Graphing Linear Equations in Slope-Intercept Form

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Exercises 1 "Rise over run" is a colloquial way to express the mathematical formula: x2​−x1​y2​−y1​​, where "rise" is the change of the y values between two points and "run" is the change in the x values between the same two points. This is the formula for the slope of a linear function.
Exercises 2 The word constant in a non-math context just means that something continues on the same as it always has and always will. A constant function is one with the same output value for the entirety of the function's existence. The slope of a constant function, because it always produces the same value, will always be 0.The above graph is an example of a constant function, y=3. No matter what the input value is, the output is constantly 3.
Exercises 3 The slope-intercept form of a linear equation is: y=mx+b, where m is the slope and b is the y-intercept. It is called the slope-intercept form because the non-variable values, not x or y, are the slope and the y-intercept.
Exercises 4 Two forms of linear equations that we know so far are the standard form, Ax+By=C, and the slope-intercept form, y=mx+b. Let's look at each equation and identify in which form they are written. ​y=-5x−12x−y=8y=x+4y=-3x+13​​⇔⇔⇔⇔​​​slope-intercept formstandard formslope-intercept formslope-intercept form​ The second equation, 2x−y=8, is written in standard form whereas the other three equation are written in slope-intercept form. It is the one that doesn't belong.
Exercises 5 Observing the given graph, we can see that the line passes through the points (-3,1) and (2,-2), creating a negative slope. One way to use a graph to find the slope of a line is to count out the change in x and y.We can see that as the graph travels from left to right, the rise, or change in y, is -3. Similarly, the run, or change in x, is 5. slope=runrise​⇔m=5-3​=-53​​ We can confirm this answer by using the slope formula. m=x2​−x1​y2​−y1​​​ In the above equation, (x1​,y1​) and (x2​,y2​) represent two points on the line. Let's substitute them by (-3,1) and (2,-2). m=x2​−x1​y2​−y1​​Substitute (-3,1) & (2,-2)m=2−(-3)-2−1​a−(-b)=a+bm=2+3-2−1​Add and subtract termsm=5-3​Put minus sign in front of fractionm=-53​ We obtained the same slope as before. We can be sure the slope is -53​.
Exercises 6 Observing the given graph, we can see that the line passes through the points (1,-1) and (4,3) and has a positive slope. One way to use a graph to find the slope of a line is to count out the change in x and y.We can see that as the graph travels from left to right, the rise, or change in y, is 4. Similarly, the run, or change in x, is 3. slope=runrise​⇔m=34​​ We can confirm this answer by using the other method, the slope formula. m=x2​−x1​y2​−y1​​​ We'll use the given points to solve for the slope. m=x2​−x1​y2​−y1​​Substitute (1,-1) & (4,3)m=4−13−(-1)​a−(-b)=a+bm=4−13+1​Add and subtract termsm=34​ A slope of m=34​ means that for every 3 steps in the positive horizontal direction, the graph travels 4 steps in the positive vertical direction.
Exercises 7 Observing the given graph, we can see that the line passes through the points (-2,-3) and (2,-3). One way to use a graph to find the slope of a line is to count out the change in x and y.We can see that as the graph travels from left to right, the rise, or change in y, is 0. Whereas, the run, or change in x, is 4. slope=runrise​⇔m=40​=0​ We can confirm this answer by using the other method, the slope formula. m=x2​−x1​y2​−y1​​​ We'll use the given points to solve for the slope. m=x2​−x1​y2​−y1​​Substitute (-2,-3) & (2,-3)m=2−(-2)-3−(-3)​a−(-b)=a+bm=2+2-3+3​Add and subtract termsm=40​Calculate quotientm=0 A slope of m=0 means that the graph is a horizontal line.
Exercises 8 Observing the given graph, we can see that the line passes through the points (0,3) and (5,-1) and has a negative slope. One way to use a graph to find the slope of a line is to count out the change in x and y.We can see that as the graph travels from left to right, the rise, or change in y, is -4. Similarly, the run, or change in x, is 5. slope=runrise​⇔m=5-4​​ We can confirm this answer by using the other method, the slope formula. m=x2​−x1​y2​−y1​​​ We'll use the given points to solve for the slope. m=x2​−x1​y2​−y1​​Substitute (0,3) & (5,-1)m=5−0-1−3​Add and subtract termsm=5-4​Put minus sign in front of fractionm=-54​ A slope of m=-54​ means that for every 5 steps in the positive horizontal direction, the graph travels 4 steps in the negative vertical direction.
