Chapter 1 Linear Functions Homework 1.1 1.a.db.cc.ad.b 3. Temperature (F) is the dependent variable and time (t) is the independent variable.The graph shows that the temperature increases as the sun rises around 6 AM (t = 0).It reaches a high in the afternoon and drops slowly as the sun begins to set.Shortly before dawn of the following day, the temperature begins to rise again. 5. The dependent variable is height (h) and the independent variable is time (t).The graph shows that the height of the airplane increased quickly after takeoff, then leveled off for the flight.Near the end of the trip, the plane descended for its landing. 7. The dependent variable is the number of people undergoing laser eye surgery (n) and the independent variable is time (t).The graph shows that the number of people having laser eye surgery is increasing steadily as time goes on from 1996. 9. The dependent variable is the percentage of major firms that perform drug testing (p) and the independent variable is time (t).The graph shows that the number of major firms performing drug testing increases over time from 1987 through the mid 1990s and then decreases over time. 11. The dependent variable is amount of gas in the tank (g) and the independent variable is time (t) in minutes since the commuter left home.The graph decreases until the commuter purchases gas, rises when the gas is purchased, and then begins to decrease again as the commuter drives to work. 13. The dependent variable is height (h) and the independent variable is age (a).The curve shows that a persons height increases from birth and levels off after a number of years.Note that height will never be 0, and that it eventually levels off at a certain age. 15. Time (T) is the dependent variable and ounces of beer (x) is the independent variable.The curve shows that the time it takes to complete a task increases as more beer is consumed.Note that 0 T since each task takes some even when beer is not consumed. Homework 1.2SSM: Intermediate Algebra 2 17. Volume (V) is the dependent variable and seconds (t) is the independent variable.The graph shows the volume of air in a persons lungs alternately increases and decreases as air is breathed in and out over time. 19. The number of times a person can lift a set of weights (n) is the dependent variable and the weight of the weights (w) is the independent variable.The graph shows that the number of repetitions decreases as the weight increases. 21. The dependent variable is n, the number of people that will buy the Honda Civic CX, and the independent variable is p, the price at which the purchase occurs.The graph shows that more people will buy less expensive Honda Civic CXs.As the price in dollars increases, fewerpeople are willing to purchase this car.The curve eventually levels off since there are only so many people who will buy the car no matter what the price. 23. Area (A) is the dependent variable and radius (r) is the independent variable.The graph shows a rapid increase in area (A) as the circles radius (r) increases.This is because the radius r is squared in the formula 2A r = . 25.Answers may vary. a.Let y represent the height of a ball x seconds after it was thrown straight up in the air. b.Let y represent the speed of a snow skier x seconds after skiing off a ramp when competing for the distance jumped. c.Let y represent the water level of a creek t days after July 10 in Georgia.The water level remained the same during a cloudy week with very little rain and it dropped again steadily as summer passed until it completely dried up. d.Let y represent the temperature of abuilding x minutes after the thermostat is adjusted to reach a warmer setting with one hour. Homework 1.2 1.1. 3. 5. SSM: Intermediate Algebra Homework 1.2 3 7. 9. 11. 13.9 3 03 93 33x yy xy x = = = 15.3 6 123 12 63 12 63 34 2y xy xy xy x == ++== + 17.2 6 14 42 6 102 10 62 25 3y xy xy xy x = =+== + 19. ( ) 3 4 2 3 5 23 4 2 3 5 104 17y x x xy x x xy x= + + = + + += + 21. ( ) ( ) 3 5 2 3 63 15 6 123 6 273 32 9y xy xy xy x = + = = = + Homework 1.2SSM: Intermediate Algebra 4 23. 25. 27.a. Yes, this graph is a line with slope 4.1 and y-intercept 8.7. b. Yes, this graph is a horizontal line passing through (0, 6). c.Answers may vary.One example is as follows. 2 3 y x = + 29.a.i. ii. iii. b.An equation of the formy b =is a horizontal line passing through (0, b). 31. 33. SSM: Intermediate Algebra Homework 1.2 5 35. 37. (0, 1) is a point on the line. 3 1 y x = + ()?1 3 0 = ?11 1 True+= 39. ( )0.83 4.98 20.83 2 4.984.98 4.980.83 24.98x yx yxy= +=+= (0, 0.40160643) is a point on the line. ( )() ( )????0.83 2 4.980.83 0 20.401606434.9820.401606434.980.40160643 0.40160643 Truexy+=+=== 41.2 10 y x = +To find the x-intercept, let y = 0 and solve for x. 0 2 1010 25xxx= + = = The x-intercept is (5, 0).To find the y-intercept, let x = 0 and solve for y. () 2 0 100 1010yyy= += += The y-intercept is (0, 10). 43.2 3 12 x y + =To find the x-intercept, let y = 0 and solve for x. () 2 3 0 122 0 122 126xxxx+ =+ === The x-intercept is (6, 0).To find the y-intercept, let x = 0 and solve for y. () 2 0 3 120 3 123 124yyyy+ =+ === The y-intercept is (0, 4). 45.3 y x =To find the x-intercept, let y = 0 and solve for x. 0 30xx== The x-intercept is (0, 0).To find the y-intercept, let x = 0 and solve for y. () 3 00yy== The y-intercept is (0, 0). 