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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,-2). This means that the value of b is -2.Finding the Slope To find the slope, we will trace along the line on the given graph until we find a lattice point, 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 used the given (1,3) as our other point. Traveling to this point from the y-intercept requires that we move 1 step horizontally in the positive direction and 5 steps vertically in the positive direction. runrise​=15​⇔m=5​Writing the Equation Now that we have the slope and the y-intercept, we can form our final equation. y=mx+by=5x+(-2)y=5x−2​
Exercises 2 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 (0,5). This means that the value of b is 5.Finding the Slope To find the slope, we will trace along the line on the given graph until we find a lattice point, 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 used the given (3,4) as our other point. Traveling to this point from the y-intercept requires that we move 3 steps horizontally in the positive direction and 1 step vertically in the negative direction. runrise​=3-1​⇔m=-31​​Writing the Equation Now that we have the slope and the y-intercept, we can form our final equation. y=mx+by=-31​x+5​
Exercises 3 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 (0,0). This means that the value of b is 0.Finding the Slope To find the slope, we will trace along the line on the given graph until we find a lattice point, 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 used the given (-2,4) as our other point. Traveling to this point from the y-intercept requires that we move 2 steps horizontally in the negative direction and 4 steps vertically in the positive direction. runrise​=-24​⇔m=-2​Writing the Equation Now that we have the slope and the y-intercept, we can form our final equation. y=mx+by=-2x+0y=-2x​
Exercises 4 We can write the equation of the line that connects the points using the point-slope form, y−y1​=m(x−x1​), where m is the slope and (x1​,y1​) is any point of the line. To write the equation, we first need to find the slope. To do this, we will use the slope formula. m=x2​−x1​y2​−y1​​Substitute (-2,5) & (1,-1)m=1−(-2)-1−5​ Calculate m a−(-b)=a+bm=1+2-1−5​Add and subtract termsm=3-6​Calculate quotient m=-2 Therefore, we have that the slope of the line is -2. We can now substitute m=-2 and either point into the point-slope form to write the equation. We will use the point (-2,5). y−5=-2(x−(-2))⇒y−5=-2(x+2) Notice that if we used the other point, (1,-1), we would get an equivalent equation that describes the same line. y−(-1)=-2(x−1)⇒y+1=-2(x−1)
Exercises 5 Using the given points, (-3,-2) and (2,-1), we are asked to write an equation in point-slope form. We will follow two steps.We will first find the slope of the equation by using the slope formula. Next, we will write the equation in point-slope form.Finding the Slope We know that the line passes through the points (-3,-2) and (2,-1). Let's substitute these points into the slope formula and find the slope. m=x2​−x1​y2​−y1​​Substitute (-3,-2) & (2,-1)m=2−(-3)-1−(-2)​a−(-b)=a+bm=51​ Thus, the slope of the line is 51​.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-3,-2) and substitute it in the point-slope formula. y−y1​=m(x−x1​)Substitute valuesy−(-2)=51​(x−(-3))a−(-b)=a+by+2=51​(x+3) Using the point (-3,-2), we found one equation for the line in point-slope form. y+2=51​(x+3)​ Please note that there are infinitely many solutions when asked to write an equation in point-slope form. Any point that lies on the line would work as (x1​,y2​).
Exercises 6 We are asked to write an equation in point-slope form, y−y1​=m(x−x1​), where m is the slope of the line and (x1​,y1​) is a point on the line. To write the equation, we'll first need to determine the slope using the slope formula. m=x2​−x1​y2​−y1​​Substitute (1,0) & (4,4)m=4−14−0​Subtract termsm=34​ The slope of the line is 34​. We can now substitute m=34​ and either point into the point-slope form to write the equation. We will use the point (1,0). y−0=34​(x−1)⇒y=34​(x−1) Note that if we used the other point, (4,4), we would get a different, but equivalent, equation that describes the same line. y−4=34​(x−4)
Exercises 7 Let's start by expressing f(0)=2 as (0,2), and f(5)=-3 as (5,-3). Remember the slope-intercept form of a line is y=mx+b, which is the same as writing f(x)=mx+b, where m is the slope and b the y-intercept. To write this equation, we must find the slope and the y-intercept. We will begin by finding the slope of the line that connects the given points, (0,2) and (5,-3), using the slope formula. m=x2​−x1​y2​−y1​​Substitute (0,2) & (5,-3)m=5−0-3−2​ Calculate m Subtract termsm=5-5​Calculate quotient m=-1 Therefore, the slope of the line is -1. Because the line crosses the y-axis at (0,2), the y-intercept b equals 2. Hence, the equation of the line written in slope-intercept form is f(x)=-1x+2⇒f(x)=-x+2.
