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Exercises 1 First, we will graph y=2x+4 for the domain x≤-1. The function has a slope of 2 and a y-intercept of 4. Since the endpoint is included, we should use a closed circle.Looking at the graph we can see that the y-values are less than or equal to 2. The range for this graph is then y≤2. Next, we will graph y=31​x−1 for the domain x>-1. The function has a slope of 31​ and a y-intercept of -1. Since the endpoint is not included, we will use an open circle.Looking at this graph, we can see that the minimum y-value is reached when x=-1. To find the corresponding y-value, we will substitute x for -1 in the equation of this piece. y=31​x−1x=-1y=31​⋅(-1)−1 Solve ca​⋅b=ca⋅b​y=31⋅(-1)​−1a(-b)=-a⋅by=3-1​−1Put minus sign in front of fractiony=-31​−1Rewrite 1 as 33​y=-31​−33​Subtract fractions y=-34​ Therefore, the range of this piece is y>-34​. Finally, let's plot both graphs on the same coordinate plane.Domain and Range of Piecewise Function Now that we've graphed both pieces together, we can look at the overall domain and range. We can see on the graph that there is neither a maximum nor a minimum y-value for the function. Therefore, the range of the piecewise function is the set of all real numbers. To find the domain, let's consider our piecewise function once more. y={2x+4,31​x−1,​if x≤-1if x>-1​ The input values of the first expression are all the values less than or equal to -1, and the input values for the second expression are all the values greater than -1. Therefore, the domain of the piecewise function is the set of all real numbers.
Exercises 2 Let's observe the given step function. f(x)=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​1,0,-1,-2,​if 0≤x<3if 3≤x<6if 6≤x<9if 9≤x<12​​Graphing the Function To think about how to draw the graph, let's look at the first piece of the function. The restriction on the domain tells us that f(x) equals 1 when x is greater than or equal to 0 and less than 3. f(x)=1 if  0≤x<3​ To graph this, we draw a horizontal line at y=1 extending from x=0 to x=3. To indicate that x=0 is contained in the solution set, we place a closed circle at that point. To indicate that x=3 is not contained in the solution set, we use an open circle.Following a similar process, we can graph the other pieces of the function.Domain and Range Now that we've graphed the function, we can describe its domain and range.Domain The domain of a function is the set of x-values for which the function is defined. From the graph (and the function rule), we can see that x can equal any value from 0 to 12, but not including 12. 0≤x<12​Range The range of a function is the set of y-values for which the function is defined. From the graph (and the function rule), we can see that y can only equal -2,-1,0, and 1. {-2,-1,0,1}​
Exercises 3 Lines written in slope-intercept form follow a certain format. y=mx+b​ In this form, m is the slope and b is the y-intercept. By substituting m=52​ and b=-7 into this equation, we can write the line's slope-intercept form equation. y=52​x+(-7) ⇒y=52​x−7​
Exercises 4 An equation in slope-intercept form follows a specific format. y=mx+b​ For an equation in this form, m is the slope and b is the y-intercept. Let's use the given points to calculate m and b. We will start by substituting the points into the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (0,6) & (3,-3)m=3−0-3−6​ Simplify right-hand side Subtract termsm=3-9​Calculate quotient m=-3 A slope of -3 means that for every 1 horizontal step in the positive direction, we take 3 vertical steps in the negative direction. Now that we know the slope, we can write a partial version of the equation. y=-3x+b​ To complete the equation, we also need to determine the y-intercept, b. Since one of the given points is (0,6), we already know that the y-intercept is 6. Now we can complete the equation. y=-3x+6​
Exercises 5 When lines are parallel, they have the same slope. y=3x−1​ Because of this, we know that all lines that are parallel to our given line will have a slope of 3. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=3x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (-2,-8). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=3x+bx=-2, y=-8-8=3(-2)+b Solve for b a(-b)=-a⋅b-8=-6+bLHS+6=RHS+6-2=bRearrange equation b=-2 Now that we have the y-intercept, we can write the equation of the line that is parallel to y=3x−1 and passes through the point (-2,-8). y=3x+(-2) ⇒y=3x−2​
