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== 10.4#h(0.6em)Graphs and Equations


    Graphs are useful tools for studying mathematical relationships. A graph provides an overview of a quantity of data, and it helps us identify trends or unexpected occurrences. Interpreting the graph can help us answer questions about the data.


    For example, here are some data showing the atmospheric pressure at different altitudes. Altitude is given in feet, and atmospheric pressure is given in inches of mercury.


    #figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([Altitude (ft)], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]], [#math.equation(block: false, alt: "20 , 000")[$20 , 000$]], [#math.equation(block: false, alt: "30 , 000")[$30 , 000$]], [#math.equation(block: false, alt: "40 , 000")[$40 , 000$]], [#math.equation(block: false, alt: "50 , 000")[$50 , 000$]]),
  [Pressure (in. Hg)], [#math.equation(block: false, alt: "29.7")[$29.7$]], [#math.equation(block: false, alt: "24.8")[$24.8$]], [#math.equation(block: false, alt: "20.5")[$20.5$]], [#math.equation(block: false, alt: "14.6")[$14.6$]], [#math.equation(block: false, alt: "10.6")[$10.6$]], [#math.equation(block: false, alt: "8.5")[$8.5$]], [#math.equation(block: false, alt: "7.3")[$7.3$]],
))


    We observe a generally decreasing trend in pressure as the altitude increases, but it is difficult to say anything more precise about this relationship. A clearer picture emerges if we plot the data. To do this, we use two perpendicular number lines called axes. We use the horizontal axis for the values of the first variable, altitude, and the vertical axis for the values of the second variable, pressure.


    The entries in the table are called #strong[ordered pairs], in which the #strong[first component] is the altitude and the #strong[second component] is the atmospheric pressure measured at that altitude. For example, the first two entries can be represented by #math.equation(block: false, alt: "open parenthesis 0 , 29.7 close parenthesis")[$( 0 , 29.7 )$] and #math.equation(block: false, alt: "open parenthesis 5000 , 24.8 close parenthesis")[$( 5000 , 24.8 )$]. We plot the points whose #strong[coordinates] are given by the ordered pairs, as shown in the figure on the left.


    #figure(figph[graphs of pressure vs altitude], alt: "graphs of pressure vs altitude", caption: none)


    We can connect the data points with a smooth curve as shown in the figure on the right. In doing this, we are assuming that one variable changes smoothly with respect to the other, and in fact this is true for many physical situations. Thus, a smooth curve will thus serve as a good model.


    
      
=== Reading a Graph


      Once we have constructed a graph, we can use it to estimate values of the variables between the known data points.


      #examplebox("Example 1")[][
From the graph of atmospheric pressure, estimate the following:


          + The atmospheric pressure measured at an altitude of #math.equation(block: false, alt: "15 , 000")[$15 , 000$] feet
+ The altitude at which the pressure is #math.equation(block: false, alt: "12")[$12$] inches of mercury


        
        #solutionbox[
+ The point with first coordinate #math.equation(block: false, alt: "15 , 000")[$15 , 000$] on the graph at right has second coordinate approximately #math.equation(block: false, alt: "17.4")[$17.4$]. We estimate the pressure at #math.equation(block: false, alt: "15 , 000")[$15 , 000$] feet to be #math.equation(block: false, alt: "17.4")[$17.4$] inches of mercury.

#figure(figph[graph of pressure vs altitude with two points highlighted], alt: "graph of pressure vs altitude with two points highlighted", caption: none)
+ The point on the graph with second coordinate #math.equation(block: false, alt: "12")[$12$] has first coordinate approximately #math.equation(block: false, alt: "25 , 000")[$25 , 000$], so an atmospheric pressure of #math.equation(block: false, alt: "12")[$12$] inches of mercury occurs at about #math.equation(block: false, alt: "25 , 000")[$25 , 000$] feet.
]
]


      We can also use the graph to obtain information about the relationship between altitude and pressure that would be difficult to see from the data alone.