47.3 y =Since y = 3 is a horizontal line, it never intersects the x-axis.Therefore, there is no x-intercept.Since the graph passes through (0, 3), this is the y-intercept. 49.xy 74 42 1 0 2 2 5 4 Answers may vary.There are other possible solutions. 51.3 y 53.2 y Homework 1.3SSM: Intermediate Algebra 6 55.6 x 57.1 x 59.a. b.To find the y-intercept, let x = 0 and solve for y. () 200 0 8000 800800yyy= += += The y-intercept is (0, 800).This means that before the person the person begins to lower the balloon, the height is 800 feet. c.To find the x-intercept, let y = 0 and solve for x. 0 200 800800 2004xxx= + = = The x-intercept is (4, 0).This means that after 4 minutes, the balloon will be lowered to the ground (height = 0). 61.The points B, C and F satisfy the equation. 63. The ordered pair (1, 3) satisfies both equations, since this point lies on the intersection of the lines. 65.Substitute for x and y in the equation and solve for b. ()25 2 75 149y x bbbb= += += + = 67.To sketch the graph of a linear equation, create a table of ordered pair solutions of the equation by substituting values of one variable and solving for the other variable.Then, plot the ordered pairs from the table and sketch the line that passes through three plotted points.The graph of an equation is all points that satisfy the equation. Homework 1.3 1.1. 2 12 19 3 625 2 3y ymx x = = = = Since m is positive 2, the line is increasing. 3. ( )2 12 110 210 2 1224 2 6 6y ymx x += = = = = Since m is negative 2, the line is decreasing. 5. ( )( )2 12 19 19 1 844 2 4 2 2y ymx x + = = = = = + Since m is positive 4, the line is increasing. 7. ( )2 12 17 47 4 3 33 8 5 5 5y ymx x + = = = = = Since m is positive 35, the line is increasing. 9. ( )( )2 12 11 21 2 318 11 8 11 3y ymx x += = = = = + Since m is positive 1, the line is increasing. 11. 2 12 11 0 111 0 1y ymx x = = = = Since m is positive 1, the line is increasing. 13. 2 12 15.4 2.6 2.81.041.2 3.9 2.7y ymx x = = = = Since m is negative 1.04, the line is decreasing. SSM: Intermediate Algebra Homework 1.3 7 15. ( )2 12 117.94 2.348.94 21.1317.94 2.34 15.61.2812.19 12.19y ymx x = = + = = Since m is positive 1.28, the line is increasing. 17. 2 12 16 6 007 2 5y ymx x = = = = Since m = 0, the line is horizontal. 19. ( )2 12 16.4 2.8 3.6 3.69.5 9.5 9.5 9.5 0y ymx x = = = = + 3.60 is undefined.Since m is undefined, the line is vertical. 21.a.The slope is defined since it is increasing.The slope is positive. b.The slope is defined since it is decreasing.The slope is negative. c.The slope is defined since it is a horizontal line.The slope is zero. d.The slope of the line is undefined.The line is vertical. 23.The points (3, 0) and (2, 2) lie on the line. 2 12 12 0 2 22 3 5 5y ymx x = = = = The slope of the line is 25 . 25.Since 1 2m m = , the lines are parallel. 27.The lines are neither parallel nor perpendicular. 29. Since 211mm= , the lines are perpendicular. 31.The lines are neither parallel nor perpendicular. 33.Since 1m= 0, 1lis a horizontal line.Since 2mis undefined, 2lis a vertical line.From geometry, any horizontal line is perpendicular to any vertical line.Therefore, the lines are perpendicular. 35.Answers may vary. 37.Answers may vary. 39.Answers may vary. 41. The line with slope 3 is steeper. 43. 1200.034000Am= =1600.0256500Bm= Road A is steeper than road B, since the slope of road A is greater than the slope of road B. Homework 1.3SSM: Intermediate Algebra 8 45. 90900.3300 300Am= = =1251250.3125450 450Bm= = =Ski run B is steeper than ski run A, since the slope of ski run B is greater than the slope of ski run A.(In determining which is steeper, ignore the sign of the slope by considering absolute value.) 47. The points (0, 0) and (10, 1) are on the line and can be used to find the slope. 2 12 11 0 110 0 10y ymx x = = = 49. The points (0, 0) and (1, 10) are on the line and can be used to find the slope. 2 12 110 0 10101 0 1y ymx x = = = = 51. 53. 55. Answers may vary.Three points that lie on the line are (1, 4), (0, 1), and (1, 2). 57.a.i. rise 22run 1= = ii. rise 33run 1= = SSM: Intermediate Algebra Homework 1.4 9 iii. rise 22run 1= = b. The slope of the line found using riserun is the same as the coefficient of x in each of the corresponding equations. 59.a. b. c. d.The graphs in parts a-c are mirror images of each other across the graph of y = x. e. 61.The line contains both points, S and Q.The slope is 23and can be written rise 2 2run 3 3= =.From point P, moving 3 units to the right and 2 units down is the equivalent of rise 2run 3= .From point P, moving 3 units to the left and 2 units up is the equivalent of rise 2run 3=. Homework 1.4 1.1.Since6 1 y x = +is in the formy mx b = + , the slope is 6 rise61 runm = = = , and the y-intercept is (0, 1). 3.Since2 7 y x = +is in the formy mx b = + , the slope is 2 rise21 runm= = = , and the y-intercept is (0, 7). Homework 1.4SSM: Intermediate Algebra 10 5. Since 524y x = is in the formy mx b = + , the slope is 5 rise4 runm = = , and the y-intercept is(0, 2). 7. Since 327y x = +is in the formy mx b = + , the slope is 3 3 rise7 7 runm= = = , and the y-intercept is (0, 2). 9. Since 513y x = is in the formy mx b = + , the slope is 5 5 rise3 3 runm= = = , and the y-intercept is (0, 1). 11.First, rewrite5 y x + =in slope-intercept form. 55y xy x+ == + The slope is 1 rise11 runm= = = , and the y-intercept is (0, 5). 