Exercises 9 Looking at the table, we can identify 4 ordered pairs. (-9,-2),(-5,0),(-1,2),(3,4)​ To calculate the slope of the line, we need only two of these. Let's find the slope by substituting (-9,-2) and (-5,0) into the slope formula. m=x2​−x1​y2​−y1​​Substitute (-9,-2) & (-5,0)m=-5−(-9)0−(-2)​a−(-b)=a+bm=-5+90+2​Add termsm=42​ba​=b/2a/2​m=21​
Exercises 10 Looking at the table, we can identify 4 ordered pairs. (-1,-6),(2,-6),(5,-6),(8,-6)​ Note that there is no change in y. We can choose any two points from the table and use the Slope Formula. m=x2​−x1​y2​−y1​​​ We will substitute the points (-1,-6) and (2,-6) for (x1​,y1​) and (x2​,y2​), respectively, to find the slope m. m=x2​−x1​y2​−y1​​Substitute (-1,-6) & (2,-6)m=2−(-1)-6−(-6)​a−(-b)=a+bm=2+1-6+6​Add termsm=30​Calculate quotientm=0
Exercises 11 From the table, we notice that there is no change in x. This means the line is vertical and passes through x=0. 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 use (0,-4) and (0,0). m=x2​−x1​y2​−y1​​Substitute (0,-4) & (0,0)m=0−0-4−0​Subtract termm=0-4​ Division by 0 is undefined, so the slope of the line is undefined.
Exercises 12 Looking at the table, we can identify 4 ordered pairs. (-4,2),(-3,-5),(-2,-12),(-1,-19)​ To calculate the slope of the line, we need only two of these. Let's find the slope by substituting (-4,2) and (-3,-5) into the slope formula. m=x2​−x1​y2​−y1​​Substitute (-4,2) & (-3,-5)m=-3−(-4)-5−2​a−(-b)=a+bm=-3+4-5−2​Add and subtract termsm=1-7​Calculate quotientm=-7
Exercises 13 The first thing we notice is that the line passes through origin, which means its y-intercept is 0. y=mx+0⇔y=mx​ To find the slope we can substitute either of the points in the function and solve for m. However, since one of our points is (1,60), we can determine the slope by just looking at the y-coordinate. Because the x-coordinate is 1 step to the right of the origin, and the y-coordinate is 60 steps up from the origin, we can deduce that the slope is 60.We can now write the equation of the line. y=60x​ We see that the slope is 60. This means that in hour the bus travels 60 miles. Therefore, the speed is 60 mph.
Exercises 14 Let's begin by plotting the 3 ordered pairs that we can identify from the table. Then, we can connect them with a straight line to see the graph of the function.The three points form a horizontal line. There is no change in y, and therefore the slope is 0. This means that no matter how much time you want to spend in the park, you will always pay $54.99 admission. In other words, the fee is fixed.
Exercises 15 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ The given equation is already in this form. Below we have highlighted the slope m and y-intercept b. y=-3x+2​ We can see that the slope is -3 and the y-intercept is 2.
Exercises 16 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ The given equation is already in this form. Below we have highlighted the slope m and y-intercept b. y=4x+(-7)​ We can see that the slope is 4 and the y-intercept is -7.
Exercises 17 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ The given equation is already in this form. Below we have highlighted the slope m and y-intercept b. y=6x+0​ We can see that the slope is 6 and the y-intercept is 0.
Exercises 18 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here, m represents the slope and b the y-intercept of the line. Let's rewrite our equation a little bit so that it more closely resembles this format. This will make it easier to identify its features. y=-1⇔y=0x+(-1)​ We see above that the slope is 0 and the y-intercept is -1.