Exercises 6 To write the equation of a line perpendicular to the given equation, we first need to determine its slope. After that, we'll write a general equation and use the given point to determine the y-intercept.Calculating the Perpendicular Line's Slope Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m1​⋅m2​=-1​ For any equation written in slope-intercept form, y=mx+b, we can identify its slope as the value of m. Looking at the given equation, we can see that its slope is 41​. y=41​x−9​ By substituting this value into our negative reciprocal equation for m1​, we can solve for the slope of the perpendicular line, m2​. m1​⋅m2​=-1m1​=41​41​⋅m2​=-1 Solve for m2​ b1​⋅a=ba​4m2​​=-1LHS⋅4=RHS⋅4 m2​=-4 With this, we can identify that any line perpendicular to the given equation will have a slope of -4.Writing the Perpendicular Line's Equation With the slope m2​=-4, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=-4x+b​ By substituting the given point (1,1) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line. y=-4x+bx=1, y=11=-4(1)+b Solve for b a⋅1=a1=-4+bLHS+4=RHS+45=bRearrange equation b=5 Now that we have the y-intercept, we can write the equation of the line that is both perpendicular to y=41​x−9 and passes through the point (1,1). y=-4x+5​
Exercises 7 Using the given point and slope, we can apply the point-slope form to write the equation. An equation in point-slope form follows a specific format. y−y1​=m(x−x1​)​ In this format, m represents the slope and the point (x1​,y1​) lies on the graph of the line. We are given the point (6,2) and the slope 10, so we have everything we need to create a point-slope form equation. y−y1​=m(x−x1​)Substitute x1​=6,y1​=2,m=10y−2=10(x−6)
Exercises 8 Equations in point-slope form follow a specific format. y−y1​=m(x−x1​)​ In this form, m is the slope of the line and (x1​,y1​) is a point on the line. Here we are given that the line passes through two known points. (-3,2)  and  (6,-1)​ To determine the slope of the line, we use the Slope Formula. m=x2​−x1​y2​−y1​​Substitute (6,-1) & (-3,2)m=-3−62−(-1)​ Simplify right-hand side a−(-b)=a+bm=-3−62+1​Add and subtract termsm=-93​ba​=b/3a/3​m=-31​Put minus sign in front of fraction m=-31​ Now that we know the slope of the line is -31​, we can write the equation of the line in point-slope form. We can use either of the given points as (x1​,y1​) in our equation. Let's use (-3,2). y−y1​=m(x−x1​)Substitute valuesy−2=-31​(x−(-3))a−(-b)=a+by−2=-31​(x+3) Please note that any point on the line can be used to form a point-slope equation. Therefore, our equation is only one of infinitely many possible equations!
Exercises 9
Exercises 10
Exercises 11
Exercises 12 Two lines are parallel if their slopes are identical. To tell if two lines are perpendicular, we check if their slopes are negative reciprocals. Let's tackle these questions one at a time.Are they parallel? For this exercise, we have been given equations that are not in slope-intercept form, so let's first rewrite all of them to identify their slopes.LineGiven EquationSlope-intercept formSlope 1y−c=axy=ax+ca 2ay=-x−by=-a1​x−ab​-a1​ 3ax+y=dy=-ax+d-a Now that we've identified the slope of each line, we can see that none of the lines have the same slope, so they are not parallel.Are they perpendicular? For lines with different slopes, we can conclude that they are not parallel. To determine whether or not they are perpendicular, we calculate the product of their slopes. Any two slopes whose product equals -1 are negative reciprocals, and therefore perpendicular. Let's start with checking lines 1 and 2. m1​⋅m2​=?-1m1​=a, m2​=-a1​a⋅(-a1​)=?-1Multiply(-aa​)=?-1aa​=1-1=-1 Therefore, lines 1 and 2 are perpendicular. We will use a similar method to check if lines 1 and 3 or 2 and 3 are perpendicular.LinesSlope 1Slope 2Product 1 & 2a-a1​-1 1 & 3a-a-a2 2 & 3-a1​-a1 We have found that only lines 1 and 2 are perpendicular to one another.
Exercises 13 Note, there is an infinite number of solutions to this exercise, this is just one possibility to help you visualize the process. First, let's think about the requirements for the domain and range:The domain must contain all real numbers. This means that our function should be continuous all the way from negative infinity to infinity. There cannot be any gaps. The range must be -3<y≤1. We need to be sure to place an open dot on y=-3. If we choose a specific value really really close to -3, there will always be one closer (-2.9→-2.99→-2.999…).One of the best strategies to ensure that your domain will continue on forever in both directions without leaving the necessary range is to set the leftmost and rightmost functions to be horizontal lines. Just make sure those lines are within the correct range! An example of this could be: y={1,0,​x≤0x≥4​ which we can graph.For our third equation, the one in the middle, we need a line that will "fill the gaps" in the range and domain. It needs to look as shown below.Notice the placement of the closed and open dots. The closed dot needs to be on the right piece while an open dot is on the middle piece. This ensures that the domain will have no gaps and the range will only include values such that y>-3. This line has a slope of -1 and a y-intercept at 1 so we can write our final piecewise function as: y=⎩⎪⎪⎨⎪⎪⎧​1,-x+1,0,​x≤00<x<4x≥4.​