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###### Exercises
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Exercises 1 Equations written in slope-intercept form follow a specific format. y=mx+b​ In this form, m is the slope of the line and b is the y-intercept. We need to identify these values using the graph. Let's start with the y-intercept.Finding the y-intercept Observe the given graph.We can see that the function intercepts the y-axis at (0,1). This means that the value of b is 1.Finding the Slope To find the slope, we will trace along the line on the given graph until we find a lattice point, which is a point that lies perfectly on the grid lines. In doing so, we will be able to identify the slope m using the rise and run of the graph.Here we've identified (4,-1) as our other point. Traveling to this point from the y-intercept requires that we move 4 steps horizontally in the positive direction and 2 steps vertically in the negative direction. runrise​=4-2​⇔m=-21​​Writing the Equation Now that we have the slope and the y-intercept, we can form our final equation. y=mx+by=-21​x+1​
Exercises 2 An equation in point-slope form follows a specific format. y−y1​=m(x−x1​)​ In this form, m is the slope and the point (x1​,y1​) lies on the line. We are given the point (4,7) and the slope -1, so we have everything we need to create a point-slope form equation. y−y1​=m(x−x1​)Substitute x1​=4,y1​=7,m=-1y−7=-1(x−4) Please note that this is only one of infinitely many possible equations for this line because any point lying on the line could be used to create an equivalent equation. However, since we were given a point, this equation was the easiest to form.
Exercises 3 An equation in slope-intercept form follows a specific format. In this case, we are asked to write our equation in function notation. f(x)=mx+b​ For an equation in this form, m is the slope and b is the y-intercept. We have also been given two points in function notation. To write these points as coordinate pairs, remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(10)=5⇔(10,5)f(2)=-3⇔(2,-3)​ Let's use the given points to calculate m and b. We will start by substituting the points into the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (10,5) & (2,-3)m=2−10-3−5​ Simplify right-hand side Subtract termsm=-8-8​Calculate quotient m=1 A slope of 1 means that for every 1 horizontal step in the positive direction, we take 1 vertical step in the positive direction. Now that we know the slope, we can write a partial version of the equation. f(x)=1x+b ⇒f(x)=x+b​ To complete the equation, we also need to determine the y-intercept, b. Since we know that the given points will satisfy the equation, we can substitute one of them into the equation to solve for b. Let's use (10,5). f(x)=x+bx=10, f(x)=55=10+b Solve for b LHS−10=RHS−10-5=bRearrange equation b=-5 A y-intercept of -5 means that the line crosses the y axis at the point (0,-5). We can now complete the equation. f(x)=1x+(-5) ⇒f(x)=x−5​
Exercises 4 An equation in slope-intercept form follows a specific format. In this case, we are asked to write our equation in function notation. f(x)=mx+b​ For an equation in this form, m is the slope and b is the y-intercept. We have also been given two points in function notation. To write these points as coordinate pairs, remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(3)=-4⇔(3,-4)f(5)=-4⇔(5,-4)​ Let's use the given points to calculate m and b. We will start by substituting the points into the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (3,-4) & (5,-4)m=5−3-4−(-4)​ Simplify right-hand side a−(-b)=a+bm=5−3-4+4​Add and subtract termsm=20​Calculate quotient m=0 A slope of 0 means that our line will be horizontal. Now that we know the slope, we can write a partial version of the equation. f(x)=0x+b ⇒f(x)=b​ To complete the equation, we also need to determine the y-intercept, b. Since we know that the given points will satisfy the equation, we can substitute one of them into the equation to solve for b. Let's use (3,-4). f(x)=bx=3, f(x)=-4-4=bRearrange equationb=-4 A y-intercept of -4 means that the line crosses the y axis at the point (0,-4). We can now complete the equation. f(x)=0x+(-4) ⇒f(x)=-4​