Exercises 19 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ Let's rewrite our equation a little bit so that it more closely resembles this format. This will make it easier to identify the features of the equation. -2x+y=4LHS+2x=RHS+2xy=2x+4 Below we have highlighted the slope m and y-intercept b. y=2x+4​ We can see that the slope is 2 and the y-intercept is 4.
Exercises 20 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ Let's rewrite our equation a little bit so that it more closely resembles this format. This will make it easier to identify the features of the equation. x+y=-6LHS−x=RHS−xy=-x−6 Below we have highlighted the slope m and y-intercept b. y=-1x+(-6)​ We can see that the slope is -1 and the y-intercept is -6.
Exercises 21 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ Let's rewrite our equation a little bit so that it more closely resembles this format. This will make it easier to identify the features of the equation. -5x=8−yLHS+y=RHS+yy−5x=8LHS+5x=RHS+5xy=5x+8 Below we have highlighted the slope m and y-intercept b. y=5x+8​ We can see that the slope is 5 and the y-intercept is 8.
Exercises 22 Equations in slope-intercept form follow a specific format, where m represents the slope of the line and b is the y-intercept. y=mx+b​ Let's rewrite our equation a little bit so that it more closely resembles this format. This will make it easier to identify the features of the equation. 0=1−2y+14xLHS+2y=RHS+2y2y=1+14xLHS/2=RHS/2y=21+14x​Calculate quotienty=7x+21​ Below we have highlighted the slope m and y-intercept b. y=7x+21​​ We can see that the slope is 7 and the y-intercept is 21​.
Exercises 23 We notice that this equation is solved for x and not y. This means it is not in slope-intercept form. Let's isolate y and then we can identify the slope and y-intercept of the graph. x=-4yChange signs-x=4yLHS/4=RHS/4-41​x=yRearrange equationy=-41​x Now, we can see that the slope of the line is -41​. Also since there is no constant, we can tell that the function intercepts the y-axis at the origin when y=0.
Exercises 24 When a function is in the slope-intercept form: y=mx+b, m is the slope and b is the y-intercept. Let's rewrite our equation so it matches the slope-intercept form. y=3x−6⇔y=3x+(-6) Now we can identify the y-intercept as -6 and the slope as 3. This means we were wrong when identifying the y-intercept.
Exercises 25 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here, m represents the slope of the line and b represents the y-intercept. We can highlight the slope and the y-intercept in our equation so that it will be easier to draw the graph. y=-x+7⇔y=-1x+7​ The slope is -1 and the y-intercept is 7.Graphing the Equation A slope of -1 means that for every unit we move in the positive horizontal direction, we move 1 unit in the negative vertical direction. -1=1-1​=runrise​​ To graph the equation, plot the y-intercept and then use the slope to find another point on the line. Then, connect the two points with a straight 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 (7,0), so the x-intercept is 7.
Exercises 26 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here m represents the slope of the line and b represents the y-intercept. We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=21​x+3​ The slope is 21​ and the y-intercept is 3.Graphing the Equation A slope of 21​ means that for every 2 units we move in the positive horizontal direction, we move 1 unit in the positive vertical direction. m=runrise​=21​​ 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 (-6,0), so the x-intercept is -6.
Exercises 27 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here m represents the slope of the line and b represents the y-intercept. We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=2x+0​ The slope is 2 and the y-intercept is 0.Graphing the Equation A slope of 2 means that for every 1 unit we move in the positive horizontal direction, we move 2 units in the positive vertical direction. m=runrise​=12​​ 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 (0,0), so the x-intercept is 0.
Exercises 28 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here m represents the slope of the line and b represents the y-intercept. We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=-1x+0​ The slope is -1 and the y-intercept is 0.Graphing the Equation A slope of -1 means that for every 1 unit we move in the positive horizontal direction, we move 1 unit in the negative vertical direction. m=runrise​=1-1​​ 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 (0,0), so the x-intercept is 0.
Exercises 29 Equations written in slope-intercept form follow a specific format. y=mx+b​ In this format, 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. 3x+y=-1LHS−3x=RHS−3xy=-3x−1 We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=-3x − 1⇔y=-3x+(-1)​ The slope is -3 and the y-intercept is -1.Graphing the Equation A slope of -3 means that for every 1 unit we move in the negative horizontal direction, we move 3 units in the positive vertical direction. m=runrise​=-13​​ 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 (-31​,0), so the x-intercept is -31​.
