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== 1.5#h(0.6em)Linear Functions


    
      
=== Slope-Intercept Form


      As we saw in Linear Models, many linear models #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] have equations of the form


      #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis starting value close parenthesis plus open parenthesis rate of change close parenthesis times x")[$bold(italic(f)) ( bold(italic(x)) ) = ( "starting value" ) + ( "rate of change" ) ⋅ bold(italic(x))$]
      The starting value, or the value of #math.equation(block: false, alt: "y")[$y$] at #math.equation(block: false, alt: "x equals 0")[$x = 0$], is the #math.equation(block: false, alt: "y")[$y$]-intercept of the graph, and the rate of change is the slope of the graph. Thus, we can write the equation of a line as


      #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals b plus m x")[$bold(italic(f)) ( bold(italic(x)) ) = bold(italic(b)) + bold(italic(m)) bold(italic(x))$]
      where the constant term, #math.equation(block: false, alt: "b")[$b$], is the #math.equation(block: false, alt: "y")[$y$]-intercept of the line, and #math.equation(block: false, alt: "m")[$m$], the coefficient of #math.equation(block: false, alt: "x")[$x$], is the slope of the line. This form for the equation of a line is called the #strong[slope-intercept form].


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Slope-Intercept Form]


        If we write the equation of a linear function in the form,


        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals b plus m x")[$f ( x ) = b + m x$]
        then #math.equation(block: false, alt: "m")[$m$] is the #strong[slope] of the line, and #math.equation(block: false, alt: "b")[$b$] is the #strong[#math.equation(block: false, alt: "y")[$y$]-intercept].
]


      (You may have encountered the slope-intercept equation in the equivalent form #math.equation(block: false, alt: "y equals m x plus b")[$y = m x + b$].)


      For example, consider the two linear functions and their graphs shown below.


      #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 10 minus 3 x")[$f ( x ) = 10 − 3 x$]


      #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$#hide($00$) x #hide($00$)$]], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$#hide($0$) f ( x ) #hide($0$)$]]),
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "10")[$10$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "7")[$7$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
))


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


      #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals minus 3 plus 2 x")[$g ( x ) = − 3 + 2 x$]


      #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$#hide($00$) x #hide($00$)$]], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$#hide($0$) f ( x ) #hide($0$)$]]),
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3")[$3$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]],
))


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


      Some observations:


      - We can see that the #math.equation(block: false, alt: "y")[$y$]-intercept of each line is given by the constant term, #math.equation(block: false, alt: "b")[$b$].
- By examining the table of values, we can also see why the coefficient of #math.equation(block: false, alt: "x")[$x$]gives the slope of the line:
- For #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$], each time #math.equation(block: false, alt: "x")[$x$] increases by #math.equation(block: false, alt: "1")[$1$] unit, #math.equation(block: false, alt: "y")[$y$] decreases by #math.equation(block: false, alt: "3")[$3$] units.
- For #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$], each time #math.equation(block: false, alt: "x")[$x$] increases by #math.equation(block: false, alt: "1")[$1$] unit, #math.equation(block: false, alt: "y")[$y$] increases by #math.equation(block: false, alt: "2")[$2$] units.


      For each graph, the coefficient of #math.equation(block: false, alt: "x")[$x$] is a scale factor that tells us how many units #math.equation(block: false, alt: "y")[$y$] changes for #math.equation(block: false, alt: "1")[$1$] unit increase in #math.equation(block: false, alt: "x")[$x$]. But that is exactly what the slope tells us about a line.


      #examplebox("Example 1")[][
Francine is choosing an Internet service provider. She paid \$30 for a modem, and she is considering three companies for service:


          - Juno charges \$14.95 per month,
- ISP.com charges \$12.95 per month,
- and peoplepc charges \$15.95 per month.


          Match the graphs in the figure to Francine’s Internet cost with each company.


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


        
        #solutionbox[
Francine pays the same initial amount, \$30 for the modem, under each plan. The monthly fee is the rate of change of her total cost, in dollars per month. We can write a formula for her cost under each plan.


          #math.equation(block: true, alt: "Juno: f open parenthesis x close parenthesis, equals 30 plus 14.95 x; ISP.com: g open parenthesis x close parenthesis, equals 30 plus 12.95 x; peoplepc: h open parenthesis x close parenthesis, equals 30 plus 15.95 x")[$"Juno: " f ( x ) & = 30 + 14.95 x \ "ISP.com: " g ( x ) & = 30 + 12.95 x \ "peoplepc: " h ( x ) & = 30 + 15.95 x$]
          The graphs of these three functions all have the same #math.equation(block: false, alt: "y")[$y$]-intercept, but their slopes are determined by the monthly fees. The steepest graph, III, is the one with the largest monthly fee, peoplepc. ISP.com, which has the lowest monthly fee, has the least steep graph, I.
]
]


      
        Delbert decides to use DSL for his Internet service.
            Earthlink charges a \$99 activation fee and \$39.95 per month, DigitalRain charges \$50 for activation and \$34.95 per month, and FreeAmerica charges \$149 for activation and \$34.95 per month.