      #examplebox("Example 2")[][
+ For what altitudes is the pressure less than #math.equation(block: false, alt: "18")[$18$] inches of mercury?
+ How much does the pressure decrease as the altitude increases from #math.equation(block: false, alt: "15 , 000")[$15 , 000$] feet to #math.equation(block: false, alt: "25 , 000")[$25 , 000$] feet?
+ For which #math.equation(block: false, alt: "10 , 000")[$10 , 000$]-foot increase in altitude does the pressure change most rapidly?


        
        #solutionbox[
+ From the graph we see that the pressure has dropped to #math.equation(block: false, alt: "18")[$18$] inches of mercury at about #math.equation(block: false, alt: "14 , 000")[$14 , 000$] feet, and that it continues to decrease as the altitude increases. Therefore, the pressure is less than #math.equation(block: false, alt: "18")[$18$] inches of mercury for altitudes greater than #math.equation(block: false, alt: "14 , 000")[$14 , 000$] feet.
+ The pressure at #math.equation(block: false, alt: "15 , 000")[$15 , 000$] feet is approximately #math.equation(block: false, alt: "17.4")[$17.4$] inches of mercury, and at #math.equation(block: false, alt: "25 , 000")[$25 , 000$] feet it is #math.equation(block: false, alt: "12")[$12$] inches. This represents a decrease in pressure of #math.equation(block: false, alt: "17.4 minus 12")[$17.4 − 12$], or #math.equation(block: false, alt: "5.4")[$5.4$], inches of mercury.
+ By studying the graph we see that the pressure decreases most rapidly at low altitudes, so we conclude that the greatest drop in pressure occurs between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "10 , 000")[$10 , 000$] feet.
]
]


    
    
      
=== Graphs of Equations


      In Example, we used a graph to illustrate data given in a table. Graphs can also help us analyze models given by equations. Let's first review some facts about solutions of equations in two variables.


      An equation in two variables, such as y = 2x + 3, is said to be satisfied if the variables are replaced by a pair of numbers that make the statement true. The pair of numbers is called a #strong[solution] of the equation and is usually written as an ordered pair #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$]. (The first number in the pair is the value of #math.equation(block: false, alt: "x")[$x$] and the second number is the value of #math.equation(block: false, alt: "y")[$y$].)


      To find a solution of a given equation, we can assign a number to one of the variables and then solve for the second variable.


      #examplebox("Example 3")[][
Find solutions to the equation #math.equation(block: false, alt: "y equals 2 x plus 3")[$y = 2 x + 3$].


        
        #solutionbox[
We choose some values for #math.equation(block: false, alt: "x")[$x$], say, #math.equation(block: false, alt: "minus 2")[$− 2$], #math.equation(block: false, alt: "0")[$0$], and #math.equation(block: false, alt: "1")[$1$]. Substitute these #math.equation(block: false, alt: "x")[$x$]-values into the equation to find a corresponding #math.equation(block: false, alt: "y")[$y$]-value for each.


          #math.equation(block: true, alt: "When x equals minus 2 ,, y equals 2 open parenthesis minus 2 close parenthesis plus 3 equals minus 1; When x equals 0 ,, y equals 2 open parenthesis 0 close parenthesis plus 3 equals 3; When x equals 1 ,, y equals 2 open parenthesis 1 close parenthesis plus 3 equals 5")[$& "When " x = − 2 , &  & y = 2 ( − 2 ) + 3 = − 1 \  & "When " x = 0 , &  & y = 2 ( 0 ) + 3 = 3 \  & "When " x = 1 , &  & y = 2 ( 1 ) + 3 = 5$]
          Thus, the ordered pairs #math.equation(block: false, alt: "open parenthesis minus 2 , minus 1 close parenthesis")[$( − 2 , − 1 )$], #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$], and #math.equation(block: false, alt: "open parenthesis 1 , 5 close parenthesis")[$( 1 , 5 )$] are three solutions of #math.equation(block: false, alt: "y equals 2 x plus 3")[$y = 2 x + 3$]. We can also substitute values for #math.equation(block: false, alt: "y")[$y$]. For example, if we let #math.equation(block: false, alt: "y equals 10")[$y = 10$], we have