Exercises 8 We have been given the following information about the function f. f(-1)=-6 and f(4)=-6 We can rewrite these equations as the ordered pairs (-1,-6) and (4,-6) because function notation tells us about the relationship f(x)=y. We can use slope-intercept form, y=mx+b⇔f(x)=mx+b, to write an equation for the line. Notice that our equation should be written in function notation to match the equations given in the exercise. We will find the slope of the line that connects the given points using the slope formula. m=x2​−x1​y2​−y1​​Substitute (-1,-6) & (4,-6)m=4−(-1)-6−(-6)​a−(-b)=a+bm=4+1-6+6​Add termsm=50​Calculate quotientm=0 The slope of the line is 0. This means that the line will not move vertically. At this point, we have the following equation. f(x)=0⋅x+b To determine the value of b, we can substitute either of the given points into the equation for x and f(x), then solve. We will use (-1,-6). f(x)=0⋅x+bx=-1, f(x)=-6-6=0⋅(-1)+bZero Property of Multiplication-6=0+bAdd terms-6=bRearrange equationb=-6 The y-intercept is (0,-6). We can now write the complete equation as f(x)=0⋅x−6. Any time we have a coefficient of 0 we can get rid of the variable using the Zero Property of Multiplication. This means that the final equation is: f(x)=-6.
Exercises 9 Using the given information, we will find two points on the graph of f. The equation of the line through these two points in slope-intercept form gives the linear function. However, we can't determine the y-intercept of the equation from the given points. Therefore, we will go through the following process.We will first find two points on the graph of f. Next, we find the slope of the line through these points by using the slope formula. Then, we will write the equation in point-slope form. Finally, we will rearrange the equation to write it in slope-intercept form.Points on the Graph We can think of function notation as f(x)=y. Therefore, the value of the function gives the y-coordinate of the point on the graph.xf(x)point on the graph -3f(-3)=-2(-3,-2) -2f(-2)=3(-2,3) Finding the Slope We know that the line passes through the points (-3,-2) and (-2,3). Let's substitute these points into the slope formula and find the slope. m=x2​−x1​y2​−y1​​Substitute (-3,-2) & (-2,3)m=-2−(-3)3−(-2)​a−(-b)=a+bm=15​1a​=am=5 Thus, the slope of the line is 5.Point-Slope Form We know the slope of the line and two points that are on the line. We can choose one of these points and write the equation of the line. Let's choose the point (-3,-2). y−(-2)=5(x−(-3))​Slope-Intercept Form We can write the equation of the line in slope-intercept form by isolating the y-variable. y−(-2)=5(x−(-3))a−(-b)=a+by+2=5(x+3)Distribute 5y+2=5x+15LHS−2=RHS−2y=5x+13 Now that the y-variable is isolated, our equation is in slope-intercept form. y=5x+13​
Exercises 10 Remember that two lines are parallel if they have the same slope and two lines are perpendicular if their slopes are negative reciprocals. Therefore, to determine if the lines are parallel or perpendicular, we need to find their slopes.Line a We can find the slope of the line by substituting the given points into the slope-formula. m=x2​−x1​y2​−y1​​Substitute (-2,2) & (2,1)m=2−(-2)1−2​ Simplify RHS a−(-b)=a+bm=2+21−2​Add and subtract termsm=4-1​Put minus sign in front of fraction m=-41​ The slope of Line a is -41​.Line b In the same way, we can calculate the slope of Line b. m=x2​−x1​y2​−y1​​Substitute (1,-8) & (3,0)m=3−10−(-8)​ Simplify RHS a−(-b)=a+bm=3−10+8​Add and subtract termsm=28​Calculate quotient m=4 Therefore, the slope of this line is 4.Line c Let's now calculate the slope of the third line. m=x2​−x1​y2​−y1​​Substitute (-4,-3) & (0,-2)m=0−(-4)-2−(-3)​ Simplify RHS a−(-b)=a+bm=0+4-2+3​Add terms m=41​ This time, the slope is 41​.Comparing the slopes We found three different slopes for the three different lines. Line a:Line b:Line c:​m=-41​m=-4m=-41​​ Line a has a slope of -41​ which is the negative reciprocal of 4. This means it is perpendicular to line b. Additionally, each slope is unique so there are no parallel lines.