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


        + Write a formula for Delbert's Internet costs under each plan. #linebreak() Earthlink: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals")[$f ( x ) =$]\_\_\_\_\_  #linebreak() DigitalRain: #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals")[$g ( x ) =$]\_\_\_\_\_  #linebreak() FreeAmerica: #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals")[$h ( x ) =$]\_\_\_\_\_
+ Match Delbert's Internet cost under each company with its graph shown above. #linebreak() Line I: \_\_\_\_\_  #linebreak() Line II: \_\_\_\_\_  #linebreak() Line III: \_\_\_\_\_


        #solutionbox[
+ Earthlink: #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 99 plus 39.95 x")[$f ( x ) = 99 + 39.95 x$]; DigitalRain: #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 50 plus 34.95 x")[$g ( x ) = 50 + 34.95 x$]; FreeAmerica: #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals 149 plus 34.95 x")[$h ( x ) = 149 + 34.95 x$]
+ DigitalRain: I; Earthlink: II; FreeAmerica: III
]


      
      
        Delbert decides to use DSL for his Internet service. Earthlink charges a \$99 activation fee and \$39.95 per month, DigitalRain charges \$50 for activation and \$34.95 per month, and FreeAmerica charges \$149 for activation and \$34.95 per month.


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


        + Write a formula for Delbert's Internet costs under each plan.
+ Match Delbert's Internet cost under each company with its graph shown above.


        #solutionbox[
+ #math.equation(block: true, alt: "Earthlink: f open parenthesis x close parenthesis, equals 99 plus 39.95 x; DigitalRain: g open parenthesis x close parenthesis, equals 50 plus 34.95 x; FreeAmerica: h open parenthesis x close parenthesis, equals 149 plus 34.95 x")[$"Earthlink:" " " " " f ( x ) & = 99 + 39.95 x \ "DigitalRain:" " " " " g ( x ) & = 50 + 34.95 x \ "FreeAmerica:" " " " " h ( x ) & = 149 + 34.95 x$]
+ DigitalRain: I; Earthlink: II; FreeAmerica: III
]


      
      
        What do the coefficients in the slope-intercept form tell you about a line?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: false, alt: "m")[$m$] is the slope; #math.equation(block: false, alt: "b")[$b$] is the #math.equation(block: false, alt: "y")[$y$]-intercept
]


      
      
        What do the coefficients in the slope-intercept form tell you about a line?


        + #math.equation(block: false, alt: "m")[$m$] is the slope; #math.equation(block: false, alt: "b")[$b$] is the #math.equation(block: false, alt: "x")[$x$]-intercept.
+ #math.equation(block: false, alt: "m")[$m$] is the slope; #math.equation(block: false, alt: "b")[$b$] is the #math.equation(block: false, alt: "y")[$y$]-intercept.
+ #math.equation(block: false, alt: "open parenthesis m , b close parenthesis")[$( m , b )$] is a point on the line.
+ #math.equation(block: false, alt: "m")[$m$] is the #math.equation(block: false, alt: "x")[$x$]-intercept; #math.equation(block: false, alt: "b")[$b$] is the #math.equation(block: false, alt: "y")[$y$]-intercept.


      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b plus m x")[$" " f ( x ) = b + m x$], we call #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "b")[$b$] #strong[parameters]. Their values are fixed for any particular linear equation; for example, in the equation #math.equation(block: false, alt: "y equals 3 plus 2 x")[$y = 3 + 2 x$], #math.equation(block: false, alt: "b equals 3")[$b = 3$] and #math.equation(block: false, alt: "m equals 2")[$m = 2$], and the variables are #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. By changing the values of #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "b")[$b$], we can write the equation for any line except a vertical line. The collection of all linear functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b plus m x")[$f ( x ) = b + m x$] is called a #strong[two-parameter] family of functions.
]


      #figure(figph[lines comparing slopes and intercepts], alt: "lines comparing slopes and intercepts", caption: none)


      
        Explain why the initial value for a linear model is often given by the #math.equation(block: false, alt: "y")[$y$]-intercept.


        \_\_\_\_\_


      
      
        Explain why the initial value for a linear model is often given by the #math.equation(block: false, alt: "y")[$y$]-intercept.


      
    
    
      
=== Slope-Intercept Method of Graphing


      Look again at the lines in the previous figure: There is only one line that has a given slope and passes through a particular point. That is, the values of #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "b")[$b$] determine the particular line. The value of #math.equation(block: false, alt: "b")[$b$] gives us a starting point, and the value of #math.equation(block: false, alt: "m")[$m$] tells us which direction to go to plot a second point. Thus, we can graph a line given in slope-intercept form without having to make a table of values.


      #examplebox("Example 2")[][
+ Write the equation #math.equation(block: false, alt: "4 x minus 3 y equals 6")[$" " 4 x − 3 y = 6 " "$] in slope-intercept form.
+ Graph the line by hand.