Exercises 30 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here 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+4y=8LHS−x=RHS−x4y=-x+8LHS/4=RHS/4y=4-x+8​Calculate quotienty=-41​x+2 We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=-41​x+2​ The slope is -41​ and the y-intercept is 2.Graphing the Equation A slope of -41​ means that for every 4 units we move in the positive horizontal direction, we move 1 unit in the negative vertical direction. m=runrise​=4-1​​ 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 (8,0), so the x-intercept is 8.
Exercises 31 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here 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. -y+5x=0LHS+y=RHS+y5x=yRearrange equationy=5x We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=5x+0​ The slope is 5 and the y-intercept is 0.Graphing the Equation A slope of 5 means that for every 1 unit we move in the positive horizontal direction, we move 5 units in the positive vertical direction. m=runrise​=15​​ 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 (0,0), so the x-intercept is 0.
Exercises 32 Equations written in slope-intercept form follow a specific format. y=mx+b​ Here 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. 2x−y+6=0LHS+y=RHS+y2x+6=yRearrange equationy=2x+6 We can highlight the slope m and y-intercept b of our equation so that it will be easier to create the graph. y=2x +6​ The slope is 2 and the y-intercept is 6.Graphing the Equation A slope of 2 means that for every 1 unit we move in the positive horizontal direction, we move 2 units in the positive vertical direction. m=runrise​=12​​ 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 (-3,0), so the x-intercept is -3.
Exercises 33 We will use the given information to determine some features of the function, starting with the slope.Slope Consider what slope represents- it's 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 4 units every time the independent variable x increases by 2 units. slope=2-4​⇔slope=-12​​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 know that the value of the function at 0 is -2. f(0)=-2​ Therefore the y-intercept is -2, and 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 (-1,0), so the x-intercept is -1.
Exercises 34 We will use the given information to determine some features of the function, starting with the slope.Slope The slope of a line is the quotient between the change in the dependent variable, usually the y-variable, and the independent variable, usually the x-variable. slope=change in xchange in y​​ We are told that the dependent variable increases by 1 unit every time the independent variable decreases by 5 units. slope=-51​⇔slope=-51​​y-intercept Think of the point where a line crosses the y-axis. The x-coordinate of that point is 0, and the y-coordinate is the y-intercept. We know that the value of the function at 0 is 3. Therefore the line intercepts the y-axis at (0,3), and thus the y-intercept is 3.Graph We now have enough information to graph the function. We will start by plotting the y-intercept. Then, we will use the slope to find another point. By connecting these points with a straight line, we will form the graph of our equation.x-intercept We can identify the x-intercept by looking at the graph.The line intercepts the x-axis at (15,0). Therefore, the x-intercept is 15.
Exercises 35 If our fingernails grow at a rate of 0.7 millimeters each week, then the rate of change can be expressed as: 1 week0.7 ml​⇒m=0.7=107​. To graph the function, we also need the y-intercept. From the equation r(0)=12, we can identify the y-intercept as (0,12). Now we have enough information to graph the function.Notice that our domain is being restricted to the first quadrant because we cannot go back in time or have a negative growth of fingernails.Interpretation of y-intercept As we previously found, the y-intercept is (0,12). We can understand what this means by thinking about what x and r(x) mean. The input x on our graph represents the number of weeks that have gone by. The output r(x) represents the length of our fingernails as a function of the number of weeks. Therefore, we have: x=0r(x)=12​⇒At the beginning of time,⇒our fingernail was 12 ml long.​