13.First, rewrite7 2 10 x y + =in slope-intercept form. 7 2 102 2 10752x yy xy x + == += + The slope is 7 rise2 runm = = , and the y-intercept is (0, 5). 15. First, rewrite( ) 3 2 9 x y =in slope-intercept form. ( ) 3 2 93 6 96 3 91 32 2x yx yy xy x = = = += The slope is 1 rise2 runm = = , and the y-intercept is 30,2 . SSM: Intermediate Algebra Homework 1.4 11 17.First, rewrite2 3 9 12 x y + =in slope-intercept form. 2 3 9 123 2 3213x yy xy x + = = += The slope is 2 rise3 runm = = , and the y-intercept is ( ) 0, 1 . 19. First, rewrite( ) 7 2 2 0 x y + =in slope-intercept form. ( ) 7 2 2 07 2 4 02 7 4722x yx yy xy x + = = = += The slope is 7 rise2 runm = = , and the y-intercept is ( ) 0, 2 . 21.Rewrite4 y x =as4 0 y x = +to obtain slope-intercept form.The slope is 4 rise41 runm = = = , and the y-intercept is( ) 0, 0 . 23.Since1.5 3 y x = +is in the formy mx b = + , the slope is 3 rise1.52 runm= = = , and the y-intercept is (0, 3). 25.Rewritey x =as1 0 y x = +to obtain slope-intercept form.The slope is 1 rise11 runm = = = , and the y-intercept is( ) 0, 0 . 27.The linear equation4 y =is a horizontal line.The slope of a horizontal line is m = 0, and there is no x-intercept. Homework 1.4SSM: Intermediate Algebra 12 29.Solve for y. 2 02yy+ == The linear equation2 y = is a horizontal line.The slope of a horizontal line is m = 0, and there is no x-intercept. 31.Set 1 is not linear.As x increases by 1, 1ydoes not change by some consistent value.It is possible that there is a line that comes close to every point since the changes in 1yare roughly the same. Set 2 is linear.As x increases by 1, 2ydecreases by 0.3. Set 3 is not linear.The value of x does not change (increase or decrease) consistently, even though 3yincreases by 5. Set 4 is linear.As x increases by 1, 4ydecreases by 10. 33. Values for Four Linear Equations Eq. 1Eq. 2Eq. 3Eq. 4 xyxyxyxy 112236917305.0 215246122314.4 318255333323.8 421264548333.2 5242737513342.6 6272829618352.0 35.a.Since this is a decreasing line, the slope, m, is negative (m < 0).The line crosses the y-axis above the origin so the y-intercept is positive (b > 0). b.Since this is as increasing line, the slope, m, is positive (m > 0).The line crosses the y-axis below the origin so the y-intercept is negative (b < 0). c.Since the line is horizontal and crosses the y-axis below the origin, the slope, m, is 0 and the y-intercept, b, is negative (b < 0). d.Since this is a decreasing line that passes through the origin, the slope, m, is negative (m < 0), and the y-intercept, b, is 0. 37.For both lines, the slope is 4.The lines are parallel. 39. The slopes are 3 8 and 8 3.The slopes are reciprocals, but not negative reciprocals.The lines are neither parallel nor perpendicular. 41.2 3 6 4 6 73 2 6 6 4 72 2 723 3 6x y x yy x y xy x y x+ = + == + = += + = + Since the slopes are both m = 23 , the lines are parallel. 43.5 3 1 3 5 23 5 1 5 3 25 1 3 23 3 5 5x y x yy x y xy x y x = + = = + = = = Since the slopes are negative reciprocals, the lines are perpendicular. 45.Since x = 2 is a vertical line, and y = 5 is a horizontal line, the lines are perpendicular. 47. 1 1y x y xy x y x= = = = The slopes are 1 and 1.Recall that the reciprocal of 1 is 1.This means that the slopes are negative reciprocals and the lines are perpendicular. SSM: Intermediate Algebra Homework 1.4 13 49.a.Substitute values for x in the equation 3 18 y x = +to solve for y. When x = 0,( ) 3 0 18 y = + 0 18 18 = + = . When x = 1,( ) 3 1 18 y = + 3 18 15 = + = . When x = 2,( ) 3 2 18 y = + 6 18 12 = + = . When x = 3,( ) 3 3 18 y = + 9 18 9 = + = . When x = 4,( ) 3 4 18 y = + 12 18 6 = + = . When x = 5,( ) 3 5 18 y = + 15 18 3 = + = . When x = 6,( ) 3 6 18 y = + 18 18 0 = + = . Amount of Gas in a Cars Gas Tank Driving Time (hours) x Amount of Gas (gallons) y 018 115 212 39 46 53 60 b.Each hour, the amount of gas in the tank decreases by 3 gallons.The slope of 3 in the equation shows this decrease of 3.As the value of the independent variable (driving time) increases by 1, the value of the dependent variable (amount of gas) decreases by 3 as determined by the slope. c.Note that in one hours time, 3 gallons of gas is used.If the person is driving at approximately 60 mph, he or she uses 3 gallons to drive 60 miles.So for 1 gallon of gas, a person can travel 20 miles ( 60 3 ). 51.a.Substitute values for x in the equation 2 26 y x = +to solve for y and complete the table. When x = 0,( ) 2 0 26 y = + 0 26 26 = + = . When x = 1,( ) 2 1 26 y = + 2 26 28 = + = . Similar calculations yield the following table. Salaries Time at Company (years) x Salary (thousands of dollars) y 026 128 230 332 434 b.Each year the persons salary increases by $2000, which corresponds to the slope of 2 26 y x = +with y in thousands of dollars.As the value of the independent variable (time at company) increases by 1, the value of the dependent variable (salary) increases by 2 as determined by the slope. 