        
        #solutionbox[
+ We solve the equation for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$].
            #math.equation(block: true, alt: "minus 3 y, equals 6 minus 4 x, Divide both sides by minus 3; y, equals the fraction 6 minus 4 x over minus 3 equals the fraction 6 over minus 3 plus the fraction minus 4 x over minus 3; y, equals minus 2 plus the fraction 4 over 3 x")[$− 3 y & = 6 − 4 x &  & "Divide both sides by " − 3 \ y & = frac(6 − 4 x, − 3) = frac(6, − 3) + frac(− 4 x, − 3) \ y & = − 2 + frac(4, 3) x$]
+ We see that the slope of the line is #math.equation(block: false, alt: "m equals the fraction 4 over 3")[$m = display(frac(4, 3))$] and its #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "b equals minus 2")[$b = − 2$]. We begin by plotting the #math.equation(block: false, alt: "y")[$y$]-intercept, #math.equation(block: false, alt: "open parenthesis 0 , minus 2 close parenthesis")[$( 0 , − 2 )$]. We then use the slope to find another point on the line. We have
        #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction 4 over 3")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(4, 3)$]
        so starting at #math.equation(block: false, alt: "open parenthesis 0 , minus 2 close parenthesis")[$( 0 , − 2 )$], we move #math.equation(block: false, alt: "4")[$4$] units in the #math.equation(block: false, alt: "y")[$y$]-direction and #math.equation(block: false, alt: "3")[$3$] units in the #math.equation(block: false, alt: "x")[$x$]-direction, to arrive at the point #math.equation(block: false, alt: "open parenthesis 3 , 2 close parenthesis")[$( 3 , 2 )$]. Finally, we draw the line through these two points.


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


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The slope of a line is a ratio and can be written in many equivalent ways. In Example, the slope is equal to #math.equation(block: false, alt: "the fraction 8 over 6")[$display(frac(8, 6))$], #math.equation(block: false, alt: "the fraction 12 over 9")[$display(frac(12, 9))$], and #math.equation(block: false, alt: "the fraction minus 4 over minus 3")[$display(frac(− 4, − 3))$]. We can use any of these fractions to locate a third point on the line as a check. If we use #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction minus 4 over minus 3")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(− 4, − 3))$], we move down #math.equation(block: false, alt: "4")[$4$] units and left #math.equation(block: false, alt: "3")[$3$] units from the #math.equation(block: false, alt: "y")[$y$]-intercept to find the point #math.equation(block: false, alt: "open parenthesis minus 3 , minus 6 close parenthesis")[$( − 3 , − 6 )$] on the line.
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Slope-Intercept Method for Graphing a Line]


        + Plot the #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$].
+ Use the definition of slope to find a second point on the line: Starting at the #math.equation(block: false, alt: "y")[$y$]-intercept, move #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] units in the #math.equation(block: false, alt: "y")[$y$]-direction and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] units in the #math.equation(block: false, alt: "x")[$x$]-direction. Plot a second point at this location.
+ Use an equivalent form of the slope to find a third point, and draw a line through the points.


        #figure(figph[line with slope-intercept method], alt: "line with slope-intercept method", caption: none)
]


      
        + Write the equation #math.equation(block: false, alt: "2 y plus 3 x plus 4 equals 0")[$2 y + 3 x + 4 = 0$] in slope-intercept form. #linebreak() #math.equation(block: false, alt: "y equals")[$y =$]\_\_\_\_\_
+ Use the slope-intercept method to graph the line.


        #solutionbox[
+ Solve the given equation for #math.equation(block: false, alt: "y")[$y$] to get: #math.equation(block: false, alt: "y equals minus 2 minus the fraction 3 over 2 x")[$y = − 2 − display(frac(3, 2)) x$]
+ A graph is shown below.
]


        A graph for part (b):


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


      
      
        + Write the equation #math.equation(block: false, alt: "2 y plus 3 x plus 4 equals 0")[$" " 2 y + 3 x + 4 = 0 " "$] in slope-intercept form.
+ Use the slope-intercept method to graph the line.


        #solutionbox[
+ Solve the given equation for #math.equation(block: false, alt: "y")[$y$] to get: #math.equation(block: false, alt: "y equals minus 2 minus the fraction 3 over 2 x")[$" " " " y = − 2 − display(frac(3, 2)) x$]
+ #figure(figph[line], alt: "line", caption: none)
]


      
      
        What is the easiest way to find the slope of the line #math.equation(block: false, alt: "18 x minus 42 y equals 60")[$" " 18 x − 42 y = 60$]?


        \_\_\_\_\_


      
      
        What is the easiest way to find the slope of the line #math.equation(block: false, alt: "18 x minus 42 y equals 60")[$" " 18 x − 42 y = 60$]?


        + Solve for #math.equation(block: false, alt: "y")[$y$] to get the slope-intercept form.
+ Find the intercepts and use them to compute the slope.
+ Graph the line and compute #math.equation(block: false, alt: "the fraction Δ y over Δ x")[$display(frac(upright(Δ) y, upright(Δ) x))$].
+ Find values of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] that make the equation true.


      
    
    
      
=== Finding a Linear Equation from a Graph


      We can also use the slope-intercept form to find the equation of a line from its graph. First, we note the value of the #math.equation(block: false, alt: "y")[$y$]-intercept from the graph, and then we calculate the slope using two convenient points.


      #examplebox("Example 3")[][
Find an equation for the line shown at right.