          #math.equation(block: true, alt: "10 equals 2 x plus 3")[$10 = 2 x + 3$]
          Solving this equation for #math.equation(block: false, alt: "x")[$x$], we find #math.equation(block: false, alt: "7 equals 2 x")[$7 = 2 x$], or #math.equation(block: false, alt: "x equals 3.5")[$x = 3.5$]. This means that the ordered pair #math.equation(block: false, alt: "open parenthesis 3.5 , 10 close parenthesis")[$( 3.5 , 10 )$] is another solution of the equation #math.equation(block: false, alt: "y equals 2 x plus 3")[$y = 2 x + 3$].
]
]


      An equation in two variables may have infinitely many solutions, so we cannot list them all. However, we can display the solutions on a graph. For this we use a #strong[Cartesian] (or #strong[rectangular]) #strong[coordinate system], as shown below left.


      #figure(figph[coordinate plane], alt: "coordinate plane", caption: none)


      #figure(figph[line in the coordinate plane], alt: "line in the coordinate plane", caption: none)


      The #strong[graph of an equation] is a picture of its solutions. A point is included in the graph if its coordinates satisfy the equation, and if the coordinates do not satisfy the equation, the point is not part of the graph. A graph of #math.equation(block: false, alt: "y equals 2 x plus 3")[$y = 2 x + 3$] is shown above right.


      This graph does not display #emph[all] the solutions of the equation, but it shows important features such as the intercepts on the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-axes. Because there is a solution corresponding to every real number #math.equation(block: false, alt: "x")[$x$], the graph extends infinitely in either direction, as indicated by the arrows.


      #examplebox("Example 4")[][
Use the graph of #math.equation(block: false, alt: "y equals 0.5 x squared minus 2")[$y = 0.5 x^(2) − 2$] shown at right to decide whether the given ordered pairs are solutions of the equation. Verify your answers algebraically.


          + #math.equation(block: false, alt: "open parenthesis minus 4 , 6 close parenthesis")[$( − 4 , 6 )$]
+ #math.equation(block: false, alt: "open parenthesis 3 , 0 close parenthesis")[$( 3 , 0 )$]


          #figure(figph[parabola in the coordinate plane], alt: "parabola in the coordinate plane", caption: none)


        
        #solutionbox[
+ Because the point #math.equation(block: false, alt: "open parenthesis minus 4 , 6 close parenthesis")[$( − 4 , 6 )$] does lie on the graph, the ordered pair #math.equation(block: false, alt: "x equals minus 4 , y equals 6")[$x = − 4 , y = 6$] is a solution of #math.equation(block: false, alt: "y equals 0.5 x squared minus 2")[$y = 0.5 x^(2) − 2$]. We can verify this by substituting #math.equation(block: false, alt: "minus 4")[$− 4$] for #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "6")[$6$] for #math.equation(block: false, alt: "y")[$y$]:
                #math.equation(block: true, alt: "0.5 open parenthesis minus 4 close parenthesis squared minus 2, equals 0.5 open parenthesis 16 close parenthesis minus 2; equals 8 minus 2, equals 6")[$0.5 ( − 4 )^(2) − 2 & = 0.5 ( 16 ) − 2 \ = 8 − 2 & = 6$]
+ Because the point #math.equation(block: false, alt: "open parenthesis 3 , 0 close parenthesis")[$( 3 , 0 )$] does not lie on the graph, the ordered pair #math.equation(block: false, alt: "x equals 3 , y equals 0")[$x = 3 , y = 0$] is not a solution of #math.equation(block: false, alt: "y equals 0.5 x squared minus 2")[$y = 0.5 x^(2) − 2$]. We substitute #math.equation(block: false, alt: "3")[$3$] for #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "0")[$0$] for #math.equation(block: false, alt: "y")[$y$] to verify this.
                #math.equation(block: true, alt: "0.5 open parenthesis 3 close parenthesis squared minus 2, equals 0.5 open parenthesis 9 close parenthesis minus 2; equals 4.5 minus 2, equals 2.5 not equal to 0")[$0.5 ( 3 )^(2) − 2 & = 0.5 ( 9 ) − 2 \ = 4.5 − 2 & = 2.5 ≠ 0$]
]
]


    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Ordered pair
- Component
- Cartesian coordinate system
- Solution
- Equation in two variables
- Satisfy an equation
- Coordinate
- Axis
- Graph


      
      
        
==== SKILLS


        Practice each skill in the exercises listed.