53.a. b. c. d.The slope of the graph in part b appears to be steeper than the graph in part a.The graph of part c appears to be less steep than the graph in part a. e.No, you cannot make the sketch of y = x appear to be a decreasing line or cross the y-axis at a point other than (0, 0) by changing the window settings.The line will always appear to be increasing and it will always lie in quadrants one and three, but it may appear to have any steepness within these constraints. Homework 1.5SSM: Intermediate Algebra 14 55.Since we have the slope, m = 2, and a point on the line, (0, 3), we can substitute in the equationy mx b = +to find the y-intercept, b. ( ) 3 2 0y mx b = + =3bb+ = We can now write the equation of the line in slope-intercept form. ( ) 2 32 3y xy x= += 57.a. b.From the graph in part a, we see that the y-intercept, b, is 7.Using the slope and the y-intercept, we can write the equation of the line in slope-intercept form. ( ) 5 75 7y xy x= += c. 59.a. b.From the graph in part a, we see that the y-intercept, b, is approximately 162.Using the slope and the y-intercept, we can write the equation of the line. 3 138 2y x = + c. 61.a.The slope is undefined for each of the lines.They are vertical lines. b.The slope for any linear equation of the form x = k is undefined when k is a constant. 63.The coefficient of x is the slope of a line, when the equation is written in slope-intercept form.The equation,2 3 6 x y + = , is not written in slope intercept form.To find the slope, rewrite the equation in slope-intercept form by solving for y. 2 3 63 2 62 63 3223x yy xy xy x+ == += += + The slope of the line is 23 . 65.Answer may vary. Homework 1.5 1.1.We are given the slope, m = 3, and a point on the line, (5, 2).Usey mx b = +to find b. ( ) 2 3 52 1513y mx bbbb= += += + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. ( ) 3 133 13y xy x= += SSM: Intermediate Algebra Homework 1.5 15 The sign of m (positive) agrees with the increasing line from the graphing calculator screen. 3.We are given the slope, m = 2, and a point on the line, (3, 9).Usey mx b = +to find b. ( ) 9 2 39 63y mx bbbb= + = + = + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. ( ) 2 32 3y xy x= += The sign of m (negative) agrees with the decreasing line from the graphing calculator screen. 5.We are given the slope, m = 1.6, and a point on the line, (2.1, 3.8).Usey mx b = +to find b. ( ) 3.8 1.6 2.13.8 3.360.44y mx bbbb= += += += Now, substitute for m and b in slope-intercept form to obtain the equation of the line. 1.6 0.44 y x = + The sign of m (positive) agrees with the increasing line shown on the graphing calculator screen. 7. We are given the slope, 35m = , and a point on the line, (20, 7).Usey mx b = +to find b. ( )37 2057 125y mx bbbb= += += + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. ( )355355y xy x= += The sign of m (positive) agrees with the increasing line shown on the graphing calculator screen. 9. We are given the slope, 16m = , and a point on the line, (2, 3).Usey mx b = +to find b. ( )13 26133y mx bbb= + = + = + 9 13 383bb = + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. 1 86 31 86 3y xy x = + = Homework 1.5SSM: Intermediate Algebra 16 The sign of m (negative) agrees with the decreasing line shown on the graphing calculator screen. 11. We are given the slope, 52m = , and a point on the line, (3, 4).Usey mx b = +to find b. ( )54 3215428 152 2232y mx bbbbb= + = + = + = + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. 5 232 25 232 2y xy x = + = The sign of m (negative) agrees with the decreasing line shown on the graphing calculator screen. 13.We are given the slope,0 m = , and a point on the line, (1, 2).We know that a line with a slope of 0 is a horizontal line.Horizontal lines are of the form y = b.In this example, y = 2. Since m is zero, the line is horizontal.This is shown on the graphing calculator screen. 15.We are given that m is undefined, and a point on the line is (3, 7).We know that a line with an undefined slope is a vertical line.Vertical lines are of the form x = a .In this example, x = 3.

Since m is undefined, the line is vertical.This is shown on the graphing calculator screen. 17.First, find the slope. 5 3 214 2 2m= = = So, y = 1x + b.Since the line contains (2, 3), substitute 2 for x and 3 for y and solve for b. ( ) 3 1 23 21bbb= += += So, the equation is y = x + 1. 19.First, find the slope. ( ) 6 4 6 4 1022 3 5 5m += = = = So, y = 2x + b.Since the line contains (2, 6), substitute 2 for x and 6 for y and solve for b. ( ) 6 2 26 42bbb= += += So, the equation is y = 2x + 2. 21.First, find the slope. ( )( )6 146 14 828 4 8 4 4m += = = = + So, y = 2x + b.Since the line contains(8, 6), substitute 8 for x and 6 for y and solve for b. ( ) 6 2 86 1622bbb = + = + = So, the equation is y = 2x 22. 23.First, find the slope. 1 0 111 0 1m= = = So, y = 1x + b.Since the line contains (0, 0) substitute 0 for x and 0 for y to solve for b. ( ) 0 1 00 00bbb= += += So, the equation is y = x. SSM: Intermediate Algebra Homework 1.5 17 25.First, find the slope. 3.9 2.2 1.70.745.1 7.4 2.3m= = =