Exercises 5 An equation in slope-intercept form follows a specific format. In this case, we are asked to write our equation in function notation. f(x)=mx+b​ For an equation in this form, m is the slope and b is the y-intercept. We have also been given two points in function notation. To write these points as coordinate pairs, remember that the input x is the x-coordinate and the output f(x) is the y-coordinate. f(x)=y⇔(x,y)f(6)=8⇔(6,8)f(9)=3⇔(9,3)​ Let's use the given points to calculate m and b. We will start by substituting the points into the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (6,8) & (9,3)m=9−63−8​ Simplify right-hand side Subtract termsm=3-5​Put minus sign in front of fraction m=-35​ A slope of -35​ means that for every 3 horizontal steps in the positive direction, we take 5 vertical step in the negative direction. Now that we know the slope, we can write a partial version of the equation. f(x)=-35​x+b​ To complete the equation, we also need to determine the y-intercept, b. Since we know that the given points will satisfy the equation, we can substitute one of them into the equation to solve for b. Let's use (6,8). f(x)=-35​x+bx=6, f(x)=88=-35​(6)+b Solve for b Multiply8=-330​+bCalculate quotient8=-10+bLHS+10=RHS+1018=bRearrange equation b=18 A y-intercept of 18 means that the line crosses the y axis at the point (0,18). We can now complete the equation. f(x)=-35​x+18​
Exercises 6 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they Parallel? For this exercise, we have been given two points on each line, so we have enough information to calculate their slopes using the Slope Formula. m=x2​−x1​y2​−y1​​​ Note that when choosing points to substitute for (x1​,y1​) and (x2​,y2​), it doesn't matter which points on the line you choose, since the result will be the same. Let's start with line a, which passes through (0,4) and (4,3). m=x2​−x1​y2​−y1​​Substitute (0,4) & (4,3)m=4−03−4​ Simplify Subtract termsm=4-1​Put minus sign in front of fraction m=-41​ We will use a similar method to identify the slopes of lines b and c.LinePoint 1Point 2x2​−x1​y2​−y1​​Slope a(0,4)(4,3)4−03−4​-41​ b(0,1)(4,0)4−00−1​-41​ c(2,0)(4,4)4−24−0​2 Now that we've identified the slope of each line, we can see that a and b have the same slope, so we know that they are parallel.Are they Perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and are therefore perpendicular. m1​⋅m2​=?-1​ Since only line c has a different slope, let's calculate the product of its slope and the slope of a and b. m1​⋅m2​=?-1m1​=-41​, m2​=2-41​⋅(2)=?-1b1​⋅a=ba​-42​=?-1ba​=b/2a/2​-21​​=-1 Therefore, lines a and b are parallel. Line c is neither parallel nor perpendicular to lines a and b, so there are no perpendicular lines.
Exercises 7 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they parallel? For this exercise, we have been given the equation of each line, so we will rewrite each of them in slope-intercept form to identify their slopes.LineGiven EquationSlope-intercept FormSlope a2x−7y=14y=72​x−272​ by=27​x−8y=27​x−827​ c2x+7y=-21y=-72​x−3-72​ Now that we've identified the slope of each line, we can see that none of the lines has the same slope, so they are not parallel.Are they perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and are therefore perpendicular. Let's start with checking lines a and b. m1​⋅m2​=?-1m1​=72​, m2​=27​72​⋅(27​)=?-1Multiply fractions1​=-1 Therefore, lines a and b are neither parallel nor perpendicular. We will use a similar method to check if lines a and c or b and c are perpendicular.LinesSlope 1Slope 2Product a & b72​27​1 a & c72​-72​-144​ b & c27​-72​-1 We have found that lines b and c are perpendicular to one another.
Exercises 8 When lines are parallel, they have the same slope. y=-4x+2​ Therefore, we know that all lines that are parallel to our given line will have a slope of -4. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=-4x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the point (1,5). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=-4x+bx=1, y=55=-4(1)+b Solve for b a⋅1=a5=-4+bLHS+4=RHS+49=bRearrange equation b=9 Now that we have the y-intercept, we can write the equation of the line that is parallel to y=-4x+2 and passes through the point (1,5). y=-4x+9​
Exercises 9 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. Then, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Perpendicular Line's Slope Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m1​⋅m2​=-1​ For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. Looking at the given equation, we can see that its slope is -2. y=-2x−3​ By substituting this value into our negative reciprocal equation for m1​, we can solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=-2-2⋅m2​=-1LHS/-2=RHS/-2m2​=-2-1​-b-a​=ba​m2​=21​ With this, we can identify that any line perpendicular to the given equation will have a slope of 21​.Writing the Perpendicular Line's Equation With the slope m2​=21​, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=21​x+b​ By substituting the given point (2,-3) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line. y=21​x+bx=2, y=-3-3=21​(2)+b Solve for b 2a​⋅2=a-3=1+bLHS−1=RHS−1-4=bRearrange equation b=-4 Now that we have the y-intercept, we can write the equation of the line that is perpendicular to y=-2x−3 and passes through the point (2,-3). y=21​x+(-4) ⇒y=21​x−4​