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


        
        #solutionbox[
The line crosses the #math.equation(block: false, alt: "y")[$y$]-axis at the point #math.equation(block: false, alt: "open parenthesis 0 , 3200 close parenthesis")[$( 0 , 3200 )$], so the #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "3200")[$3200$]. To calculate the slope of the line, we locate another point, say #math.equation(block: false, alt: "open parenthesis 20 , 6000 close parenthesis")[$( 20 , 6000 )$], and compute:


          #math.equation(block: true, alt: "m, equals the fraction Δ y over Δ x equals the fraction 6000 minus 3200 over 20 minus 0; equals the fraction 2800 over 20 equals 140")[$m & = frac(upright(Δ) y, upright(Δ) x) = frac(6000 − 3200, 20 − 0) \  & = frac(2800, 20) = 140$]
          The slope-intercept form of the equation, with #math.equation(block: false, alt: "m equals 140")[$m = 140$] and #math.equation(block: false, alt: "b equals 3200")[$b = 3200$], is


          #math.equation(block: true, alt: "y equals 3200 plus 140 x")[$y = 3200 + 140 x$]
]
]


      
        Find an equation for the line shown at right.


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


        #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_


        #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_


        #math.equation(block: false, alt: "y equals")[$y =$]\_\_\_\_\_


        #solutionbox[
We can read the #math.equation(block: false, alt: "y")[$y$]-intercept from the graph, so #math.equation(block: false, alt: "b equals 80")[$b = 80$]. Another point is #math.equation(block: false, alt: "open parenthesis 20 , 30 close parenthesis")[$( 20 , 30 )$], and from those two points we can compute the slope  #math.equation(block: false, alt: "m equals the fraction minus 5 over 2")[$m = display(frac(− 5, 2))$]. So the slope-intercept form for the equation of the line is  #math.equation(block: false, alt: "y equals 80 minus the fraction 5 over 2 x")[$y = 80 − display(frac(5, 2)) x$]
]


      
      
        Find an equation for the line shown.


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


        #solutionbox[
We can read the #math.equation(block: false, alt: "y")[$y$]-intercept from the graph, so #math.equation(block: false, alt: "b equals 80")[$b = 80$]. Another point is #math.equation(block: false, alt: "open parenthesis 20 , 30 close parenthesis")[$( 20 , 30 )$], and from those two points we can compute the slope,  #math.equation(block: false, alt: "m equals the fraction minus 5 over 2")[$m = display(frac(− 5, 2))$]. So the slope-intercept form for the equation of the line is


          #math.equation(block: true, alt: "y equals 80 minus the fraction 5 over 2 x")[$y = 80 − display(frac(5, 2)) x$]
]


      
      
        How can you find the equation of a line from its graph?


        \_\_\_\_\_


      
      
        How can you find the equation of a line from its graph?


      
    
    
      
=== Point-Slope Form


      We can find the equation for a line if we know its slope and #math.equation(block: false, alt: "y")[$y$]-intercept. What if we do not know the #math.equation(block: false, alt: "y")[$y$]-intercept, but instead know some other point on the line? There is only one line that passes through a given point and has a given slope, so we should be able to find its equation.


      For example, we can graph the line of slope #math.equation(block: false, alt: "the fraction minus 3 over 4")[$display(frac(− 3, 4))$] that passes through the point #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$]. 
We first plot the given point, #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$], as shown in the figure below.


      Then we use the slope to find another point on the line. The slope is


      #math.equation(block: true, alt: "m equals the fraction minus 3 over 4 equals the fraction Δ y over Δ x")[$m = frac(− 3, 4) = frac(upright(Δ) y, upright(Δ) x)$]
      so we move down #math.equation(block: false, alt: "3")[$3$] units and then #math.equation(block: false, alt: "4")[$4$] units to the right, starting from #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$]. This brings us to the point #math.equation(block: false, alt: "open parenthesis 5 , minus 7 close parenthesis")[$( 5 , − 7 )$]. We can then draw the line through these two points.


      #figure(figph[graph line with point slope], alt: "graph line with point slope", caption: none)


      We can also find an equation for the line, as shown in Example.


      #examplebox("Example 4")[][
Find an equation for the line that passes through the point #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$] and has slope #math.equation(block: false, alt: "the fraction minus 3 over 4")[$display(frac(− 3, 4))$].


        
        #solutionbox[
We will use the formula for slope,


          #math.equation(block: true, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1")[$m = frac(y_(2) − y_(1), x_(2) − x_(1))$]
          We substitute #math.equation(block: false, alt: "the fraction minus 3 over 4")[$display(frac(− 3, 4))$] for the slope, #math.equation(block: false, alt: "m")[$m$], and #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$] for #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$]. For the second point, #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$], we use the variable point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$]. Substituting these values into the slope formula gives us


          #math.equation(block: true, alt: "the fraction minus 3 over 4 equals the fraction y minus open parenthesis minus 4 close parenthesis over x minus 1 equals the fraction y plus 4 over x minus 1")[$frac(− 3, 4) = frac(y − ( − 4 ), x − 1) = frac(y + 4, x − 1)$]
          To solve for #math.equation(block: false, alt: "y")[$y$] we first multiply both sides by #math.equation(block: false, alt: "x minus 1")[$x − 1$].