Exercises 36 If the amount of milk sold on the farm decreases by 500 gallons for each increase of $1 in price, then the rate of change can be expressed as: 1 dollar-500 gallons​⇒m=-500. To graph the function, we also need the y-intercept. From the equation m(0)=3000, we can identify the y-intercept as (0,3000). Now we have enough information to graph the function.Notice that our domain is being restricted to the first quadrant because we cannot sell a negative number of gallons of milk and we cannot sell things for a negative amount of money.Interpretation of y-intercept As we previously found, the y-intercept is (0,3000). We can understand what this means by thinking about what x and m(x) mean. The input x on our graph represents the cost of each gallon of milk. The output m(x) represents the number of gallons of milk sold as a function of the cost. Therefore, we have: x=0m(x)=3000​⇒If the milk is free,⇒3000 gallons would be taken.​Interpretation of x-intercept We can interpret the x-intercept, (6,0), in a similar manner. x=6m(x)=0​⇒If the milk costs $6 per gallon,⇒0 gallons would be purchased.​
Exercises 37 aLet's start with determining the slope and the d-intercept of the given function to graph it. d(t)=21​t+6​ We can notice that the given function is in slope-intercept form. y=mx+b​ In this form, m tells us the slope and b states the y-intercept. However, for the given equation, b stands for the d-intercept. m=21​b=6​ Now that we know the slope and d-intercept of the function, our first step is to plot the d-intercept and find a second point using the slope. Notice that the d-axis represents the depth and t-axis represents the time.Next, we will draw the line that represents the function. However, we have two restrictions. Since the time cannot be negative, t will be greater than or equal to 0. Moreover, because the function models the first 9 hours of the storm, t will be less than or equal to 9.Now, we are able to determine the domain and range of the function. In order to do that, we will draw vertical and horizontal lines at the endpoints of the line segment to create a rectangle view of the domain and range.The intersection points of the vertical lines with the t-axis determine the domain. Domain: 0<t<9​ The range is determined by the points where the horizontal lines intersect the d-axis. Range: 6<t<10.5​bNow, let's interpret the slope and d-intercept.In Part A, we found that the slope is 21​. Since the function models the depth of the snow, the slope tells us the depth of the snow increases 21​ inch every hour. The slope of the function represents the depth of the snow just before the storm starts. This means that the depth of the snow is 6 inches at t=0.
Exercises 38 aWe have been given a linear equation and are asked to graph the line as well as find its domain and range. Since the equation is already in slope-intercept form, let's jump right into graphing. We need to identify the slope and y-intercept of the equation, c(x)=0.5x+70. Slope: y-intercept: ​0.570​ We can begin graphing by plotting the y-intercept.Next, we will use the slope to find a second point on the graph. Thinking of our slope as 21​ will make it is easier to find the "rise" and the "run." Based on the dimensions of our graph so far, it will be even easier if we think of this as 2010​. Therefore, we "rise" 10 units as we "run" 20 units. Once we have second point, we can draw a line through the points.To find the domain and range it is important to think about how many miles one can drive. We can drive zero or more miles, but we cannot drive a negative number of miles. This means that the domain is all real numbers greater than 0. The range is then all real numbers greater than $70. Domain: Range: ​70≤x70≤c​bAs a unit rate, the slope is 21​. This means that, as we drive 2 miles, the cost of renting a truck c rises $1 above $70. 21​ = 2 miles driven$1 in truck rental fees​ The c-intercept tells us that the base price for renting the truck is $70.
Exercises 39 Since it is a linear function, we can write it in slope-intercept form: y=mx+b. The y-intercept is the point at which x=0. From the the table we see that this occurs at y=40. Let's calculate the slope by substituting two points from the table in the slope formula. m=x2​−x1​y2​−y1​​Substitute (100,120) & (50,80)m=100−50120−80​ Simplify RHS Subtract termsm=5040​ba​=b/10a/10​m=54​Write as a decimal m=0.8 The slope-intercept form of this function is y=0.8x+40. Let's graph it along with the three points in the table in a coordinate plane.The equation from Exercise 38 was c(x)=0.5x+70. Let's add this function to our coordinate plane so that we can compare the two graphs.Comparing the two functions, we can see that function y has a greater slope but a lower y-intercept than c(x). In other words, the starting fee for this company is lower but it costs more per hour to rent the truck.
Exercises 40 Let's rewrite the given equation in slope-intercept form first. y+1=3x⇔y=3x−1 Now we can identify the y-intercept as -1, which has been graphed correctly. Additionally, the slope is 3. This means that for every step we take to the right, the function increases by 3 steps up.Graphing the function Let's graph the function and see if it overlaps with the given graph.The graphs don't overlap. The error committed was not graphing the slope properly.
