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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. (0,1),(5,6),(7,9)​ Examining the ordered pairs, we can see that there is no x-value with multiple y-values. Therefore, the relation is a function.
Exercises 2 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 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. 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 4 aSince the amount y of money in your bank account depends on the number x of hours for which you babysit, we can identify that y is the dependent variable and x is the independent variable.bThe minimum number of minutes for which you can work is 0. To determine the amount of money that corresponds with this time, we can substitute x=0 into the equation and solve for y. y=10x+100x=0y=10(0)+100Zero Property of Multiplicationy=100 The minimum amount of money in your bank account is 100 dollars. Now, let's calculate the maximum amount. To do that, we can substitute x=4 (maximum number of hours you can work) into the equation and solve for y. y=10x+10x=4y=10(4)+100Distribute 10y=40+100Add termsy=140 Since y=140 when x=4, the maximum amount of money in your bank account is 140 dollars. From our work above, we have established that we can have from 100 to 140 dollars in our bank account. These values represent the lower and upper boundaries of our domain and range. Domain: Range: ​0≤x≤4100≤y≤140​
Exercises 5 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.Since we can use a straight edge to connect all of the given points, they lie on the same line in the coordinate plane. Therefore, the function is linear.
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 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 7 aThe given function represents the amount y (in dollars) of money we have after buying x tickets. y=60−8x​ Since the minimum number of tickets we can buy is 0, the minimum value of the domain is 0. Let's find the amount of money if we buy 0 tickets. To do that we will substitute 0 for x and solve the function. y=60−8xx=0y=60−8(0)Zero Property of Multiplicationy=60−0Subtract termy=60 As a result, our initial amount is $60. How many tickets can we buy if we spend $60? Notice that if we spend all of our money, the remaining amount will be 0. Thus, we will substitute 0 for y and solve the equation for x. y=60−8xy=00=60−8xLHS+8x=RHS+8x8x=60LHS/8=RHS/8x=7.5 We cannot buy part of a ticket so x must be whole number. Therefore, the maximum number of tickets that we can buy is 7. Finally, we can determine the domain of the function. Domain: {0, 1, 2, 3, 4, 5, 6, 7}​ We also conclude that the domain is discrete because it consists of only certain numbers.bIn order to graph the function, we will first make an input-output table to find the ordered pairs.Input, xy=60−8xOutput, y(x,y) 0y=60−8(0)60(0,60) 1y=60−8(1)52(1,52) 2y=60−8(2)44(2,44) 3y=60−8(3)36(3,36) 4y=60−8(4)28(4,28) 5y=60−8(5)20(5,20) 6y=60−8(6)12(6,12) 7y=60−8(7)4(7,4) Now that we know the ordered pairs, we will plot them on the coordinate. Because the domain is discrete, the graph will consist of individual points.
Exercises 8 Evaluating a function for a certain value of x means substituting it into the function and simplifying until f(x) equals a number. Let's look at each of these cases one at a time.Evaluating f(-3) f(x)=x+8x=-3f(-3)=-3+8Add termsf(-3)=5 When x=-3, the value of the function is 5. Therefore, f(-3)=5.Evaluating f(0) f(x)=x+8x=0f(0)=0+8Identity Property of Additionf(0)=8 The value of the function is 8 when x=0. Therefore, f(0)=8.Evaluating f(5) f(x)=x+8x=5f(5)=5+8Add termsf(5)=13 Finally, if x=5 the value of the function is 13. Therefore, f(5)=13.
Exercises 9 To evaluate a function for a certain value of x means that we substitute the x-value into the function and simplify. Let's look at what happens when x=-3 in the given function. g(x)=4−3xx=-3g(-3)=4−3(-3)(-a)(-b)=a⋅bg(-3)=4+9Add termsg(-3)=13 When x=-3, the function's value is 13. Using the same reasoning, we can evaluate this function for x=0 and x=5. We have shown this in the table below.x4−3xg(x) 04−3⋅04 54−3⋅5-11
Exercises 10 To find the value of x that will make the function equal to 49, we will substitute 49 for k(x) in the given function rule. Then we can solve for x. k(x)=7xk(x)=4949=7xLHS/7=RHS/77=xRearrange equationx=7 When x=7, the function equals 49. ⧼ebox-type-answer-check⧽ Verifying Our Solution We can verify our solution by substituting x=7 into the function and evaluating. k(x)=7xx=7k(x)=7(7)Multiplyk(x)=49 When x=7, k(x)=49, so our solution is correct.