        + Read values from a graph: \#1–4
+ Find solutions to an equation in two variables: \#5–8
+ Make a table of values from an equation: \#9–12
+ Make a table of values from a graph: \#13–16
+ Estimate values from a graph: \#17–20
+ Use the Trace feature on a calculator: \#21–24
+ Find solutions to an equation in two variables from a graph: \#25–32


      
    
    
      
=== Exercises A.4


      For Problems 1-4, answer the questions about the graph.


      
        The graph shows the temperatures recorded during a winter day in Billings, Montana.


        #figure(figph[temperature vs time], alt: "temperature vs time", caption: none)


        + What were the high and low temperatures recorded during the day?
+ During what time intervals is the temperature above #math.equation(block: false, alt: "5 degrees")[$5^(∘)$]F? Below #math.equation(block: false, alt: "minus 5 degrees")[$− 5^(∘)$]F?
+ Estimate the temperatures at 7 a.m. and 2 p.m. At what time(s) is the temperature approximately #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]F? Approximately #math.equation(block: false, alt: "minus 12 degrees")[$− 12^(∘)$]F?
+ How much did the temperature increase between 3 a.m. and 6 a.m.? Between 9 a.m. and noon? How much did the temperature decrease between 6 p.m. and 9 p.m.?
+ During which 3-hour interval did the temperature increase most rapidly? Decrease most rapidly?


        
          + High: #math.equation(block: false, alt: "7 degrees")[$7^(∘)$]F; Low: #math.equation(block: false, alt: "minus 19 degrees")[$− 19^(∘)$]F
+ Above #math.equation(block: false, alt: "5 degrees")[$5^(∘)$]F from noon to 3 p.m.; Below #math.equation(block: false, alt: "minus 5 degrees")[$− 5^(∘)$]F from midnight to 9 a.m. and from 7 p.m. to midnight
+ 7 a.m.: #math.equation(block: false, alt: "minus 10 degrees")[$− 10^(∘)$]F; 2 p.m.: #math.equation(block: false, alt: "6 degrees")[$6^(∘)$]F; 10 a.m. and 5 p.m.: #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]F; 6 a.m. and 10 p.m.: #math.equation(block: false, alt: "minus 12 degrees")[$− 12^(∘)$]F
+ Between 3 a.m. and 6 a.m.: #math.equation(block: false, alt: "6 degrees")[$6^(∘)$]F; Between 9 a.m. and noon: #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]F; Between 6 p.m. and 9 p.m.: #math.equation(block: false, alt: "9 degrees")[$9^(∘)$]F
+ Increased most rapidly: 9 a.m. to noon; Decreased most rapidly: 6 p.m. to 9 p.m.


        
      
      
        The graph shows the altitude of a commercial jetliner during its flight from Denver to Los Angeles.


        #figure(figph[altitude vs time], alt: "altitude vs time", caption: none)


        + What was the highest altitude the jet achieved? At what time(s) was this altitude recorded?
+ During what time intervals was the altitude greater than 10,000 feet? Below 20,000 feet?
+ Estimate the altitudes 15 minutes into the flight and 35 minutes into the flight. At what time(s) was the altitude approximately 16,000 feet? 32,000 feet?
+ How many feet did the jet climb during the first 10 minutes of flight? Between 20 minutes and 30 minutes? How many feet did the jet descend between 100 minutes and 120 minutes?
+ During which 10-minute interval did the jet ascend most rapidly? Descend most rapidly?


      
      
        The graph shows the gas mileage achieved by an experimental model automobile at different speeds.


        #figure(figph[miles per gallon vs speed], alt: "miles per gallon vs speed", caption: none)


        + Estimate the gas mileage achieved at 43 miles per hour.
+ Estimate the speed at which a gas mileage of 34 miles per gallon is achieved.
+ At what speed is the best gas mileage achieved? Do you think that the gas mileage will continue to improve as the speed increases? Why or why not?
+ The data illustrated by the graph were collected under ideal test conditions. What factors might affect the gas mileage if the car were driven under more realistic conditions?