Exercises 10 We want to determine the roasting time for a 12-pound turkey from the given graph.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution18697_1_1295179066_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-6.5, 4.75, 38, -0.75],{desktopSize:'medium','style':'usa',xScale:8}); b.coordPlane([4,0],[0.5,0],'x','y',{x_pos:true,y_pos:true,paddingRight:2, labelX:{text:"Weight (pounds)",pos:[19,-0.55],fontSize:1.2}, labelY:{text:"Roasting time (hours)",pos:[-5.5,2.375],rotate:90,fontSize:1.2}, yaxis:{crunch:0.5,crunchStart:0,start:2.5,stepSize:0.5}}); b.point(6,0.75,{fillColor:'dodgerblue',size:0.25}); b.point(8,1,{fillColor:'dodgerblue',size:0.25}); b.point(12,2,{fillColor:'dodgerblue',size:0.25}); b.point(14,2.5,{fillColor:'dodgerblue',size:0.25}); b.point(18,3.25,{fillColor:'dodgerblue',size:0.25}); b.point(20,3.5,{fillColor:'dodgerblue',size:0.25}); b.point(24,3.75,{fillColor:'dodgerblue',size:0.25});} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution18697_1_1295179066_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution18697_1_1295179066_l", "Solution18697_1_1295179066_p", 1, code); }); } ); } window.JXQtable["Solution18697_1_1295179066_l"] = true;From here, we will begin by drawing a horizontal line from the point that has an x-value of 12.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution18697_2_1380653410_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-6.5, 4.75, 38, -0.75],{desktopSize:'medium','style':'usa',xScale:8}); b.coordPlane([4,0],[0.5,0],'x','y',{x_pos:true,y_pos:true,paddingRight:2, labelX:{text:"Weight (pounds)",pos:[19,-0.55],fontSize:1.2}, labelY:{text:"Roasting time (hours)",pos:[-5.5,2.375],rotate:90,fontSize:1.2}, yaxis:{crunch:0.5,crunchStart:0,start:2.5,stepSize:0.5}}); b.point(6,0.75,{fillColor:'dodgerblue',size:0.25}); b.point(8,1,{fillColor:'dodgerblue',size:0.25}); var p1 = b.point(12,2,{fillColor:'crimson',size:0.25}); b.point(14,2.5,{fillColor:'dodgerblue',size:0.25}); b.point(18,3.25,{fillColor:'dodgerblue',size:0.25}); b.point(20,3.5,{fillColor:'dodgerblue',size:0.25}); b.point(24,3.75,{fillColor:'dodgerblue',size:0.25}); var n1 = b.node(0,2,{}); b.segment(p1,n1,{strokeColor:'crimson',dash:2});} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution18697_2_1380653410_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution18697_2_1380653410_l", "Solution18697_2_1380653410_p", 1, code); }); } ); } window.JXQtable["Solution18697_2_1380653410_l"] = true;The line crosses the y-axis at y=4. Therefore, it takes 4 hours to roast a 12-pound turkey.
Exercises 11 To write an equation that models the roasting time as a function of the weight of a turkey, we will first draw a line that appears to fit the given data closely. There should be approximately as many points above the line as below it.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution18698_1_400257598_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-6.5, 4.75, 38, -0.75],{desktopSize:'medium','style':'usa',xScale:8}); b.coordPlane([4,0],[0.5,0],'x','y',{x_pos:true,y_pos:true,paddingRight:2, labelX:{text:"Weight (pounds)",pos:[19,-0.55],fontSize:1.2}, labelY:{text:"Roasting time (hours)",pos:[-5.5,2.375],rotate:90,fontSize:1.2}, yaxis:{crunch:0.5,crunchStart:0,start:2.5,stepSize:0.5}}); b.point(6,0.75,{fillColor:'dodgerblue',size:0.25}); b.point(8,1,{fillColor:'dodgerblue',size:0.25}); b.point(12,2,{fillColor:'dodgerblue',size:0.25}); b.point(14,2.5,{fillColor:'dodgerblue',size:0.25}); b.point(18,3.25,{fillColor:'dodgerblue',size:0.25}); b.point(20,3.5,{fillColor:'dodgerblue',size:0.25}); b.point(24,3.75,{fillColor:'dodgerblue',size:0.25}); b.func('0.1875*x-0.25',{xMin:1.5,firstArrow:false,strokeColor:"crimson"});} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution18698_1_400257598_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution18698_1_400257598_l", "Solution18698_1_400257598_p", 1, code); }); } ); } window.JXQtable["Solution18698_1_400257598_l"] = true;Now, we can start to write an equation for the line of fit. Notice that the points (12,4) and (20,5.5) lie on the line. With these points, we can find the slope of the line by using the slope formula. m=x2​−x1​y2​−y1​​Substitute (12,4) & (20,5.5)m=12−204−5.5​Subtract termsm=-8-1.5​-b-a​=ba​m=81.5​ba​=b⋅2a⋅2​m=163​ We found that the slope of the line is 163​. Next, we will find the y-intercept of the line. Therefore, we should write the equation in slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept. y=163​x+b​ By substituting either of the point on the line into the above equation, we can find b. Let's substitute (12,4)! y=163​x+bx=12, y=44=163​(12)+b Solve for b ca​⋅b=ca⋅b​4=1636​+bCalculate quotient4=2.25+bLHS−2.25=RHS−2.251.75=bRearrange equation b=1.75 With this, we can complete the equation. y=163​x+1.75​ In the context, since the slope is positive, the roasting time increases as the weight of the turkey increases. 163​ hours=11.25 minutes​ More specifically, the roasting time increases 11.25 minutes for each pound of turkey. On the other hand, because there cannot be any roasting time for 0-pound of turkey, the y-intercept does not make sense.