Exercises 41 Let's rewrite the given equation in slope-intercept form: -4x+y=-2⇔y=4x−2 The y-intercept is -2 and the slope is 4. These values seemed to have been switched around when the graph was drawn. Let's graph the correct line on the same coordinate plane.
Exercises 42 aLet's rewrite each graph in slope-intercept form: 3y=−x−32y−14=4x4x−3−y=0x−12=-3y​⇔y=-31​x−1⇔y=2x+7​⇔y=4x−3⇔y=-31​x+4​ Now we can graph the lines on a coordinate plane by identifying the y-intercept and slope of each equation. We will also be able to identify the enclosed shape formed by the lines.Looking at the graph we identify a quadrilateral. Since the blue and purple line have the same slope means they are parallel. This means the quadrilateral is, in fact, a trapezoid.bWe can tell that the green and the red lines are not parallel as they are getting closer to each other in the first quadrant. However, the purple and blue line are parallel as they have the same slope
Exercises 43 aIf the width of the first rectangle is y and the length is x, we can calculate the perimeter P as 2x+2y=P. Let's isolate x+y to get an equation that describes the sum of the rectangle's width and length. 2x+2y=P Isolate x+y LHS/2=RHS/222x+2y​=2P​Write as a sum of fractions22x​+22y​=2P​Calculate quotient x+y=2P​ Rewriting the equation from the graph, we can determine that the sum of the rectangle's length and width is 20 inches: y=20−x⇔x+y=20. If we substitute this into the equation x+y=2P​, we can solve for the perimeter. x+y=2P​x+y=2020=2P​ Solve for P LHS⋅2=RHS⋅240=PRearrange equation P=40 The first rectangle's perimeter is 40 inches. If the second rectangle's perimeter is 10 inches less than the first's, it has a perimeter of: 40−10=30 inches. This means that the sum of the rectangle's length and width is half of this, 15 inches. We can now write a relationship that describes the second rectangle's width and length. y+x=15⇔y=15−x Let's graph this. Notice that the endpoints are left open as neither the width nor the length can be 15 inches.bLet's graph both of the lines on the same coordinate plane and compare.Comparing the graphs, we can see that the slopes are the same. We can also see that the blue graph's endpoints are closer to the origin.
Exercises 44 aIf the base of the first triangle is x and the length of the two equal sides is y, we can calculate the perimeter P as x+2y=P. We can use the equation provided on the graph, y=6−21​x, to solve for P. x+2y=Py=6−21​xx+2(6−21​x)=P Solve for P Distribute 2x+12−x=PSubtract term12=PRearrange equation P=12 The first triangle's perimeter is 12 meters. If the second rectangle's perimeter is 8 meters more than the first's, it has a perimeter of: 12+8=20 meters. We can now write an equation that describes the perimeter for the second triangle: x+2y=20. Let's rewrite this equation into slope-intercept form so that we may graph it. x+2y=20LHS−x=RHS−x2y=-x+20LHS/2=RHS/2y=-21​x+10 Let's graph this. Notice that the endpoints are left open as neither the width nor the length can be equal to the full perimeter. This would mean that the other sides have a length of 0, which is not possible.bLet's graph both of the lines on the same coordinate plane and compare.Comparing the graphs, we can see that the slopes are the same. We can also see that the blue graph's endpoints are farther from the origin.