(rounded to nearest hundredth) So, y = 0.74x + b.Since the line contains ( ) 5.1, 3.9 , substitute 5.1 for x and 3.9 for y to solve for b. ( ) 3.9 0.74 5.13.9 3.7747.67bbb= += += (rounded to nearest hundredth) So, the equation is y = 0.74x + 7.67. 27.First, find the slope. 4.9 1.2 3.71.096.8 3.4 3.4m= = = (rounded to nearest hundredth) So, y = 1.09x + b.Since the line passes through (3.4, 1.2), substitute 3.4 for x and 1.2 for y to solve for b. ( ) 1.2 1.09 3.41.2 3.7062.51bbb= += + = (rounded to nearest hundredth) So, the equation is y = 1.09x 2.51. 29.First, find the slope. 5 1 47 2 5m= = So, y = 45 x + b.Since the line passes through ( ) 2,1 , substitute 2 for x and 1 for y to solve for b. ( )41 258155 85 535bbbb= += += + = So, the equation is 4 35 5y x = 31.First, find the slope. ( )( )7 27 2 5 55 3 5 3 2 2m + = = = = + So, y = 52 x + b.Since the line passes through (5, 7), substitute 5 for x and 7 for y to solve for b. ( )57 52257214 252 2112bbbb = + = + = += So, the equation is 5 112 2y x = + . 33.First, find the slope. ( ) 2 5 2 5 7 74 2 6 6 6m += = = = So, y = 76 x + b.Since the line passes through (4, 2), substitute 4 for x and 2 for y to solve for b. ( )72 461423bb= += + 143 383bb6= + = So, the equation is 7 86 3y x = . 35.First, find the slope. 5 5 004 2 2m= = = So, y = 0x + b or y = b.Since this is a horizontal line passing through (2, 5), the equation is y = 5. 37.First, find the slope. ( )4 6 10 undefined3 3 0m = = The slope is undefined.This is a vertical line passing through (3, 4), so the equation is 3 x = . Homework 1.5SSM: Intermediate Algebra 18 39.The slope of the given line is 3.A line parallel to the given line also has a slope of 3 and an equation3 y x b = + .Since the point (4, 5) lies on the parallel line, substitute 4 for x and 5 for y to solve or b. ( ) 5 3 45 127bbb= += + = The parallel lines equation isy = 3x 7. 41.The slope of the given line is 2.A line parallel to this has a slope of 2 and an equation 2 y x b = + .Substitute 3 for x and 8 for y to solve for b since the parallel line contains ( ) 3,8 . ( ) 8 2 38 62bbb= += += The parallel lines equation isy = 2x + 2. 43. The slope of the given line is 12.A line parallel to this has a slope of 12 and an equation 12y x b = + .Substitute 4 for x and 1 for y to solve for b since the parallel line contains (4, 1). ( )11 421 21bbb= += + = The parallel lines equation is 112y x = or 0.5 1 y x = . 45.To find the slope, isolate y. 3 4 124 3 12334x yy xy x = = += The slope is 34.A line parallel to this has the same slope and an equation 34y x b = + .Substitute 3 for x and 4 for y to solve for b since the parallel line contains (3, 4). ( )34 3494416 94 474bbbb= += += += The parallel lines equation 3 74 4y x = +or 0.75 1.75 y x = + . 47.To find the slope, isolate y. 6 76 71 76 6y xy xy x = = = The slope is 16.A line parallel to this has the same slope and an equation 16y x b = + .Substitute 3 for x and 2 for y to solve for b since the parallel line contains( ) 3, 2 . ( )12 36122bb = + = + 4 12 232bb = + = The parallel lines equation is 1 36 2y x = . 49.The line6 y =is horizontal and has a slope of 0.A line parallel to y = 6 is also horizontal.Since the parallel line contains (2, 3) and horizontal lines are of the form y = b, the equation of the line is3 y = . 51.The line x = 2 is vertical and has undefined slope.A line parallel to x = 2 is also vertical.Since vertical lines are of the form x = a, and the parallel line contains (5, 4), the equation of the line is x = 5. SSM: Intermediate Algebra Homework 1.5 19 53.The slope of the given line is 2.A line perpendicular to the given line must then have a slope of 12and an equation 12y x b = + .Substitute 3 for x and 8 for y to solve for b since the line contains (3, 8). ( )18 3238216 32 2192bbbb= += += += The equation of the line is 1 192 2y x = +or 0.5 9.5 y x = + . 55.The slope of the given line is 3.A line perpendicular to the given line must then have a slope of 13 and an equation 13y x b = + .Substitute 1 for x and 7 for y to solve for b since the line contains (1, 7). ( )17 13173bb= += + 21 13 3223bb= += The equation of the line is 1 223 3y x = + . 57. The slope of the given line is 25 .A line perpendicular to the given line must then have a slope of 52 and an equation 52y x b = + .Substitute 2 for x and 7 for y to solve for b since the line contains (2, 7). ( )57 227 52bbb= += += The equation of the line is 522y x = +or 2.5 2 y x = + . 59.To find the slope of the given line, isolate y. 4 5 75 4 74 75 5x yy xy x = = += The slope of this line is 45.A line perpendicular to this line must have a slope of 54and an equation 54y x b = + .Substitute 10 for x and 3 for y to solve for b since the line contains ( ) 10, 2 . ( )53 10425326 252 2312bbbb= += += += The equation of the line is 5 314 2y x = +or 1.25 15.5 y x = + . 61.To find the slope of the given line, isolate y. 2 52 5x yy x + == + The slope of the line is 2.A line perpendicular to this line must have a slope of 12and an equation 12y x b = + .Substitute 3 for x and 1 for y to solve for b since the line contains ( ) 3, 1 . ( )11 32312bb = + = + Homework 1.5SSM: Intermediate Algebra 20 2 32 252bb = + = The equation of the line is 1 52 2y x = or 0.5 2.5 y x = . 63.The slope of the equation x = 5 is undefined.The graph of the equation is a vertical line.A line perpendicular to x = 5 is a horizontal line with a slope of 0.Since this perpendicular line contains (2, 3) and the y-value at this point is 3, the equation of the line is y = 3. 65.The slope of the equation y = 3 is 0.The graph of the equation is a horizontal line.A line perpendicular to y = 3 is a vertical line with an undefined slope.Since this perpendicular line contains( ) 2, 8and the x-value at this point is 2, the equation of the line is x = 2. 67.Choose any two points to find the slope. 17 19 221 0 1m = = = So, y = 2x + b.Since the point (0, 19) is a solution to the equation, substitute 0 for x and 19 for y to solve for b. ( ) 19 2 019 019bbb= += += The equation describing the relationship between x and y is y = 2x + 19. 69.Find the slope by choosing two points on the line, such as (2, 0) and (5, 1). 