        
          + 28 mpg
+ 50 mph
+ Best gas mileage at 70 mph. The graph seems to be leveling off for higher speeds; any improvement in mileage probably would not be significant, and the mileage might in fact deteriorate.
+ Road condition, weather conditions, traffic, weight in the car


        
      
      
        The graph shows the fish population of a popular fishing pond.


        #figure(figph[population vs time], alt: "population vs time", caption: none)


        + During what months do the young fish hatch?
+ During what months is fishing allowed?
+ When does the park service restock the pond?


      
      For Problems 5-8, find five solutions (ordered pairs) for the equation.


      
        #math.equation(block: true, alt: "y equals 4 minus the fraction x over 3")[$y = 4 − display(frac(x, 3))$]


        
          #math.equation(block: true, alt: "open parenthesis minus 6 , 6 close parenthesis , open parenthesis minus 3 , 5 close parenthesis , open parenthesis 0 , 4 close parenthesis , open parenthesis 3 , 3 close parenthesis , open parenthesis 6 , 2 close parenthesis")[$( − 6 , 6 ) , " " " " ( − 3 , 5 ) , " " " " ( 0 , 4 ) , " " " " ( 3 , 3 ) , " " " " ( 6 , 2 )$]


        
      
      
        #math.equation(block: true, alt: "the fraction x minus 5 over 2 plus 1 equals y")[$display(frac(x − 5, 2)) + 1 = y$]


      
      
        #math.equation(block: true, alt: "3 x squared minus 1 equals y")[$3 x^(2) − 1 = y$]


        
          #math.equation(block: true, alt: "open parenthesis minus 2 , minus 11 close parenthesis , open parenthesis minus 1 , 2 close parenthesis , open parenthesis 0 , minus 1 close parenthesis , open parenthesis 1 , 2 close parenthesis , open parenthesis 2 , 11 close parenthesis")[$( − 2 , − 11 ) , " " " " ( − 1 , 2 ) , " " " " ( 0 , − 1 ) , " " " " ( 1 , 2 ) , " " " " ( 2 , 11 )$]


        
      
      
        #math.equation(block: true, alt: "y equals 9 minus open parenthesis x minus 2 close parenthesis squared")[$y = 9 − ( x − 2 )^(2)$]


      
      For Problems 9-12, fill in the table of values for the given equation.


      
        #math.equation(block: true, alt: "3 x plus 2 y equals 1")[$3 x + 2 y = 1$]


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 3")[$− 3$]], [],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "1")[$1$]], [],
  [], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
  [], [#math.equation(block: false, alt: "minus 4")[$− 4$]],
))


        
          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "5")[$5$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]],
  [#math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "the fraction 5 over 3")[$frac(5, 3)$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "minus 4")[$− 4$]],
))


        
      
      
        #math.equation(block: true, alt: "5 y minus 3 x equals 1")[$5 y − 3 x = 1$]


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 2")[$− 2$]], [],
  [], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [#math.equation(block: false, alt: "1")[$1$]], [],
  [], [#math.equation(block: false, alt: "3")[$3$]],
))


      
      
        #math.equation(block: true, alt: "y equals 1 minus the fraction x over 4")[$y = 1 − display(frac(x, 4))$]


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 4")[$− 4$]], [],
  [], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "3")[$3$]], [],
  [], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "5")[$5$]], [],
  [], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
))


        
          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "2")[$2$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "the fraction minus 1 over 4")[$frac(− 1, 4)$]],
  [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
))


        
      
      
        #math.equation(block: true, alt: "the fraction x plus 7 over 3 equals y")[$display(frac(x + 7, 3)) = y$]


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$" " y " "$]]),
  [#math.equation(block: false, alt: "minus 2")[$− 2$]], [],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [], [#math.equation(block: false, alt: "3")[$3$]],
  [#math.equation(block: false, alt: "5")[$5$]], [],
  [], [#math.equation(block: false, alt: "5")[$5$]],
  [], [#math.equation(block: false, alt: "7")[$7$]],
))


      
      For Problems 13-16, fill in the table of values for the graph.