          #math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis the fraction minus 3 over 4, equals the fraction y plus 4 over x minus 1 open parenthesis x minus 1 close parenthesis; the fraction minus 3 over 4 open parenthesis x minus 1 close parenthesis, equals y plus 4, Apply the distributive law.; the fraction minus 3 over 4 x plus the fraction 3 over 4, equals y plus 4, Subtract 4 from both sides.; the fraction minus 3 over 4 x minus the fraction 13 over 4, equals y, the fraction 3 over 4 minus 4 equals the fraction 3 over 4 minus the fraction 16 over 4 equals the fraction minus 13 over 4")[$( bold(italic(x)) − 1 ) frac(− 3, 4) & = frac(y + 4, x − 1) ( bold(italic(x)) − 1 ) &  &  \ frac(− 3, 4) ( x − 1 ) & = y + 4 &  & "Apply the distributive law." \ frac(− 3, 4) x + frac(3, 4) & = y + 4 &  & "Subtract 4 from both sides." \ frac(− 3, 4) x − frac(13, 4) & = y &  & frac(3, 4) − 4 = frac(3, 4) − frac(16, 4) = frac(− 13, 4)$]
          The equation of the line is  #math.equation(block: false, alt: "y equals the fraction minus 13 over 4 minus the fraction 3 over 4 x")[$y = display(frac(− 13, 4)) − display(frac(3, 4)) x$]
]
]


      
        What will be wrong with your answer if you accidentally compute the slope as #math.equation(block: false, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 1 minus x sub 2")[$m = display(frac(y_(2) − y_(1), x_(1) − x_(2)))$]?


        \_\_\_\_\_


        #solutionbox[
It will have the wrong sign.
]


      
      
        What will be wrong with your answer if you accidentally compute the slope as #math.equation(block: false, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 1 minus x sub 2")[$m = display(frac(y_(2) − y_(1), x_(1) − x_(2)))$]?


        + The number will be too big.
+ The line will be decreasing.
+ That is the slope of the perpendicular line.
+ It will have the wrong sign.


      
      When we use the slope formula in this way to find the equation of a line, we substitute a variable point #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] for the second point. This version of the formula,


      #math.equation(block: true, alt: "m equals the fraction y minus y sub 1 over x minus x sub 1")[$m = frac(y − y_(1), x − x_(1))$]
      is called the #strong[point-slope form] for a linear equation. It is sometimes stated in another form obtained by clearing the fraction to get


      #math.equation(block: true, alt: "open parenthesis x minus x sub 1 close parenthesis m, equals the fraction y minus y sub 1 over x minus x sub 1 open parenthesis x minus x sub 1 close parenthesis, Multiply both sides by open parenthesis x minus x sub 1 close parenthesis; open parenthesis x minus x sub 1 close parenthesis m, equals y minus y sub 1, Clear fractions and solve for y .; y, equals y sub 1 plus m open parenthesis x minus x sub 1 close parenthesis")[$( bold(italic(x)) − bold(italic(x))_(1) ) m & = frac(y − y_(1), x − x_(1)) ( bold(italic(x)) − bold(italic(x))_(1) ) &  & "Multiply both sides by " ( bold(italic(x)) − bold(italic(x))_(1) ) \ ( x − x_(1) ) m & = y − y_(1) &  & "Clear fractions and solve for " bold(italic(y)) "." \ y & = y_(1) + m ( x − x_(1) )$]
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Point-Slope Form]


        The equation of the line that passes through the point #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and has slope #math.equation(block: false, alt: "m")[$m$] is


        #math.equation(block: true, alt: "y equals y sub 1 plus m open parenthesis x minus x sub 1 close parenthesis")[$y = y_(1) + m ( x − x_(1) )$]
]


      
        Use the point-slope form to find the equation of the line that passes through the point #math.equation(block: false, alt: "open parenthesis minus 3 , 5 close parenthesis")[$( − 3 , 5 )$] and has slope #math.equation(block: false, alt: "minus 1.4")[$− 1.4$].


        #math.equation(block: true, alt: "y, equals y sub 1 plus m open parenthesis x minus x sub 1 close parenthesis, Substitute minus 1.4 for m and open parenthesis minus 3 , 5 close parenthesis for open parenthesis x sub 1 , y sub 1 close parenthesis .; Simplify: Apply the distributive law.")[$y & = y_(1) + m ( x − x_(1) ) &  & "Substitute " − 1.4 " for " bold(italic(m)) " and " ( − 3 , 5 ) " for " ( bold(italic(x))_(1) , bold(italic(y))_(1) ) . \  &  &  & "Simplify: Apply the distributive law."$]
        #math.equation(block: false, alt: "y equals")[$y =$]\_\_\_\_\_


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


      
      
        Use the point-slope form to find the equation of the line that passes through the point #math.equation(block: false, alt: "open parenthesis minus 3 , 5 close parenthesis")[$( − 3 , 5 )$] and has slope #math.equation(block: false, alt: "minus 1.4")[$− 1.4$].