Exercises 11 To find the value of x that will make the function equal to 19, we will substitute 19 for r(x) in the given function rule. Then we can solve for x. r(x)=-5x−1r(x)=1919=-5x−1LHS+1=RHS+120=-5xLHS/-5=RHS/-5-4=xRearrange equationx=-4 When x=-4, the function equals 19. ⧼ebox-type-answer-check⧽ Verifying Our Solution We can verify our solution by substituting x=-4 into the function and evaluating. r(x)=-5x−1x=-4r(-4)=-5(-4)−1Multiplyr(-4)=20−1Add termsr(-4)=19 When x=-4, r(x)=19, so our solution is correct.
Exercises 12 Functions written in slope-intercept form follow a specific format. f(x)=mx+b​ In this form, m represents the slope of the line and b represents the y-intercept. Below we highlight the slope m and y-intercept b. f(x)=-2x − 3⇔f(x)=-2x+(-3)​ The slope is -2. The y-intercept is -3, so the graph crosses the y-axis at the point (0,-3).Graphing the Equation A slope of -2 means that for every 1 unit we move in the positive horizontal direction, we move 3 units in the negative vertical direction. m=1-2​⇔runrise​=1-2​​ To graph the equation, plot the y-intercept and then use the slope to find another point on the line.
Exercises 13 Functions written in slope-intercept form follow a specific format. y=mx+b​ In this form, m represents the slope of the line and b represents the y-intercept.Identifying Slope and Intercept Since the given equation is not in slope-intercept form, let's rewrite it so that it will be easier to identify the slope and y-intercept. y=4−3xRearrange termsy=-3x+4 Below we highlight the slope m and y-intercept b. f(x)=-3x + 4​ The slope is -3. The y-intercept is 4, so the graph crosses the y-axis at the point (0,4).Graphing the Equation A slope of -3 means that for every 1 unit we move in the positive horizontal direction, we move 3 units in the negative vertical direction. m=1-3​⇔runrise​=1-3​​ To graph the equation, plot the y-intercept and then use the slope to find another point on the line.
Exercises 14 In order to graph the given equation, we need to first identify each of the intercepts.x-intercept The x-intercept occurs when the function crosses the x-axis. To find the x-intercept, we substitute y for 0 and solve for x. 8x−4y=16y=08x−4⋅0=16Use the Zero Product Property8x−0=16Subtract terms8x=16LHS/8=RHS/888x​=816​Calculate quotientx=2 The function's x-intercept is at (2,0).y-intercept Similarly, the y-intercept occurs when the function crosses the y-axis. We can find it by substituting x for 0 and solving for y. 8x−4y=16x=08⋅0−4y=16Use the Zero Product Property0−4y=16Subtract terms-4y=16LHS/(-4)=RHS/(-4)-4-4y​=-416​Calculate quotienty=-4 The function's y-intercept is at (0,-4).Graph We can plot the intercepts.Finally, we draw the line passing through the two plotted points.
Exercises 15 In order to graph the given equation, we need to first identify each of the intercepts.x-intercept The x-intercept occurs when the function crosses the x-axis. To find the x-intercept, we substitute 0 for y and solve for x. -12x−3y=36y=0-12x−3⋅0=36Zero Property of Multiplication-12x=36LHS/(-12)=RHS/(-12)x=-3 The function's x-intercept is at (-3,0).y-intercept In the same way, the y-intercept occurs when the function crosses the y-axis. We can find it by substituting 0 for x. -12x−3y=36x=0-12⋅0−3y=36Zero Property of Multiplication-3y=36LHS/(-3)=RHS/(-3)y=-12 The function's y-intercept is at (0,-12).
Exercises 16 In this exercise, we are asked to graph the function y=-5. Notice that this equation does not have an x variable. This means that for every point on the line x can be any value, while y must always equal -5. As a consequence, we will have a horizontal line 5 units below the x-axis.
Exercises 17 In this exercise, we are asked to graph the function x=6. Notice that this equation does not have a second variable y. This means that for every point on the line, y can be any value, while x must always equal 6. When graphed, we will see a vertical line 6 units to the right of the y-axis.