1 0 15 2 3m= = So, 13y x b = + .Since the line contains (2, 0), substitute 2 for x and 0 for y to solve for b. ( )10 2320323bbb= += + = The equation for the line is 1 23 3y x = . 71.Choose two points on the line to find the slope, such as (3, 2) and (5, 1). ( ) 2 1 2 1 3 33 5 2 2 2m += = = = So, 32y x b = + .Since the line contains (3, 2), substitute 3 for x and 2 for y to solve for b. ( )32 329224 92 2132bbbb= += += += The equation for the line is 3 132 2y x = + . 73.a. It is possible for a line to have no x-intercepts. Horizontal lines of the form y b =(where b is a constant not equal to 0) have no x-intercepts. 73.b.It is possible for a line to have exactly one x-intercept.One example is y = x + 1, where the x-intercept is (1, 0).Other lines of the form y = mx + b have one x-intercept, as long as0 m . 73.c.It is not possible for a line to have exactly two x-intercepts.A line can never intersect the x-axis at exactly two points. 73.d.It is possible for a line to have an infinite number of x-intercepts.The line y = 0 lies on the xaxis and therefore, intersects the x-axis at an infinite number of points. 75.Yes, a line contains all these points.Using ( ) 4,15 and (1, 9), note that the slope is 2.So, y = 2x + b.Substitute 4 for x and 15 for y to solve for b since the line contains (4, 15). ( ) 15 2 415 87bbb= += += The equation of the line is y = 2x + 7.When each of the other points is substituted into this equation, we see that the line contains all of the given points. SSM: Intermediate Algebra Homework 1.6 21 77.a. Answers may vary.Consider the following example. Table for y = 3x 6 xy 06 13 20 33 46 59 612 77.b. Answers may vary.Consider the following example. Table of Points Lying Close to y = 3x 6xy 05 14 21 32 47 510 613 77.c.Answers may vary. 79.a. Any equation of the form4 y x b = + , where0 and/or0 x y , will have a slope of 4.One example is4 1 y x = + 79.b. Any equation of the form 37y mx = +where 0 y , will have a y-intercept of 30,7 . 79.c.Since the line must contain the point (2, 8), substitute 2 for x and 8 for y to find an equation. ( ) 8 28 2y mx bm bm b= += += + Choose any slope and then solve for b, the y-intercept. Then, use the slope and b to write the equation of the line.Choose m = 2, for example. ( ) 8 2 28 412bbb= += += An equation of a line that passes through the point (2, 8) is2 12 y x = + .Use TRACE to verify your equation. 79.d.The line that has a slope of 4 and y-intercept 30,7 is 347y x = + .Check to see if the point (2, 8) satisfies this equation. ( )38 4 2738 87307= += += This is false.There is no such equation. 81.Answers may vary.One possible answer follows.First, use the two points to find the slope of the line.Next, use one of the points to find b, by substituting in the equationy mx b = + .Finally, substitute for m and b in the equation y = mx + b.To verify that the equation contains the two points, substitute the values of x and y for each point to see if a true statement results. Homework 1.6 1.1.Relation 1 is not a function.The input x = 3 yields two outputs y = 5 and y = 7. Relation 2 could possibly be a function since each input yields only one output. Relation 3 could possibly be a function since each input yields only one output. Relation 4 is not a function.The input x = 8 yields two outputs y = 40 and y = 50. 3.No, the relation is not a function since an input yields more than one output. 5.Yes, it is possible that the relation is a function.Two inputs can yield the same output, but one input cannot yield two outputs. Homework 1.6SSM: Intermediate Algebra 22 7.a. This graph is a function since it passes the vertical line test. 7.b.This graph is not a function since a vertical line can intersect the graph at more than one point. 7.c.This graph is a function since it passes the vertical line test. 7.d.This graph is not a function since a vertical line can intersect the graph at more than one point. 9.Begin by sketching the graph. Note that this line passes the vertical line test. Therefore, it is a function. 11.First, isolate y. 2 5 105 2 10225x yy xy x = = += Now, sketch the graph. Note that this line passes the vertical line test.Therefore, it is a function. 13.y = 2 is a horizontal line and passes the vertical line test.It is a function. 15.x = 3 is a vertical line and does not pass the vertical line test.This is not a function. 17.First, isolate y. 7 2 213 6 212 7x y y xy xy x = + + = += Now, sketch the graph. Note that this line passes the vertical line test.Therefore, it is a function. 19.Yes, any nonvertical line is a function since it passes the vertical line test. 21.No, a circle is not the graph of a function since a vertical line may intersect the circle at more than one point. 23.Answers may vary. 25.Answers may vary.One example follows. Suppose that when x = 2, y = 1 and y = 4.Then suppose when x = 6, y = 0.Sketch these points. This relation is not a function since it does not pass the vertical line test. 27. The relationy x =is a function since it passes the vertical line test. 29.Consider the input x = 16.Substitute 16 for x and solve for y. 4162yy== Since the input x = 16 yields two outputs, the relation, 4y x = , is not a function. SSM: Intermediate AlgebraChapter 1 Review Exercises 23 31. Sketch the graph of 3y x = . Note that this graph passes the vertical line test.Therefore, 3y x =is a function. 