      
        #figure(figph[line], alt: "line", caption: none)


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 6")[$− 6$]], [],
  [], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [], [#math.equation(block: false, alt: "0")[$0$]],
  [], [#math.equation(block: false, alt: "1")[$1$]],
))


        
          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 6")[$− 6$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "1")[$1$]],
))


        
      
      
        #figure(figph[line], alt: "line", caption: none)


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [], [#math.equation(block: false, alt: "5")[$5$]],
  [#math.equation(block: false, alt: "minus 1")[$− 1$]], [],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [], [#math.equation(block: false, alt: "minus 5")[$− 5$]],
))


      
      
        #figure(figph[curve], alt: "curve", caption: none)


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [], [#math.equation(block: false, alt: "3")[$3$]],
  [], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "0")[$0$]], [],
  [#math.equation(block: false, alt: "5")[$5$]], [],
))


        
          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "3")[$3$]],
  [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
))


        
      
      
        #figure(figph[curve], alt: "curve", caption: none)


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "minus 3")[$− 3$]], [],
  [#math.equation(block: false, alt: "minus 1")[$− 1$]], [],
  [], [#math.equation(block: false, alt: "minus 3")[$− 3$]],
  [], [#math.equation(block: false, alt: "4")[$4$]],
))


      
      For Problems 17-20, estimate from the graph, any values of #math.equation(block: false, alt: "x")[$x$] with the given value of #math.equation(block: false, alt: "y")[$y$].


      
        #math.equation(block: true, alt: "y equals minus 3")[$y = − 3$]


        #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


        
          #math.equation(block: true, alt: "minus 4")[$− 4$]


        
      
      
        #math.equation(block: true, alt: "y equals minus 3")[$y = − 3$]


        #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


      
      
        #math.equation(block: true, alt: "y equals 0")[$y = 0$]


        #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


        
          #math.equation(block: true, alt: "minus 4 , 2")[$− 4 , " " 2$]


        
      
      
        #math.equation(block: true, alt: "y equals minus 2")[$y = − 2$]


        #figure(figph[piecewise linear], alt: "piecewise linear", caption: none)


      
      For Problems 21-24, graph the equation in the given friendly window. Use the calculator's #emph[Trace] feature to make a table of values. (See Using a Graphing Calculator for help with entering expressions.) Round #math.equation(block: false, alt: "y")[$y$]-values to three decimal places.


      
        #math.equation(block: true, alt: "y equals vertical bar vertical bar x plus 2 vertical bar minus vertical bar x minus 2 vertical bar vertical bar")[$y = | | x + 2 | − | x − 2 | |$]


        #math.equation(block: true, alt: "Xmin equals minus 4.7 ; Xmax equals 4.7")[$"Xmin" = − 4.7 "; Xmax" = 4.7$]


        #math.equation(block: true, alt: "Ymin equals minus 6.2 ; Ymax equals 6.2")[$"Ymin" = − 6.2 "; Ymax" = 6.2$]


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3.2")[$− 3.2$]], [#math.equation(block: false, alt: "minus 1.5")[$− 1.5$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "1.9")[$1.9$]], [$#hide($000$)$], [$#hide($000$)$]),
  [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [$#hide($000$)$], [$#hide($000$)$],
))


        
          #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3.2")[$− 3.2$]], [#math.equation(block: false, alt: "minus 1.5")[$− 1.5$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "1.9")[$1.9$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3")[$3$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "3.8")[$3.8$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "4")[$4$]],
))


        
      
      
        #math.equation(block: true, alt: "y equals vertical bar x squared minus x minus 2 vertical bar")[$y = | x^(2) − x − 2 |$]


        #math.equation(block: true, alt: "Xmin equals minus 4.7 ; Xmax equals 4.7")[$"Xmin" = − 4.7 "; Xmax" = 4.7$]


        #math.equation(block: true, alt: "Ymin equals minus 9.3 ; Ymax equals 9.3")[$"Ymin" = − 9.3 "; Ymax" = 9.3$]


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3.1")[$− 3.1$]], [#math.equation(block: false, alt: "minus 1.5")[$− 1.5$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [$#hide($000$)$], [$#hide($000$)$]),
  [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [$#hide($000$)$], [$#hide($000$)$],
))