Exercises 18 Looking at the table, we can identify 4 ordered pairs: (6,9), (11,15), (16,21), and (21,27). Although, to calculate the slope of the line, we need only two of these. Let's find it by substituting (6,9) and (11,15) into the slope formula. m=x2​−x1​y2​−y1​​Substitute (6,9) & (11,15)m=11−615−9​Subtract termsm=56​ Therefore, the slope of the line is 56​.
Exercises 19 From the table, we notice that there is no change in x. This means the line is vertical and passes through x=3. Therefore, the slope of this line is undefined. We can verify this by substituting two points from the table into the slope formula. m=x2​−x1​y2​−y1​​ Let's take (3,5) and (3,8). m=x2​−x1​y2​−y1​​Substitute (3,5) & (3,8)m=3−38−5​Subtract termm=03​ Division by 0 is undefined, so the slope of the line is undefined.
Exercises 20 Looking at the table, we can identify 4 ordered pairs. (-4,-1),(-3,-1),(1,-1),(9,-1)​ To calculate the slope of the line, we need only two of these. Let's find the slope by substituting (1,-1) and (9,-1) into the slope formula. m=x2​−x1​y2​−y1​​Substitute (1,-1) & (9,-1)m=9−1-1−(-1)​Subtract termsm=80​Calculate quotientm=0 The slope of this line is 0, or a horizontal line.
Exercises 21 The line is already written in slope-intercept form, y=mx+b, where m is the slope and b the y-intercept. For our line, we can immediately identify that 2 is the slope and 4 is the y-intercept. We will start by plotting the y-intercept.Now we will use the slope to find another point on the line.Finally, we draw the line which passes through our two points.The line crosses the x-axis at (-2,0). Therefore, the x-intercept is -2.
Exercises 22 The x-intercept occurs when the function crosses the x-axis, when y=0. To find the x-intercept, we substitute 0 for y and solve for x. -5x+y=-10y=0-5x+0=-10Identity Property of Addition-5x=-10LHS/(-5)=RHS/(-5)x=2 The function's x-intercept is at x=2. In order to graph the function we need another point, so why not the y-intercept? The y-intercept occurs when x=0. In any function in slope-intercept form, the constant always shows the y-intercept. Let's rewrite our function in slope-intercept form. -5x+y=-10⇒y=5x − 10 In our function the constant b is -10. This means that the y-intercept occurs at (0,-10). By graphing both the x- and y-intercepts and drawing a line through these points, we can graph the function.
Exercises 23 Equations written in slope-intercept form follow a specific format. y=mx+b​ In this form, m represents the slope of the line and b represents the y-intercept.Writing the Equation in Slope-Intercept Form Since the given equation is not in slope-intercept form, let's rewrite it. x+3y=9LHS−x=RHS−x3y=-x+9LHS/3=RHS/3y=-31​x+3 We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=-31​x + 3​ The slope is -31​ and the y-intercept is 3.Graphing the Equation A slope of -31​ means that for every 3 units we move in the positive horizontal direction, we move 1 unit in the negative vertical direction. m=runrise​=3-1​​ To graph the equation, plot the y-intercept and then use the slope to find another point on the line.Identify the x-intercept Now that we've graphed the given function, we can use our graph to identify the x-intercept. This is the point at which the line intercepts the x-axis.The line crosses the axis at (9,0) so the x-intercept is 9.
Exercises 24 We will use the given information to determine some features of the function, starting with the slope.Slope Consider what slope represents. It is a change in the x- and y-values of the graph. slope=change in xchange in y​​ We are told that the dependent variable y decreases by 2 units every time the independent variable x increases by 3 units. slope=3-2​⇔slope=-32​​y-intercept Think of the point where the graph of an equation crosses the y-axis. The x-value of that (x,y) coordinate pair is 0, and the y-value is the y-intercept. We are given this value. h(0)=2​ The y-intercept is 2, so the graph intercepts the y-axis at the point (0,2).Graph We now have enough information to graph the function. To start, plot the y-intercept and one other point using the slope we found above. By connecting these points with a line, we will form the graph of our equation.x-intercept Looking at our graph, we can also identify the x-intercept of this function.The x-intercept of the function lies at the point (3,0), so the x-intercept is 3.