        #solutionbox[
#math.equation(block: true, alt: "y, equals y sub 1 plus m open parenthesis x minus x sub 1 close parenthesis, Substitute minus 1.4 for m and open parenthesis minus 3 , 5 close parenthesis for open parenthesis x sub 1 , y sub 1 close parenthesis .; y, equals 5 minus 1.4 open parenthesis x plus 3 close parenthesis, Simplify: Apply the distributive law.; y, equals 0.8 minus 1.4 x")[$y & = y_(1) + m ( x − x_(1) ) &  & "Substitute " − 1.4 " for " bold(italic(m)) " and " ( − 3 , 5 ) " for " ( bold(italic(x))_(1) , bold(italic(y))_(1) ) . \ y & = 5 − 1.4 ( x + 3 ) &  & "Simplify: Apply the distributive law." \ y & = 0.8 − 1.4 x$]
]


      
      The point-slope form is useful for modeling linear functions when we don't know the initial value but do know some other point on the line.


      #examplebox("Example 5")[][
Under a proposed graduated income tax system, single taxpayers would owe \$1500 plus 20% of the amount of their income over \$13,000. (For example, if your income is \$18,000, you would pay \$1500 plus 20% of \$5000.)


          + Complete the table of values for the tax, #math.equation(block: false, alt: "T")[$T$], on various incomes, #math.equation(block: false, alt: "I")[$I$].#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "I")[$" " I " "$]], [#math.equation(block: false, alt: "15 , 000")[$15 , 000$]], [#math.equation(block: false, alt: "20 , 000")[$20 , 000$]], [#math.equation(block: false, alt: "22 , 000")[$22 , 000$]]),
  [#math.equation(block: false, alt: "T")[$" " T " "$]], [], [], [],
))
+ Write a linear equation in point-slope form for the tax, #math.equation(block: false, alt: "T")[$T$], on an income #math.equation(block: false, alt: "I")[$I$].
+ Write the equation in slope-intercept form.


        
        #solutionbox[
+ On an income of \$15,000, the amount of income over \$13,000 is \$15,000 - \$13,000 = \$2000, so you would pay \$1500 plus 20% of \$2000, or
        #math.equation(block: true, alt: "T equals 1500 plus 0.20 open parenthesis 2000 close parenthesis equals 1900")[$T = 1500 + 0.20 ( 2000 ) = 1900$]
        You can compute the other function values in the same way.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "I")[$" " I " "$]], [#math.equation(block: false, alt: "15 , 000")[$15 , 000$]], [#math.equation(block: false, alt: "20 , 000")[$20 , 000$]], [#math.equation(block: false, alt: "22 , 000")[$22 , 000$]]),
  [#math.equation(block: false, alt: "T")[$" " T " "$]], [#math.equation(block: false, alt: "1900")[$1900$]], [#math.equation(block: false, alt: "2900")[$2900$]], [#math.equation(block: false, alt: "3300")[$3300$]],
))
+ On an income of #math.equation(block: false, alt: "I")[$I$], the amount of income over \$13,000 is #math.equation(block: false, alt: "I minus $ 13 , 000")[$I − upright(\$) 13 , 000$], so you would pay \$1500 plus 20% of #math.equation(block: false, alt: "I minus 13 , 000")[$I − 13 , 000$], or
            #math.equation(block: true, alt: "T equals 1500 plus 0.20 open parenthesis I minus 13 , 000 close parenthesis")[$T = 1500 + 0.20 ( I − 13 , 000 )$]
+ Simplify the right side of the equation to get
            #math.equation(block: true, alt: "T, equals 1500 plus 0.20 I minus 2600; T, equals minus 1100 plus 0.20 I")[$T & = 1500 + 0.20 I − 2600 \ T & = − 1100 + 0.20 I$]
]
]


      
        A healthy weight for a young woman of average height, 64 inches, is 120 pounds. To calculate a healthy weight for a woman taller than 64 inches, add 5 pounds for each inch of height over 64.


        + Write a linear equation in point-slope form for the healthy weight, #math.equation(block: false, alt: "W")[$W$], for a woman of height, #math.equation(block: false, alt: "H")[$H$], in inches. #linebreak() #math.equation(block: false, alt: "W equals")[$W =$]\_\_\_\_\_
+ Write the equation in slope-intercept form. #linebreak() #math.equation(block: false, alt: "W equals")[$W =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "W equals 120 plus 5 open parenthesis H minus 64 close parenthesis")[$W = 120 + 5 ( H − 64 )$]
+ #math.equation(block: false, alt: "W equals func2")[$W = "func2"$]
]


      
      
        A healthy weight for a young woman of average height, 64 inches, is 120 pounds. To calculate a healthy weight for a woman taller than 64 inches, add 5 pounds for each inch of height over 64.


        + Write a linear equation in point-slope form for the healthy weight, #math.equation(block: false, alt: "W")[$W$], for a woman of height, #math.equation(block: false, alt: "H")[$H$], in inches.
+ Write the equation in slope-intercept form.