Exercises 11 Lines are parallel if they have the same slope and are perpendicular if their slopes are negative reciprocals. Let's start by finding the slope of each line by rewriting the equations in slope-intercept form, y=mx+b, where m is slope and b is the y-intercept. Since the second equation is already in slope-intercept form, we need to rewrite only lines a and c.Line a We can rewrite the equation using inverse operations. 2x+6y=-12 Solve for y LHS−2x=RHS−2x6y=-2x−12LHS/6=RHS/6y=6-2x−12​Write as a difference of fractionsy=6-2x​−612​ca⋅b​=ca​⋅by=6-2​x−612​Put minus sign in front of fractiony=-62​x−612​Calculate quotient y=-31​x−2 The slope is -31​.Line c Let's rewrite this equation in a similar way. 3x−2y=-4 Solve for y LHS+2y=RHS+2y3x=2y−4LHS+4=RHS+43x+4=2yLHS/2=RHS/223x+4​=yWrite as a sum of fractions23x​+24​=yCalculate quotient23x​+2=yca⋅b​=ca​⋅b23​x+2=yRearrange equation y=23​x+2 The slope is 23​.Comparing slopes We found three slopes. Line a:Line b:Line c:​m=-31​m= 23​m= 23​​ The slopes of line b and c are equal, so these lines are parallel. However, the slope of the line a is not equal to the slope of line b or c, so we know it is not parallel to lines b and c. To find out if a is perpendicular to b and c, we check if the slopes are negative reciprocals. ma​⋅mb,c​=?-1ma​=-31​, mb,c​=23​-31​⋅23​=?-1Multiply fractions-3⋅21⋅3​=?-1Multiply-63​=?-1ba​=b/3a/3​-21​≠-1 Line a is neither parallel nor perpendicular to line b and line c.
Exercises 12 aTo write the equation of a line, we can use the slope-intercept form, y=mx+b, where m is the slope and b the y-intercept. Parallel lines always have the same slope. Therefore, in order to find the slope of the asked line we need to find the slope of the given line. Looking at the graph, we can see that the given line passes through the points (1,-1) and (2,2). To find the slope we will use the slope formula. m=x2​−x1​y2​−y1​​Substitute (1,-1) & (2,2)m=2−12−(-1)​ Calculate m a−(-b)=a+bm=2−12+1​Add and subtract termsm=13​Calculate quotient m=3 The slope of the given line and, consequently, of the parallel line is 3. Because the parallel line passes through the point (6,2), to find the y-intercept, we substitute x for 6 and y for 2. y=3x+bx=6, y=22=3⋅6+b Solve for b Multiply2=18+bLHS−18=RHS−18-16=bRearrange equation b=-16 Hence, equation of the parallel line is: y=3x−16.bNow we want the lines to be perpendicular. This means their slopes have to be negative reciprocals. From part A, we know that the slope of the given line is 3. To find the slope of the perpendicular line, we can use the following formula. m1​⋅m2​=-1 We will use the slope 3 for m2​ and solve for m1​. m1​⋅m2​=-1m2​=3m1​⋅3=-1 Solve for m1​ LHS/3=RHS/3m1​=3-1​Put minus sign in front of fraction m1​=-31​ For the lines to be perpendicular, the slope has to equal -31​. Let's substitute the known point in the slope-intercept formula. y=-31​x+bx=6, y=22=-31​⋅6+b Solve for b ca​⋅b=ca⋅b​2=-31⋅6​+bMultiply2=-36​+bCalculate quotient2=-2+bLHS+2=RHS+24=bRearrange equation b=4 Therefore, the equation of the perpendicular line is: y=-31​x+4.