Exercises 25 First, let's use a table of values to find points on the graph of f(x)=3x+4.x3x+4f(x) 03⋅0+44 23⋅2+410 43⋅4+416 We can plot these points and connect them with a straight line to have the graph of f(x).Function h(x) Now, let's look at how the function h(x)=f(x+3) differs from f(x).xx+3f(x+3)h(x) -303⋅0+44 -123⋅2+410 143⋅4+416 If we plot these points on the same coordinate plane as f(x), we can see that each x value is being translated 3 units to the left. Therefore, we have a horizontal translation.Our final graph with the original and transformed lines is shown below.
Exercises 26 First, let's use a table of values to find points from the graph of f(x)=3x+4.x3x+4f(x) 03⋅0+44 13⋅1+47 23⋅2+410 We can plot these points and connect them with a straight line to form the graph of f(x).Function h(x) Now, let's look at how the function h(x)=f(x)+1 differs from f(x).xf(x)f(x)+1h(x) 044+15 177+18 21010+111 If we plot these points on the same coordinate plane as f(x), we can see that each y value is being translated 1 unit up. This is a vertical translation.
Exercises 27 First, let's use a table of values to find points on the graph of f(x)=3x+4.x3x+4f(x) 03⋅0+44 23⋅2+410 43⋅4+416 We can plot these points and connect them with a straight line to have the graph of f(x).Function h(x) Now, let's look at how the function h(x)=f(-x) differs from f(x).x-xf(-x)h(x) 003⋅0+44 -223⋅2+410 -443⋅4+416 If we plot these points on the same coordinate plane as f(x), we can see that the sign of each x-value is being changed (note that if we change the sign of 0 it will still be 0). This is a reflection in the y-axis.Our final graphs of the given and transformed lines are shown below.
Exercises 28 First, let's use a table of values to find points graph of f(x)=3x+4.x3x+4f(x) 03⋅0+44 13⋅1+47 23⋅2+410 We can plot these points and connect them with a straight line to have the graph of f(x).Function h(x) Now let's look at how the function h(x)=-f(x) differs from f(x).xf(x)-f(x)h(x) 04-1⋅4-4 17-1⋅7-7 210-1⋅10-10 If we plot these points on the same coordinate plane as f(x), we can see that each y value is being reflected about the x-axis. This is a reflection in the x-axis.
Exercises 29 First, let's use a table of values to find points on the graph of f(x)=3x+4.x3x+4f(x) 03(0)+44 13(1)+47 23(2)+410 We can plot these points and connect them with a straight line to have the graph of f(x).Function h(x) Now, let's look at how the function g(x)=3f(x) differs from f(x).xf(x)3f(x)g(x) 043(4)12 173(7)21 2103(10)30 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched three times as far from the x-axis as they were before.This is called a vertical stretch by a factor of 3.
Exercises 30 First, let's use a table of values to find points on the graph of f(x)=3x+4.x3x+4f(x) 03(0)+44 13(1)+47 23(2)+410 We can plot these points and connect them with a straight line to have the graph of f(x).Function g(x) Now, let's look at how the function g(x)=f(6x) differs from f(x).xf(x)f(6x)g(x) 043(6(0))+44 173(6(1))+422 2103(6(2))+440 If we plot these points on the same coordinate plane as f(x), we can see that our points have shrunk six times closer to the y-axis than they were before.This is called a horizontal shrink by a factor of 61​.
Exercises 31 Let's begin by graphing both functions. Then we can compare the differences between them.First, look at how the slope changes between f(x) and h(x).Looking at the slope triangles, we can see that the rise in h(x) is 5 times greater than the rise of f(x). This means that h(x) is moving away from the x-axis 5 times as quickly as f(x). slope of f(x): slope of h(x): ​1 ⇔runrise​=11​5⇔runrise​=15​​ When a function is pulled away from or pushed towards the x-axis, this is either a vertical stretch or shrink. Because the transformed function is farther from the x-axis, it is a vertical stretch by a factor of 5. The next thing to notice is how the y-intercept changes.The y-intercept moved up 1 unit. This is a horizontal translation up by 1.