        #solutionbox[
+ #math.equation(block: false, alt: "W equals 120 plus 5 open parenthesis H minus 64 close parenthesis")[$W = 120 + 5 ( H − 64 )$]
+ #math.equation(block: false, alt: "W equals minus 200 plus 5 H")[$W = − 200 + 5 H$]
]


      
      
        What do you get when you substitute the point #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$] into the point-slope formula?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "y equals m x plus b")[$y = m x + b$]
]


      
      
        What do you get when you substitute the point #math.equation(block: false, alt: "open parenthesis 0 , b close parenthesis")[$( 0 , b )$] into the point-slope formula?


        + #math.equation(block: false, alt: "x equals 0")[$x = 0$]
+ #math.equation(block: false, alt: "y equals m x plus b")[$y = m x + b$]
+ #math.equation(block: false, alt: "y equals b")[$y = b$]
+ #math.equation(block: false, alt: "a x plus b y equals 0")[$a x + b y = 0$]


      
      
        Explain the difference between the slope-intercept form and the point-slope form for a linear equation.


        \_\_\_\_\_


      
      
        Explain the difference between the slope-intercept form and the point-slope form for a linear equation.


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Slope-intercept form
- Point-slope form
- Parameter


      
      
        
==== CONCEPTS


        + Linear functions form a two-parameter family, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b plus m x")[$f ( x ) = b + m x$].
+ The initial value of a linear function and the #math.equation(block: false, alt: "y")[$y$]-intercept of its graph are given by #math.equation(block: false, alt: "b")[$b$]. The rate of change of the function and the slope of its graph are given by #math.equation(block: false, alt: "m")[$m$].
+ The slope-intercept form, #math.equation(block: false, alt: "y equals b plus m x")[$y = b + m x$], is useful when we know the initial value and the rate of change.
+ The point-slope form, #math.equation(block: false, alt: "y equals y sub 1 plus m open parenthesis x minus x sub 1 close parenthesis")[$y = y_(1) + m ( x − x_(1) )$], is useful when we know the rate of change and one point on the line.


      
      
        
==== STUDY QUESTIONS


        + How can you put a linear equation into slope-intercept form?
+ What do the coefficients in the slope-intercept form tell you about the line?
+ Explain how to graph a line using the slope-intercept method.
+ Explain how to find an equation for a line from its graph.
+ Explain how to use the point-slope form for a linear equation.
+ Francine says that the slope of the line #math.equation(block: false, alt: "y equals 4 x minus 6")[$y = 4 x − 6$] is #math.equation(block: false, alt: "4 x")[$4 x$]. Is she correct? Explain your answer
+ Delbert says that the slope of the line #math.equation(block: false, alt: "3 x minus 4 y equals 8")[$3 x − 4 y = 8$] is #math.equation(block: false, alt: "3")[$3$]. Is he correct? Explain your answer.


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Write a linear equation in slope-intercept form: \#1–14
+ Identify the slope and #math.equation(block: false, alt: "y")[$y$]-intercept: \#1–10
+ Graph a line by the slope-intercept method: \#11–14
+ Find a linear equation from its graph: \#21–26, 29–32, 53–56
+ Interpret the slope and #math.equation(block: false, alt: "y")[$y$]-intercept: \#21–28, 63 and 64
+ Find a linear equation from one point and the slope: \#33–50


      
    
    
      
=== Homework 1.5


      In Problems 1–10,


      + Write each equation in slope-intercept form.
+ State the slope and #math.equation(block: false, alt: "y")[$y$]-intercept of the line.


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


        
          + #math.equation(block: false, alt: "y equals the fraction 1 over 2 minus the fraction 3 over 2 x")[$y = display(frac(1, 2)) − display(frac(3, 2)) x$]
+ Slope #math.equation(block: false, alt: "the fraction minus 3 over 2")[$display(frac(− 3, 2))$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]


        
      
      
        #math.equation(block: true, alt: "5 x minus 4 y equals 0")[$5 x − 4 y = 0$]


      
      
        #math.equation(block: true, alt: "the fraction 1 over 4 x plus the fraction 3 over 2 y equals the fraction 1 over 6")[$display(frac(1, 4)) x + display(frac(3, 2)) y = display(frac(1, 6))$]


        
          + #math.equation(block: false, alt: "y equals the fraction 1 over 9 minus the fraction 1 over 6 x")[$y = display(frac(1, 9)) − display(frac(1, 6)) x$]
+ Slope #math.equation(block: false, alt: "the fraction minus 1 over 6")[$display(frac(− 1, 6))$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "the fraction 1 over 9")[$display(frac(1, 9))$]


        
      
      
        #math.equation(block: true, alt: "the fraction 7 over 6 x minus the fraction 2 over 9 y equals 3")[$display(frac(7, 6)) x − display(frac(2, 9)) y = 3$]


      
      
        #math.equation(block: true, alt: "4.2 x minus 0.3 y equals 6.6")[$4.2 x − 0.3 y = 6.6$]


        
          + #math.equation(block: false, alt: "y equals minus 22 plus 14 x")[$y = − 22 + 14 x$]
+ Slope #math.equation(block: false, alt: "14")[$14$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "minus 22")[$− 22$]


        
      