Exercises 13 aThe slopes of parallel lines are the same. Therefore, to find the slope of the parallel line, we need to know the slope of the given line. Let's calculate it by substituting known points into the slope formula. For this exercise, we can see the y-intercept, (0,2), and a point through which the line passes, (4,0). m=x2​−x1​y2​−y1​​Substitute (4,0) & (0,2)m=4−00−2​ Solve for m Subtract termsm=4-2​Put minus sign in front of fractionm=-42​ba​=b/2a/2​ m=-21​ The slope of the given line and any line parallel to it is -21​. By substituting the slope and the given point (-2,-3) into the slope-intercept form, we can find the equation of the parallel line. y=mx+bSubstitute values-3=-21​(-2)+b Solve for b -a(-b)=a⋅b-3=21​⋅2+b2a​⋅2=a-3=1+bLHS−1=RHS−1-4=bRearrange equation b=-4 The equation of the parallel line is: y=-21​x−4.bWhen lines are perpendicular, their slopes are negative reciprocals. From Part A, we know that the slope of the given line is m=-21​. By substituting this into the formula m1​⋅m2​=-1 we can find the perpendicular slope. m1​⋅m2​=-1m2​=-21​m1​(-21​)=-1 Solve for m1​ LHS⋅-2=RHS⋅-2m1​(-21​)(-2)=-1(-2)-a(-b)=a⋅bm1​⋅21​⋅2=1⋅2Multiplym1​⋅1=2a⋅1=a m1​=2 For the lines to be perpendicular, the slope has to equal 2. Let's substitute the given point (-2,-3) and perpendicular slope into the slope-intercept form to find the y-intercept. y=mx+bSubstitute values-3=2(-2)+b Solve for b a(-b)=-a⋅b-3=-2⋅2+bMultiply-3=-4+bLHS+4=RHS+41=bRearrange equation b=1 Now that we have the slope and the y-intercept, we can write the equation of the perpendicular line. y=2x+1
Exercises 14 aFirst, we can trace along the given line until we find two points that lie perfectly on grid lines. The slopes of parallel lines are the same. Therefore, to find the slope of a parallel line, we need to know the slope of the given line. Let's calculate it by substituting the points that lie on the given line into the slope formula. m=x2​−x1​y2​−y1​​Substitute (-3,3) & (-2,-1)m=-2−(-3)-1−3​a−(-b)=a+bm=-2+3-1−3​Add and subtract termsm=1-4​1a​=am=-4 The slope of the given line and, consequently, of the parallel line is -4. Next, by substituting the slope and the given point, (-4,0), into the slope-intercept form, we can find the y-intercept. y=mx+bSubstitute values0=-4(-4)+b Solve for b -a(-b)=a⋅b0=16+bLHS−16=RHS−16-16=bRearrange equation b=-16 We can now use the slope and y-intercept to write the equation of the parallel line. y=-4x−16​bNow we want to find the equation of a perpendicular line that passes through the given point. When two lines are perpendicular, their slopes are negative reciprocals. This means that the product of the slopes must be -1. m1​⋅m2​=-1​ From Part A, we know that the given line's slope is m=-4. We can substitute this into the above equation to look for the slope of the perpendicular line. m1​⋅m2​=-1m1​=-4(-4)⋅m2​=-1LHS/(-4)=RHS/(-4)m2​=-4-1​-b-a​=ba​m2​=41​ The slope of the perpendicular line is 41​. Once again, by substituting the slope and the given point, (-4,0), into the slope-intercept form, we can find the y-intercept. y=mx+bSubstitute values0=41​(-4)+b Solve for b a(-b)=-a⋅b0=-1+bLHS+1=RHS+11=bRearrange equation b=1 We can now use the slope and y-intercept to write the equation of the perpendicular line. y=41​x+1​
Exercises 15 aLet's assume x is the number of months and y is the total cost. Since the total cost changes at a constant rate with respect to the number of months, the equation given below will help us to write a linear equation. y=mx+b​ The value of m is the constant rate of change and b is the initial or starting value of y. According to the given information, the constant rate of change is \$44 and the initial value is \$48. Thus, we can write our equation. y=44x+48​bIf we want to set up a website and maintain it for 6 months, we can find the total cost by substituting 6 for x into the equation that we wrote in Part A. Let's do it! y=44x+48x=6y=44(6)+4Multiplyy=264+48Add termsy=312 As a result, the total cost is \$312.cIn this part, we have another information about a different company. This company charges \$62 per month to maintain a website but there is no initial fee. Let's write an equation of the total cost for this company. y=62x​ If we have \$620, let's determine at which company we can set up and maintain a website for the greatest amount of time. We will start with the first company. y=44x+48y=620620=44x+48LHS−48=RHS−48572=44xLHS/44=RHS/4413=xRearrange equationx=13 For the first company, we can maintain a website for 13 months. Next, we will find number of months for the second company. y=62xy=620620=62xLHS/62=RHS/6210=xRearrange equationx=10 Thus, 10 months is maximum time for us to maintain a website at second company if we have \$620. Therefore, the first company would be a better option for us.
Exercises 16 Let's find the rate of change for the consecutive data pairs in the table.Rate of change between...x2​−x1​y2​−y1​​m Column 5 & Column 416−14135−140​-2.5 Column 4 & Column 314−12140−145​-2.5 Column 3 & Column 212−10145−150​-2.5 Column 2 & Column 110−8150−155​-2.5Since the rate of change is constant, the data in the table can be modeled by a linear equation with a slope of -2.5. Using one of the points, let's say (8,155), we can write a linear model in point-slope form. y−155=-2.5(x−8)​ In this model, x is time in minutes and y is the amount of water in the tank in gallons. Since we are asked to express the amount of water as a function of time, we need to solve this equation for y. y−155=-2.5(x−8)Distribute -2.5y−155=-2.5x+20LHS+155=RHS+155y=-2.5x+175 Thus, after x minutes there are y=-2.5x+175 gallons of water in the tank.