33.No, the students conclusion is not correct.In a function, two inputs can yield the same output, but one input cannot yield two outputs. 35.Answers may vary. Chapter 1 Review Exercises 1.1.The length of the candle (L) is the dependent variable and minutes (t) is the independent variable. 2.The weight of a person (w) is the dependent variable and age in years (t) is the independent variable. 3.The amount of time (T) to cook a marshmallow is the dependent variable and the distance (d) from the campfire is the independent variable. 4.The monthly car insurance payment (M) is the dependent variable and the amount the car is worth (d) is the independent variable. 5.The number of points a student scores on an exam (p) is the dependent variable and the number of hours the student reviews (t) is the independent variable. 6.Answers may vary. 7.When x = 5, y = 4. 8. When x = 0, y is approximately 34 . 9.When y = 2, x = 4. 10.When y = 0, x = 1. 11. 7 4 312 5 3m= = = 12. ( ) 2 5 3 33 2 5 5m = = = 13. ( )( )7 34 19 1 8 2m = = = 14. 2.8 8.9 6.13.81251.6 3.2 1.6m = = = 15. ( )( )5 5003 2 1m = = = 16. 1 3 44 4 0m = = undefined 17.Use the slope, 2, and the y-intercept, 3, to graph the equation. Chapter 1 Review ExercisesSSM: Intermediate Algebra 24 18.Use the slope, 3, and the y-intercept, 10, to graph the equation. 19.First, isolate y. 2 02y xy x+ == Use the slope, 2, and the y-intercept, 0, to graph the equation. 20.7 y =is a horizontal line with a y-intercept of 7. 21.First, isolate y. 3 2 122 3 12362x yy xy x = = += Use the slope, 32, and the y-intercept, 6, to graph the equation. 22.First, isolate y. ( ) 2 2 1 72 4 1 74 2 8122x yx yy xy x = = = += Use the slope, 12, and the y-intercept, 2, to graph the equation. 23. Use the slope, 23 , and the y-intercept, 5, to graph the equation. SSM: Intermediate AlgebraChapter 1 Review Exercises 25 24.First, isolate y. 0.2 0.6 0.32 6 32 6 3332y xy xy xy x = == += + Use the slope, 3, and the y-intercept, 32, to graph the equation. 25.First, isolate y. ( ) 3 2 2 93 6 2 93 2 15253y xy xy xy x + = + = + = += Use the slope, 23 , and the y-intercept, 5, to graph the equation. 26. Use the slope, 10.254 = , and the y-intercept, 2, to graph the equation. 27.Solve the first equation for y to find the slope. 2 5 75 2 72 75 5x yy xy x+ == += + The slopes are 2 2 and 5 5 .They are reciprocals, but not negative reciprocals.The lines are neither parallel nor perpendicular. 28.Since x = 2 is a vertical line, and y = 4 is a horizontal line, the lines are perpendicular. 29.We are given the slope, m = 4, and a point on the line, (3, 7).Usey mx b = +to find b. ( ) 7 4 37 125y mx bbbb= += += + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. ( ) 4 54 5y xy x= += 30. We are given the slope, m = 23 , and a point on the line, (5, 4).Usey mx b = +to find b. ( )24 53104312 103 323y mx bbbbb= + = + = + = + = Now, substitute for m and b in slope-intercept form to obtain the equation of the line. 2 23 32 23 3y xy x = + = 31.First, find the slope. 8 5 333 4 1m= = = So, y = 3x + b.Since the line contains (3, 8), substitute 3 for x and 8 for y and solve for b. Chapter 1 Review ExercisesSSM: Intermediate Algebra 26 ( ) 8 3 38 917bbb= += += So, the equation is3 17 y x = + . 32.First, find the slope. 9 1 827 3 4m= = = So, y = 2x + b.Since the line contains (3, 1), substitute 3 for x and 1 for y and solve for b. ( ) 1 2 31 65bbb= += + = So, the equation is2 5 y x = . 33.First, find the slope. 3.7 1.1 2.60.722.8 6.4 3.6m= = = (rounded to the nearest hundredth) So, y = 0.72x + b.Since the line contains (2.8, 3.7), substitute 2.8 for x and 3.7 for y and solve for b. ( ) 3.7 0.72 2.83.7 2.0165.72bbb= += + = (rounded to the nearest hundredth) So, the equation is0.72 5.72 y x = . 34.First, find the slope. ( )( )6 282 3 5m= = So, y = 85 x + b.Since the line contains (2, 6), substitute 2 for x and 6 for y and solve for b. ( )86 25166530 165 5145bbbb= += += += So, the equation is 8 145 5y x = + . 35.First, find the slope. ( ) 6 2 8 44 2 6 3m= = = So, y = 43 x + b.Since the line contains(2, 2), substitute 2 for x and 2 for y and solve for b. ( )42 236 83 323bbb = + = += So, the equation is 4 23 3y x = + . 36.First, find the slope. ( ) 5 2 7 undefined3 3 0m= = A line with undefined slope is a vertical line of the form x = a.Since (3, 5) is a point on the line and the value of x is 3 at that point, the equation of the line is x = 3. 37.Line 1:Two points on line 1 are (1, 2) and(1, 2).Use these points to find the slope. ( ) 2 2421 1 2m= = = So, y = 2x + b.We can see from the graph that the y-intercept is 0.So, the equation of the line is y = 2x + 0 = 2x. Line 2:Two points on line 1 are (1, 0) and(0, 2).Use these points to find the slope. ( )2 0 220 1 1m= = = So, y = 2x + b.We can see from the graph that the y-intercept is 2.So, the equation of the line is y = 2x + 2. Line 3:Since line 3 is a horizontal line, it is of the form y = b.Since the y-intercept is 3, the equation of the line is y = 3. SSM: Intermediate AlgebraChapter 1 Review Exercises 27 38.Find the slope of the line to determine the rate of change. Points on a Line xy 220 316 412 58 64 72 39.Line 1 is parallel to0.5 6 y x = + .Since parallel lines have the same slope, the slope of line 1 is 0.5.The equation of line 1 is0.5 y x b = + .Line 1 has the same y-intercept as the line 1.5 3 y x = + .The y-intercept is b = 3.The equation of line 1 is0.5 3 y x = + . 