Exercises 45 Let's note the important features of each graph and then compare those requirements to the given equations.As we can see, a. has a positive slope as it's slanting upwards. Also, the graph intercepts the y-axis at a positive value.As we can see, b. also has a positive slope. In this case, the graph intercepts the y-axis at a negative value.In this case, the graph is slanting downwards which means it has a negative slope. Also, since the graph intercepts the y-axis above the x-axis, its y-intercept is positive.Here, the graph is slanting downwards and therefore it has a negative slope. Also, since the graph intercepts the y-axis below the x-axis, its y-intercept is negative. Now, we can notice the same qualities about the given equations and see which ones are a match. We will rewrite some of them to clearly see their characteristics.FunctionSlopey-interceptMatches y=-3x+8NegativePositivec. y=-7x⇕y=-7x+0Negative0None y=47​x−41​PositiveNegativeb. y=-4x−9NegativeNegatived. y=-x−34​⇕y=-1x−34​NegativeNegatived. y=2x−4PositiveNegativeb. y=31​x+5PositivePositivea. y=6⇕y=0x+60PositiveNone Let's summarize what we have found. Graph a:Graph b:Graph c:Graph d:​y=31​x+5y=47​x−41​ or y=2x−4y=-3x+8y=-4x−9 or y=-x−34​​
Exercises 46 Most straight lines can be written in slope-intercept form. But what about a perfectly vertical line? A vertical line is considered to be "undefined," it has more than one output for only one input. These types of lines are written as x=a which does not follow the slope-intercept form. Note, there also exist lines that aren't straight, such as absolute value functions, parabolas, and hyperbolas which you may or may not have heard about before, but they are also not written in slope-intercept form.
Exercises 47 There are several ways to think about the definition of slope. The most basic way to find slope when looking at a graph is to find integer values that the function crosses through and count how many steps up or down and how many steps to the right it took to get from one point to the next. The most mathematical way to define this process is: m=x2​−x1​y2​−y1​​, where (x1​,y1​) and (x2​,y2​) are two points on the line. This subtraction is often described verbally as: m=the change in xthe change in y​. This is because the difference between y2​ and y1​ is how much the line changes in the vertical direction and the difference between x2​ and x1​ is how much the line changes in the horizontal direction. The most informal, but arguably most memorable, description of slope is: m=runrise​, which describes how we need to divide the amount the line is rising by the amount the line is running.
Exercises 48 Here we will look at two possible solutions, it's okay if you think of something else!First solution We can let the time t that it took to get to the beach be shown on the x-axis and the distance d from the beach be on the y-axis. Then we have the following graph.The line segment begins at (0,300) because no time has passed when we first begin the journey and we are 300 miles away from the beach. It continues down and to the right until our journey has finished at (6,0), 6 hours have gone by and we are now 0 miles away from the beach.Second solution We can let the time t that it took to get to the beach be shown on the x-axis and the the distance d from home be on the y-axis. Then we have the following graph.The line segment begins at (0,0) because no time has passed when we first begin the journey and we are 0 miles from home. It continues up and to the right until our journey has finished at (6,300), 6 hours have gone by and we are now 300 miles from home.
Exercises 49 aSince there are no numbers on the coordinate plane, we must choose which function matches which graph based on the steepness of the slopes. We are given two functions with the following slopes: ​g(x)=6x+a,m=6h(x)=2x+b,m=2​ Because slope tells us how many units to move up for each unit we move to the right, a function with a slope of 6 will rise much faster than a function with a slope of 2. Therefore, the functions can be assigned as shown below.bIf we look at the slope-intercept form, y=mx+b, we can see that a in g(x) and b in h(x) are the y-intercepts of the functions. Slope-intercept form: .....yGiven function: g(x)Given function: h(x)​=mx+b=6x+a=2x+b​cThe first thing we are instructed to do is mark point C, which is on the line for function g and 2 units to the right of (p,q).Now we can do the same thing for function h to mark point D.We are asked to find the difference between the y-coordinates of points C and D. In the graphs above, we can see that point C was an increase of 12 with respect to the y-axis and point D was an increase of 4. Therefore, C is 12−4=8 units greater than D.
Exercises 50 aOn the graph, there are three unique sections. These include the time you are walking to the bus stop, the time spent waiting for the bus at the bus stop, and the time riding the bus.It takes you 10 minutes to walk the 0.5 miles to the bus stop. You then stand still while waiting for the bus for 4 minutes. Finally, you get on the bus and ride for 4 minutes and go 2 miles.bIn this example, the slope of each segment is found by calculating timedistance​ because distance is on the y-axis and time is on the x-axis. Let's calculate this for each segment.Segmenttimedistance​m Walking100.5​201​ Waiting40​0 Riding the bus42​21​The first slope means that you can walk 1 mile in 20 minutes, which is 3 mph. The second slope means that, when standing still, your distance does not change at all and you are going 0 mph. The third slope means that, when riding the bus, you go 1 mile every 2 minutes or 30 mph.
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