0) represents the balance in the account on Day 0, B(4) represents the balance in the account on Day 4, B(2) represents the balance in the account on Day 2, All together, the statement is saying that the balance on Day 0 is less than the balance on Day 4 and on Day 4 it's less than on Day 2. From this, we can gather than the balance on Day 2 is the greatest and the balance on Day 0 is the least.Example situation The following is one of infinitely many possible situations that can represent the given statement. On Wednesday, the balance in your bank account is 200 dollars. On Friday, you deposit your paycheck of 500 dollars into the account. On Sunday, you withdraw 300 to pay bills and buy groceries. The table below gives the account balance for each day.DaytB(t)Balance B Wednesday0B(0)200 Friday2B(2)700 Sunday4B(4)400 From the table, we can see that B(0)<B(4)<B(2). | |

Exercises 33 | |

Exercises 34 | |

Exercises 35 | |

Exercises 36 Let's test the statement f(a+b)=f(a)+f(b) with the linear function f(x)=3x+3, and two arbitrary values, a=1 and b=2. f(a+b)f(a)+f(b)⇒f(1+2)⇒f(1)+f(2) Let's start by calculating f(1+2). To do so, we will substitute (1+2) for x in f(x)=3x+3. f(x)=3x+3x=1+2f(1+2)=3(1+2)+3 Simplify Add termsf(3)=3(3)+3Multiplyf(3)=9+3Add terms f(3)=12 Let's now calculate the value of f(1)+f(2).f(x)=3x+3 x=1x=2 f(1)=3(1)+3f(2)=3(2)+3 f(1)=3+3f(2)=6+3 f(1)=6f(2)=9 Therefore, f(1)+f(2) equals 6+9=15. f(1+2)=12f(1)+f(2)=15 Since 12 does not equal 15, we have that f(1+2)=f(1)+f(2). Therefore, we know that it is not always true that f(a+b)=f(a)+f(b). The statement is false. | |

Exercises 37 First, let's split the compound inequality into separate inequalities. Compound Inequality:-2≤xFirst Inequality:-2≤xSecond Inequality: x−11≤6−11−11≤6 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." -2≤x−11andx−11≤6 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. -2≤x−11LHS+11≤RHS+119≤x This inequality tells us that 9 is less than or equal to all values that satisfy the inequality.Note that the point on 9 is closed because it is included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. x−11≤6LHS+11≤RHS+11x≤17 This inequality tells us that all values less than or equal to 17 will satisfy the inequality.Note that the point on 17 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:9≤xSecond solution set:xIntersecting solution set: 9≤x≤17≤17 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 38 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 5a<-35LHS/5<RHS/5a<-7 This inequality tells us that all values less than -7 will satisfy the inequality.Note that the point on -7 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. a−14>1LHS+14>RHS+14a>15 This inequality tells us that all values greater than 15 will satisfy the inequality.Note that the point on 15 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:a<-7a<-7 or a>15a<-7 or a>15 Finally, we'll graph the solution set to the compound inequality. | |

Exercises 39 First, let's split the compound inequality into separate inequalities. Compound Inequality: -16<6kFirst Inequality: -16<6kSecond Inequality: 6k+2<0+2+2<0 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." -16<6k+2and6k+2<0 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. -16<6k+2LHS−2<RHS−2-18<6kLHS/6<RHS/6-3<k This inequality tells us that all values greater than -3 will satisfy the inequality.Note that the point on -3 is open because it is not included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. 6k+2<0LHS−2<RHS−26k<-2LHS/6<RHS/6k<-62ba=b/2a/2k<-31 This inequality tells us that all values less than -31 will satisfy the inequality.Note that the point on -31 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:-3<kSecond solution set:kIntersecting solution set: -3<k<-31<-31 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 40 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2d+7<-9LHS−7<RHS−72d<-16LHS/2<RHS/2d<-8 This inequality tells us that all values less than -8 will satisfy the inequality.Note that the point on -8 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. 4d−1>-3LHS+1>RHS+14d>-2LHS/4>RHS/4d>-42ba=b/2a/2d>-21 This inequality tells us that all values greater than -21 will satisfy the inequality.Note that the point on -21 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:d<-8d<-8 or d>-21d<-8 or d>-21 Finally, we'll graph the solution set to the compound inequality. | |

Exercises 41 First, let's split the compound inequality into separate inequalities. Compound Inequality:5≤3yFirst Inequality:5≤3ySecond Inequality: 3y+8<17+8+8<17 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." 5≤3y+8and3y+8<17 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. 5≤3y+8LHS−8≤RHS−8-3≤3yLHS/3≤RHS/3-1≤y This inequality tells us that -1 is less than or equal to all values that satisfy the inequality.Note that the point on -1 is closed because it is included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. 3y+8<17LHS−8<RHS−83y<9LHS/3<RHS/3y<3 This inequality tells us that all values less than 3 will satisfy the inequality.Note that the point on 3 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:-1≤ySecond solution set:yIntersecting solution set: -1≤y<3<3 Finally, we'll graph the solution set to the compound inequality on a number line. | |

Exercises 42 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 4v+9≤5LHS−9≤RHS−94v≤-4LHS/4≤RHS/4v≤-1 This inequality tells us that all values less than or equal to -1 will satisfy the inequality.Note that the point on -1 is closed because it is included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. -3v≥-6Divide by -3 and flip inequality signv≤2 This inequality tells us that all values less than or equal to 2 will satisfy the inequality.Note that the point on 2 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:v≤-1v≤2v≤2 Finally, we'll graph the solution set to the compound inequality. The union of these solution sets is v≤2. |

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##### Other subchapters in Graphing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Functions
- Linear Functions
- 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