Exercises 12 Let's first recall the model given in the example. y=0.50x−23.5​ We will verify whether it is a good fit for the data by making a scatter plot of the residuals. We will begin by finding the residuals.Height, xShoe size, yy=0.50x−23.5y-value from the modelResidual 6490.50(64)−23.58.59−8.5=0.5 6270.50(62)−23.57.57−7.5=-0.5 70120.50(70)−23.511.512−11.5=0.5 6380.50(63)−23.588−8=0 72130.50(72)−23.512.513−12.5=0.5 689.50.50(68)−23.510.59.5−10.5=-1 6690.50(66)−23.59.59−9.5=-0.5 7413.50.50(74)−23.513.513.5−13.5=0 68100.50(68)−23.510.510−10.5=-0.5 596.50.50(59)−23.566.5−6=0.5 Now, we can make the scatter plot.As we can see, the points are evenly dispersed about the horizontal axis. Therefore, the equation y=0.50x−23.5 is a good fit.
Exercises 13
Exercises 14 When two variables have a causal relationship it means that one factor directly influences the other. Height and shoe size are definitely related to one another, taller people tend to wear larger shoe sizes. However, getting taller does not automatically imply that your feet will also get larger. The opposite is also true, your feet changing size does not mean that you will get taller.Examples Pregnant women and training athletes often experience a, usually temporary, growth in shoe size just due to swelling and/or gravity. Some men continue to grow well into their early 20s although their feet have stopped growing in their mid-teens. When you are flying your feet sometimes swells a bit. You have just been sitting too long, and all the blood has sunk to your feet. This doesn't imply getting taller.
Exercises 15 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=11. Let's observe the other terms to determine the common difference d. 11→−110→−19→−18…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=-1, a1​=11an​=11+(n−1)(-1)Distribute (-1)an​=11−n+1Add termsan​=12−nCommutative Property of Multiplicationan​=-n+12 This equation can be used to find any term in the given sequence. To find a30​, the 30th term in the sequence, we substitute 30 for n. an​=-n+12n=30a30​=-30+12Add termsa30​=-18 The 30th term in the sequence is -18.
Exercises 16 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=6. Let's observe the other terms to determine the common difference d. 6→+612→+618→+624…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=6, a1​=6an​=6+(n−1)(6)Distribute 6an​=6+6n−6Subtract terman​=6n This equation can be used to find any term in the given sequence. To find a30​, the 30th term in the sequence, we substitute 30 for n. an​=6nn=30a30​=6(30)Multiplya30​=180 The 30th term in the sequence is 180.
Exercises 17 Explicit equations for arithmetic sequences follow a specific format. an​=a1​+(n−1)d​ In this form, a1​ is the first term in a given sequence, d is the common difference from one term to the next, and an​ is the nth term in the sequence. For this exercise, the first term is a1​=-9. Let's observe the other terms to determine the common difference d. -9→+3-6→+3-3→+30…​ By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an​=a1​+(n−1)dd=3, a1​=-9an​=-9+(n−1)(3)Distribute 3an​=-9+3n−3Subtract terman​=3n−12 This equation can be used to find any term in the given sequence. To find a30​, the 30th term in the sequence, we substitute 30 for n. an​=3n−12n=30a30​=3(30)−12Multiplya30​=90−12Subtract terma30​=78 The 30th term in the sequence is 78.