40.To find the x-intercept, set y = 0 and solve for x. ( ) 3 5 0 x 173 17173xx=== To find the y-intercept, set x = 0 and solve for y. ( ) 3 0 5 175 17175yyx = == The x-intercept is 173.The y-intercept is 175 . 41.a.B, C and F b.C and E c.C d.A and D 42. Use the slope, 32 , and the point on the line, (3, 8), to find the y-intercept. ( )38 3298216 92 272bbbb= += += += The equation of the line is 3 72 2y x = + . 43.Since the line is parallel to the line3 6 y x = , it will have a slope of 3.Use the slope, 3, and the point on the line, (2, 5), to find the y-intercept. ( ) 5 3 25 611bbb= += += The equation of the line is3 11 y x = + . 44.A line with an infinite number of x-intercepts is the line that lies on the x-axis.This is the horizontal line y = 0. 45.Relations 1 and 3 could possibly be functions, since there is only one output for each input.Relations 2 and 4 could not possibly be functions, since there are two or more outputs for a single input. 46.Answers may vary.Any graph that passes the vertical line test is the graph of a function. 47.The graph of4 7 y x = is a nonvertical line.Any nonvertical line is a function since it passes the vertical line test. 48.First, isolate y. 5 6 36 5 35 16 2x yy xy x = = += The graph of 5 16 2y x = is a nonvertical line.Any nonvertical line is a function since it passes the vertical line test. 49.x = 9 is a vertical line.It does not pass the vertical line test and is not a function. 50.y = 1 is a horizontal line.It passes the vertical line test and is a function. Chapter 1 TestSSM: Intermediate Algebra 28 51. Sketch the graph of 2y x = . Note that this graph passes the vertical line test.Therefore, 2y x =is a function. 52. Sketch the graph of 2y x = . Note that this graph does not pass the vertical line test.Therefore, 2y x =is not a function. Chapter 1 Test 1.1.Answers may vary.One example is as follows. 2.Answers may vary. (Hint: See Section 1 for examples. 3.Line 1:Two points on line 1 are (4, 0) and(2, 5).Use these points to find the slope. 5 0 5 52 4 2 2m= = = So, y = 52 x + b.Use the slope, 52 , and a point on the line to find the y-intercept. ( )50 420 1010y mx bbbb= += += += The equation of line 1 is y = 52 x + 10. Line 2:Two points on line 1 are (0, 2) and(7, 4).Use these points to find the slope. 4 2 27 0 7m= = So, y = 27 x + b.We can see from the graph that the y-intercept is 2.So, the equation of the line is y = 27 x + 2. Line 3:Since line 3 is a vertical line, it is of the form x = a.Since the x-intercept is 3, the equation of the line is x = 3. 4.a.k m > , since the liney kx c = +is steeper than the liney mx b = + . b.b c > , since the y-intercept of the line y kx c = +is greater than the y-intercept of the liney mx b = + . 5. 85850.31270 270Am= = 1401400.29475 475Bm= = Ski run A is steeper than ski run B, since the slope of ski run A is greater than the slope of ski run B.(In determining which is steeper, ignore the sign of the slope by considering absolute value.) 6.Answers may vary. 7.The table is completed as follows. Points on a Line xy 425 529 633 737 841 945 8. Use the slope, 15 , and the y-intercept, 4, to graph the line. SSM: Intermediate Algebra Chapter 1 Test 29 9.First, isolate y. 2 3 93 2 9233x yy xy x = = += Use the slope, 23, and the y-intercept 3, to graph the line. 10. ( ) 2 8 10 53 5 8 4m= = = 11.First, find the slope. 8 6 2 22 5 3 3m= = = Use the slope and a point on the line, (2, 8), to find b. ( )28 2348324 43 3283y mx bbbbb= += += += += The equation of the line is 2 283 3y x = + .To find points on the line, substitute values for x to find y.For example, let x = 8. ( )2 28 16 28 128 43 3 3 3 3y = + = + = =The point, (8, 4) is on the line.Similar calculations will show that (11, 2) and (14, 0), among others, also lie on the line. 12.Use the slope, 3, and the point, (2, 5), to find the y-intercept, b. ( ) 5 3 25 61y mx bbbb= += += + = The equation of the line is3 1 y x = . 13.First, find the slope. ( ) 7 5 123 2 5m= = Use the slope and a point on the line, (2, 5), to find b. ( )125 2525 245 515y mx bbbb= + = + = + = The equation of the line is 12 15 5y x = . 14.All of the points except (0, 2) lie on a line. Choose two of the points to find the slope. ( ) 5 3822 2 4m= = = Use the slope and a point on the line to find the y-intercept, b. ( ) 5 2 25 41y mx bbbb= += += += The equation of the line is2 1 y x = + . 15.First, find the slope of the line by isolating y. 3 5 205 3 20345x yy xy x = = += Chapter 1 TestSSM: Intermediate Algebra 30 The slope of the line is 35.The slope of a line perpendicular to this line is 53 .Use the slope, 53 , and a point on the line, (4, 1), to find the y-intercept. ( )51 4320133 203 3173y mx bbbbb= + = + = + = += The equation of the line is 5 173 3y x = + . 16.To find the x-intercept, set y = 0 and solve for x. ( ) ( ) 2 0 5 4 1 30 5 4 4 36 46432xxxxx+ = ++ = +=== To find the y-intercept, set x = 0 and solve for y. ( ) 2 5 4 0 1 32 5 4 32 63yyyy+ = ++ = += = The x-intercept is 32.The y-intercept is3 . 17.Answers may vary.To show that the relation is a function, show that for each input, there is only one output. 18.Answers may vary.Any graph that does not pass the vertical line test is not a function. 19. Sketch the graph ofy x = . Note that this graph does not pass the vertical line test.Therefore,y x = is not a function. 20.The graph of2 5 y x = +is a nonvertical line.Any nonvertical line is a function since it passes the vertical line test.