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###### Exercises

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Exercises 1 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.Is the Relation a Function? Let's look at the given values.DomainRange -16 05 19 214 Since each input x has only one output, the relation in the table is a function.Is the Relation Linear? To determine whether this function is linear or nonlinear, we observe the changes in x and y. If both of these rates of change are constant, we can conclude that the function is linear.Change in xxyChange in y-16+105-1 +119+4 +1214+5 We can see that as x increases by 1, y increases by different amounts. The rate of change is not constant. So, the function is nonlinear. | |

Exercises 2 A given relation is a function if any input x will result in no more than one output y.Is the Relation a Function? To test the given relation, we will substitute some arbitrary values for x and simplify.xy=-2x+3y -4y=-2(-4)+3y=11 -2y=-2(-2)+3y=7 0y=-2(0)+3y=3 2y=-2(2)+3y=-1 4y=-2(4)+3y=-5 We can see that there are no x-values that give more than one y-value. It appears that the given relation is a function. We can confirm this conclusion by using the points from our table to graph the function and perform a vertical line test.The vertical line test confirms for us that the given relation is a function.Is the Relation Linear? To tell if this relation is linear, we will use a table to compare the change in x with the change in y. If both values change at a constant rate, the relation is linear.Change in xxy=-2x+3yChange in y-4y=-2(-4)+3y=11+2-2y=-2(-2)+3y=7-4 +20y=-2(0)+3y=3-4 +22y=-2(2)+3y=-1-4 +24y=-2(4)+3y=-5-4 Every time x increases by 2, y decreases by 4. These changes are consistent, so the relation is a linear function. | |

Exercises 3 A given relation is a function if it pairs each input x with exactly one output y. We know that every function passes a vertical line test. Let's graph our relation x=-2 and perform this test.We can see that the line doesn't pass the vertical line test. Therefore, the relation x=-2 is not a function. | |

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Exercises 7 The arrows on the graph indicate that the function will continue on forever in the downward and upward direction. Therefore, we can receive any real number y as an output. This also tells us that there are no gaps in the function, and that we can input any real number for x.Therefore, we can write our domain and range as an interval that contains all real numbers. Domain: -∞<x<∞Range: -∞<y<∞ Because the given graph is an unbroken line, we can say that the function is continuous. | |

Exercises 8 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. (-2,7),(-1,6),(0,1),(1,-2),(2,-5),(3,-8) 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,-1,0,1,2,3}{7,6,1,-2,-5,-8} Because the given function is a set of disconnected points, we can say that the function is discrete. | |

Exercises 9 Let's begin by drawing the graph of f(x)=x. This line is already written in slope-intercept form, y=mx+b, where 1 is the slope and 0 is the y-intercept.If we look at the difference between f(x) and g(x), we can see that {f(x)=xg(x)=-x+3⇒g(x)=-f(x)+3. This means that we start by applying the negative sign to f(x) and then we add 3. Let's draw the graph of y=-x. This transformation changes the sign of each x-value of all the points lying on the line. In other words, it is a reflection of the line in the x-axis.Finally, we will translate the graph of our previous line, the green one, vertically 3 units. This will give us the graph of g(x)=-x+3.In conclusion, the transformation from the graph of f(x) to the graph of g(x) is a reflection in the x-axis and a vertical translation 3 units up. | |

Exercises 10 First, let's look at the graph of the parent function, f(x)=∣x∣.When we are graphing the transformation of an absolute value function, we should work our way from the "inside" to the "outside." This means that we should start with the transformations closest to x and work our way to the transformations farthest away. In this exercise, the given function is: g(x)=∣2x+4∣. Before we start our transformation, let's rewrite our function in vertex form, h(x)=a⋅f(bx−c)+d, where a determines vertical stretching, shrinking, and reflection, b determines horizontal stretching, shrinking, and reflection, c determines horizontal translation, and d determines vertical translation. g(x)=∣2x+4∣Factor out 2g(x)=∣2(x+2)∣ Since we have our function in vertex form, g(x)=∣2(x+2)∣, where 2 determines a horizontal shrink and +2 determines a horizontal translation, we can begin doing the transformations. The "inside" transformation is +2, a horizontal translation 2 units to the left.The "outside" transformation is a horizontal shrink by a factor of 21. | |

Exercises 11 Let's look at each of these functions individually.Function A We have a jar full of quarters and want to know how much money we have based on the number of quarters we have. Each and every quarter will always be worth 25 cents or $0.25. If we let the number of quarters be x, then the total amount of money, in dollars, can be expressed as: 0.25x. We can only have whole quarters, there will never be a fraction of a quarter. Therefore, the input values for x will only be whole numbers. The domain will then be all positive integers.Function B When thinking about your distance from home over time, you need to consider that the distance changes based on how much time has passed. You can measure time as precisely as you want because time is continually continuing. But, it can never run backwards. Therefore, the function's domain is continuous and can be any value greater than or equal to zero, t≥0, zero being whenever you started keeping track of the amount of time that has passed. | |

Exercises 12 aWe are told that the equation of the function shown on the graph is: f(x)=125x+50. We can compare this to the slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept by first remembering that f(x) is just the function notation version of y. We can rewrite the equation to be: y=125x+50. Now it is easy to see that m=125 is the slope and b=50 is the y-intercept. We can understand what this slope tells us by looking at the units given on the graph. We know that slope is: change in xchange in y=1 hour125 ft. Therefore, the climber is going up 125 feet in elevation each hour. Similarly, the y-intercept being 50 tells us that when 0 hours have passed, the climber is at an elevation of 50. So he began the climb at 50 feet above sea level.bWe can find f(3) by looking at the graph, finding the y value that corresponds to x=3, or algebraically, by substituting x=3 into the equation. Since the graph goes up by 0.4 on the x-axis, it is not easy to find the correct x value let alone the exact corresponding y value. Solving algebraically is going to be easier and more accurate. f(x)=125x+50x=3f(3)=125⋅3+50Multiplyf(3)=375+50Add termsf(3)=425 f(3)=425 tells us that after 3 hours have passed, the climber will be at an elevation of 425 feet.cOnce again, we could solve this either graphically or algebraically. In this case, however, graphically is going to be easier. Algebraically would require us to substitute f(x)=500 and solve for x: 500=125x+50. When looking at the graph though, y=500 feet lies perfectly on an intersection of graph lines. We can follow the line down to the x-axis from the point marking the end of his climb. This gives us that it took x=3.6 hours for the climber to reach the top. | |

Exercises 13 We are given the function f(x)=x+1. When we transform this function to be g(x)=f(2x), we need to multiply only the values of x, the input, by 2. We then have: g(x)=f(2x)=2x+1. Note, this is a horizontal shrink by a factor of 21. We can easily identify the slopes and y-intercepts by comparing the equations to the slope-intercept form. The x-intercept is not as simple, we need to set each equation equal to 0 and solve for x. 0=x+10=2x+1⇔x=-1⇔x=-21 The key features of these graphs are then:Featuref(x)g(x) Slope12 y-intercept11 x-intercept-1-21 This transformation doubled the slope, halved the x-intercept, and left the y-intercept unchanged. | |

Exercises 14 |

##### Other subchapters in Graphing Linear Functions

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Functions
- 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
- Cumulative Assessment