Exercises 32 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values for function f.x∣x∣f(x) -1∣-1∣1 0∣0∣0 1∣1∣1 Now, let's do the same with function m.x∣x∣+6Simplifym(x) -1∣-1∣+61+67 0∣0∣+60+66 1∣1∣+61+67 Finally, we can take these ordered pairs and plot them on one coordinate plane.Comparison We can see that plot of function m is function f moved up 6 units. This is a vertical translation up by 6 units.Domain and range To find the domain and range of an absolute value function, we need to think about where the vertex is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 6, and it will continue increasing indefinitely. Range: 6≤y<∞​ The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​
Exercises 33 To graph the functions without going through the process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values for function f.x∣x∣f(x) -1∣-1∣1 0∣0∣0 1∣1∣1 Now, let's do the same with function p.x∣x∣+6Simplifyp(x) 3∣3−4∣∣-1∣1 4∣4−4∣∣0∣0 5∣5−4∣∣1∣1 Finally, we can take these ordered pairs and plot them on one coordinate plane.Comparison We can see that the plot of function p is the same as the plot of function f moved right by 4. This is a horizontal translation to the right by 4 units.Domain and range To find the domain and range of an absolute value function, we need to think about where the vertex is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0, and it will continue increasing indefinitely. Range: 0≤y<∞​ The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​
Exercises 34 To graph the functions without going through the process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values for function f.x∣x∣f(x) -1∣-1∣1 0∣0∣0 1∣1∣1 Now, let's do the same with function q.x∣x∣+6Simplifyq(x) -14∣-1∣4⋅14 04∣0∣4⋅00 14∣1∣4⋅14 Finally, we can take these ordered pairs and plot them on one coordinate plane.If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched four times as far from the x-axis as they were before. This is called a vertical stretch by a factor of 4.Domain and range To find the domain and range of an absolute value function, we need to think about where the vertex is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0, and it will continue increasing indefinitely. Range: 0≤y<∞​ The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​
Exercises 35 To graph the functions without going through the process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values for function f.x∣x∣f(x) -1∣-1∣1 0∣0∣0 1∣1∣1 Now, let's do the same with function r.x-41​∣x∣Simplifyr(x) -4-41​∣-4∣-41​⋅4-1 0-41​∣0∣4⋅00 4-41​∣4∣-41​⋅4-1 Finally we can take these ordered pairs and plot them on one coordinate plane.We can rewrite r(x) as -41​f(x), which is a horizontal shrink by a factor of 4 and a reflection in the x-axis. Notice that the y-intercept is the same for both graphs.Domain and range To find the domain and range of an absolute value function, we need to think about where the vertex is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The maximum of the given function is 0, and it will continue decreasing indefinitely. Range: -∞<y<0​ The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​
Exercises 36 To graph the functions without going through the process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values for function f.x∣x−2∣+4Simplifyf(x) 1∣1−2∣+4∣-1∣+45 2∣2−2∣+4∣0∣+44 3∣3−2∣+4∣1∣+45 Now, let's do the same with function g.x∣3x−2∣+4Simplifyg(x) -1∣3(-1)−2∣+4∣-5∣+49 0∣3(0)−2∣+4∣-2∣+46 1∣3(1)−2∣+4∣1∣+45 2∣3(2)−2∣+4∣4∣+48 Finally, we can take these ordered pairs and plot them on one coordinate plane.Comparison We can rewrite g(x) as f(3x), which is a horizontal shrink by a factor of 31​. Notice that the y-intercept is the same for both graphs.
Exercises 37 aWe can look at each of the transformations individually and work our way from the "inside" to the "outside." g(x)=31​∣x−1∣−2​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Let's start with the transformation closest to x and end with the farthest from x. First, we have a horizontal translation 1 unit to the right, ∣x∣⇒∣x−1∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Next, we have a vertical shrink by a factor of 31​, ∣x−1∣⇒31​∣x−1∣Finally, we have a vertical translation 2 units down, 31​∣x−1∣⇒31​∣x−1∣−2.bTo graph the function without going through the process of transforming the parent function, we can make a table of values. Then, we only need to plot the ordered pairs that we find.x31​∣x−1∣−2Simplifyg(x) -231​∣(-2)−1∣−231​∣-3∣−2-1 131​∣1−1∣−231​∣0∣−2-2 431​∣4−1∣−231​∣3∣−2-1 Now, we can plot these points and connect them to create our graph of g(x).