Exercises 18
Exercises 19 First, we will graph y=x+6 for the domain x≤0. The function has a slope of 1 and a y-intercept of 6. Since the endpoint is included, we should use a closed circle.Looking at the graph we can see that the y-values are less than or equal to 6. The range for this graph is then y≤6. Next, we will graph y=-3x for the domain x>0. The function has a slope of -3 and a y-intercept of 0. Since the endpoint is not included, we will use an open circle.Looking at this graph we can see that all y-values are less than 0. Therefore, the range for this graph is y<0. Finally, let's plot both graphs on the same coordinate plane.Domain and Range of Piecewise Function Now that we've graphed both pieces together, we can look at the overall domain and range. We can see on the graph that the maximum y-value for the function is 6. Also, because the range of the piece containing this maximum is (-∞,6], the range of the entire function is all values of y such that y≤6. To find the domain, let's consider our piecewise function once more. y={x+6,-3x,​if x≤0if x>0​ The input values of the first expression are all the values less than or equal to 0, and the input values for the second expression are all the values greater than 0. Therefore, the domain of the piecewise function is the set of all real numbers.
Exercises 20 We have been given the following piecewise function. y={4x+2,2x−6,​if x<-4if x≥-4​ We will begin by graphing y=4x+2 for the domain, x<-4. The function has a slope of 4 and a y-intercept of 2. Since the endpoint is not included, we should keep the circle open.From the graph, we can see that it can produce all y-values that are less than -14. The range for this graph is thus y<-14. Next, we will graph y=2x−6 for the domain x≥-4. Since the endpoint is included, we should close the circle.From the graph we can see all y-values that are greater than or equal to -14. The range for this graph is thus y≥-14. When we combine the graphs, the domain and the range are both all real numbers.
Exercises 21 We are asked to write the absolute value function ∣x∣+15 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b​ In this form, m is the slope of the line and b is the y-intercept. Let's start with the line to the left of the vertex. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=15 and the slope is m=-1. Therefore, the equation is y=-x+15. We will now do the same with the line to the right of the vertex.For the second line, the y-intercept is b=15 and the slope is m=1. Therefore, the equation is y=x+15. Knowing the equations for both lines, we can now write the absolute value function ∣x∣+15 as the combination of two functions. y={-x+15-x+15​​ The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=0, so this is where our domain needs to change. The point at x=0 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-x+15-x+15​ifif​x≤0x>0​ory={-x+15-x+15​ifif​x<0x≥0​​
Exercises 22 We are asked to write the absolute value function 4∣x+5∣ as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b​ In this form, m is the slope of the line and b is the y-intercept. Let's start with the line to the left of the vertex. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-20 and the slope is m=-4. Therefore, the equation is y=-4x−20. We will now do the same with the line to the right of the vertex.For the second line, the y-intercept is b=20 and the slope is m=4. Therefore, the equation is y=4x+20. Knowing the equations for both lines, we can now write the absolute value function 4∣x+5∣ as the combination of two functions. y={-4x−20-4x+20​​ The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-5, so this is where our domain needs to change. The point at x=-5 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-4x−20-4x+20​ifif​x≤-5x>-5​ory={-4x−20-4x+20​ifif​x<-5x≥-5​​
Exercises 23 We are asked to write the absolute value function 2∣x+2∣−3 as a piecewise function. Let's start by looking at the graph of our given function.As we can see, the behavior of the absolute value function changes at the vertex. Therefore, this is a good point to separate the graph into two pieces, both being straight lines. Our next step will then be to find the equation for both lines. For this, let's recall the slope-intercept form of an equation. y=mx+b​ In this form, m is the slope of the line and b is the y-intercept. Let's start with the line to the left of the vertex. We can identify the y-intercept and slope from the graph shown below.For this line, the y-intercept is b=-7 and the slope is m=-2. Therefore, the equation is y=-2x−7. We will now do the same with the line to the right of the vertex.For the second line, the y-intercept is b=1 and the slope is m=2. Therefore, the equation is y=2x+1. Knowing the equations for both lines, we can now write the absolute value function 2∣x+2∣−3 as the combination of two functions. y={-2x−7-2x+1​​ The final step will be deciding the domain of each piece. In this case, the pieces meet and change directions at x=-2, so this is where our domain needs to change. The point at x=-2 can belong to either piece, as long as it belongs to only one of them. This means that we can correctly write the piecewise function in two different ways. y={-2x−7-2x+1​ifif​x≤-2x>-2​ory={-2x−7-2x+1​ifif​x<-2x≥-2​​
Exercises 24 We are told that the charge is \$65 for the first day and \$35 for each additional day. Let's use a table to organize the given information.Number of daysTotal cost in dollars 0<x≤165 1<x≤2100 2<x≤3135 On the basis of the table we write a piecewise function, where x is the number of days and y is the total cost. f(x)=⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧​65,100,135,​if 0<x≤1if 1<x≤2if 2<x≤3​​ The last step is to graph our function interval by interval.