      
      
        #math.equation(block: true, alt: "y equals the fraction x minus 2 over x plus 2")[$y = display(frac(x − 2, x + 2))$]


        #math.equation(block: true, alt: "Xmin equals minus 4.7 ; Xmax equals 4.7")[$"Xmin" = − 4.7 "; Xmax" = 4.7$]


        #math.equation(block: true, alt: "Ymin equals minus 9.3 ; Ymax equals 9.3")[$"Ymin" = − 9.3 "; Ymax" = 9.3$]


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2.2")[$− 2.2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "4")[$4$]], [$#hide($000$)$], [$#hide($000$)$]),
  [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [$#hide($000$)$], [$#hide($000$)$],
))


        
          #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2.2")[$− 2.2$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]),
  [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "21")[$21$]], [#math.equation(block: false, alt: "undefined")[$"undefined"$]], [#math.equation(block: false, alt: "0.33 bar")[$0.3 limits(3)^(―)$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.6")[$0.6$]],
))


        
      
      
        #math.equation(block: true, alt: "y equals the square root of x squared minus 1.96")[$y = sqrt(x^(2) − 1.96)$]


        #math.equation(block: true, alt: "Xmin equals minus 4.7 ; Xmax equals 4.7")[$"Xmin" = − 4.7 "; Xmax" = 4.7$]


        #math.equation(block: true, alt: "Ymin equals minus 6.2 ; Ymax equals 6.2")[$"Ymin" = − 6.2 "; Ymax" = 6.2$]


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3.0")[$− 3.0$]], [#math.equation(block: false, alt: "minus 1.4")[$− 1.4$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "1.9")[$1.9$]], [$#hide($000$)$], [$#hide($000$)$]),
  [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [$#hide($000$)$], [$#hide($000$)$],
))


      
      For Problems 25–32,


      + Use the graph to find the missing component in each solution of the equation.
+ Verify your answers algebraically.


      
        #math.equation(block: true, alt: "s equals 2 t plus 4")[$s = 2 t + 4$]


        #figure(figph[line], alt: "line", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 3 , ? close parenthesis open parenthesis 1 , ? close parenthesis open parenthesis ? , 0 close parenthesis open parenthesis ? , 4 close parenthesis")[$( − 3 , " ?" ) " " " " " " ( 1 , " ?" ) " " " " " " ( "?" , " " 0 ) " " " " " " ( "?" , " " 4 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 3 , minus 2 close parenthesis , open parenthesis 1 , 6 close parenthesis , open parenthesis minus 2 , 0 close parenthesis , open parenthesis 0 , 4 close parenthesis")[$( − 3 , − 2 ) , " " " " ( 1 , 6 ) , " " " " ( − 2 , 0 ) , " " " " ( 0 , 4 )$]


        
      
      
        #math.equation(block: true, alt: "s equals minus 2 t plus 4")[$s = − 2 t + 4$]


        #figure(figph[line], alt: "line", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 2 , ? close parenthesis open parenthesis 3 , ? close parenthesis open parenthesis ? , 0 close parenthesis open parenthesis ? , 4 close parenthesis")[$( − 2 , " ?" ) " " " " " " ( 3 , " ?" ) " " " " " " ( "?" , " " 0 ) " " " " " " ( "?" , " " 4 )$]


      
      
        #math.equation(block: true, alt: "w equals v squared plus 2")[$w = v^(2) + 2$]


        #figure(figph[parabola], alt: "parabola", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 2 , ? close parenthesis open parenthesis 2 , ? close parenthesis open parenthesis ? , 3 close parenthesis open parenthesis ? , 2 close parenthesis")[$( − 2 , " ?" ) " " " " " " ( 2 , " ?" ) " " " " " " ( "?" , " " 3 ) " " " " " " ( "?" , " " 2 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 2 , 6 close parenthesis , open parenthesis 2 , 6 close parenthesis , open parenthesis 1 , 3 close parenthesis or open parenthesis minus 1 , 3 close parenthesis , open parenthesis 0 , 2 close parenthesis")[$( − 2 , 6 ) , " " ( 2 , 6 ) , " " ( 1 , 3 ) " or " ( − 1 , 3 ) , " " ( 0 , 2 )$]


        
      
      
        #math.equation(block: true, alt: "w equals v squared minus 4")[$w = v^(2) − 4$]


        #figure(figph[parabola], alt: "parabola", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 1 , ? close parenthesis open parenthesis 3 , ? close parenthesis open parenthesis ? , 0 close parenthesis open parenthesis ? , minus 4 close parenthesis")[$( − 1 , " ?" ) " " " " " " ( 3 , " ?" ) " " " " " " ( "?" , " " 0 ) " " " " " " ( "?" , " " − 4 )$]


      
      
        #math.equation(block: true, alt: "p equals the fraction 1 over m minus 1")[$p = display(frac(1, m − 1))$]


        #figure(figph[hyperbola], alt: "hyperbola", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 1 , ? close parenthesis open parenthesis the fraction 1 over 2 , ? close parenthesis open parenthesis ? , the fraction 1 over 3 close parenthesis open parenthesis ? , minus 1 close parenthesis")[$( − 1 , " ?" ) " " " " " " ( display(frac(1, 2)) , " ?" ) " " " " " " ( "?" , " " display(frac(1, 3)) ) " " " " " " ( "?" , − 1 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 1 , minus the fraction 1 over 2 close parenthesis , open parenthesis the fraction 1 over 2 , minus 2 close parenthesis , open parenthesis 4 , the fraction 1 over 3 close parenthesis open parenthesis 0 , minus 1 close parenthesis")[$( − 1 , − display(frac(1, 2)) ) , " " ( display(frac(1, 2)) , − 2 ) , " " ( 4 , " " display(frac(1, 3)) ) " " ( 0 , − 1 )$]


        
      
      
        #math.equation(block: true, alt: "p equals the fraction 1 over m plus 1")[$p = display(frac(1, m + 1))$]


        #figure(figph[hyperbola], alt: "hyperbola", caption: none)


        #math.equation(block: true, alt: "open parenthesis the fraction minus 3 over 2 , ? close parenthesis open parenthesis 3 , ? close parenthesis open parenthesis ? , minus 1 close parenthesis open parenthesis ? , 2 close parenthesis")[$( display(frac(− 3, 2)) , " ?" ) " " " " " " ( 3 , " ?" ) " " " " " " ( "?" , − 1 ) " " " " " " ( "?" , " " 2 )$]


      
      
        #math.equation(block: true, alt: "y equals x cubed")[$y = x^(3)$]


        #figure(figph[cubic], alt: "cubic", caption: none)


        #math.equation(block: true, alt: "open parenthesis minus 2 , ? close parenthesis open parenthesis the fraction 1 over 2 , ? close parenthesis open parenthesis ? , 0 close parenthesis open parenthesis ? , minus 1 close parenthesis")[$( − 2 , " ?" ) " " " " " " ( display(frac(1, 2)) , " ?" ) " " " " " " ( "?" , " " 0 ) " " " " " " ( "?" , − 1 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 2 , minus 8 close parenthesis open parenthesis the fraction 1 over 2 , the fraction 1 over 8 close parenthesis open parenthesis 0 , 0 close parenthesis open parenthesis minus 1 , minus 1 close parenthesis")[$( − 2 , − 8 ) " " " " " " ( display(frac(1, 2)) , display(frac(1, 8)) ) " " " " " " ( 0 , " " 0 ) " " " " " " ( − 1 , − 1 )$]


        
      
      
        #math.equation(block: true, alt: "y equals the square root of x plus 4")[$y = sqrt(x + 4)$]


        #figure(figph[half-parabola], alt: "half-parabola", caption: none)


        #math.equation(block: true, alt: "open parenthesis 0 , ? close parenthesis open parenthesis 5 , ? close parenthesis open parenthesis ? , 0 close parenthesis open parenthesis ? , 1 close parenthesis")[$( 0 , " ?" ) " " " " " " ( 5 , " ?" ) " " " " " " ( "?" , " " 0 ) " " " " " " ( "?" , " " 1 )$]