      
        #math.equation(block: true, alt: "0.8 x plus 0.004 y equals 0.24")[$0.8 x + 0.004 y = 0.24$]


      
      
        #math.equation(block: true, alt: "y plus 29 equals 0")[$y + 29 = 0$]


        
          + #math.equation(block: false, alt: "y equals minus 29")[$y = − 29$]
+ Slope #math.equation(block: false, alt: "0")[$0$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "minus 29")[$− 29$]


        
      
      
        #math.equation(block: true, alt: "y minus 37 equals 0")[$y − 37 = 0$]


      
      
        #math.equation(block: true, alt: "250 x plus 150 y equals 2450")[$250 x + 150 y = 2450$]


        
          + #math.equation(block: false, alt: "y equals the fraction 49 over 3 minus the fraction 5 over 3 x")[$y = display(frac(49, 3)) − display(frac(5, 3)) x$]
+ Slope #math.equation(block: false, alt: "the fraction minus 5 over 3")[$display(frac(− 5, 3))$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "the fraction 49 over 3")[$display(frac(49, 3))$]


        
      
      
        #math.equation(block: true, alt: "80 x minus 360 y equals 6120")[$80 x − 360 y = 6120$]


      
      In Problems 11–14,


      + Sketch by hand the graph of the line with the given slope and #math.equation(block: false, alt: "y")[$y$]-intercept.
+ Write an equation for the line.
+ Find the #math.equation(block: false, alt: "x")[$x$]-intercept of the line.


      
        #math.equation(block: false, alt: "m equals 3")[$m = 3$] and #math.equation(block: false, alt: "b equals minus 2")[$b = − 2$]


        
          + #figure(figph[m=3 and b=-2], alt: "m=3 and b=-2", caption: none)
+ #math.equation(block: false, alt: "y equals minus 2 plus 3 x")[$y = − 2 + 3 x$]
+ #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$]


        
      
      
        #math.equation(block: false, alt: "m equals minus 4")[$m = − 4$] and #math.equation(block: false, alt: "b equals 1")[$b = 1$]


      
      
        #math.equation(block: false, alt: "m equals minus the fraction 5 over 3")[$m = − display(frac(5, 3))$] and #math.equation(block: false, alt: "b equals minus 6")[$b = − 6$]


        
          + #figure(figph[m = -5/3 and b = -6], alt: "m = -5/3 and b = -6", caption: none)
+ #math.equation(block: false, alt: "y equals minus 6 plus the fraction 5 over 3 x")[$y = − 6 + display(frac(5, 3)) x$]
+ #math.equation(block: false, alt: "the fraction minus 18 over 5")[$display(frac(− 18, 5))$]


        
      
      
        #math.equation(block: false, alt: "m equals the fraction 3 over 4")[$m = display(frac(3, 4))$] and #math.equation(block: false, alt: "b equals minus 2")[$b = − 2$]


      
      
        The point #math.equation(block: false, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$] lies on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 3 x plus b")[$f ( x ) = − 3 x + b$]. Find #math.equation(block: false, alt: "b")[$b$].


        
          #math.equation(block: true, alt: "5")[$5$]


        
      
      
        The point #math.equation(block: false, alt: "open parenthesis minus 3 , minus 8 close parenthesis")[$( − 3 , − 8 )$] lies on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over 3 x plus b")[$f ( x ) = display(frac(2, 3)) x + b$]. Find #math.equation(block: false, alt: "b")[$b$].


      
      
        The point #math.equation(block: false, alt: "open parenthesis 8 , minus 5 close parenthesis")[$( 8 , − 5 )$] lies on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals m x minus 3")[$f ( x ) = m x − 3$]. Find #math.equation(block: false, alt: "m")[$m$].


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


        
      
      
        The point #math.equation(block: false, alt: "open parenthesis minus 5 , minus 6 close parenthesis")[$( − 5 , − 6 )$] lies on the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals m x plus 2")[$f ( x ) = m x + 2$]. Find #math.equation(block: false, alt: "m")[$m$].


      
      
        Find the slope and intercepts of the line #math.equation(block: false, alt: "A x plus B y equals C")[$A x + B y = C$]


        
          #math.equation(block: false, alt: "m equals the fraction minus A over B")[$m = display(frac(− A, B))$], #math.equation(block: false, alt: "x")[$x$]-intercept #math.equation(block: false, alt: "open parenthesis the fraction C over A , 0 close parenthesis")[$( display(frac(C, A)) , 0 )$], #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "open parenthesis 0 , the fraction C over B close parenthesis")[$( 0 , display(frac(C, B)) )$]


        
      
      
        Find the slope and intercepts of the line #math.equation(block: false, alt: "the fraction x over a plus the fraction y over b equals 1")[$display(frac(x, a)) + display(frac(y, b)) = 1$]


      
      In Problems 21–26,


      + Find a formula for the function whose graph is shown.
+ Say what the slope and the vertical intercept tell us about the problem.


      
        The graph shows the altitude, #math.equation(block: false, alt: "a")[$a$] (in feet), of a skier #math.equation(block: false, alt: "t")[$t$] minutes after getting on a ski lift.


        #figure(figph[altitude of skier on lift], alt: "altitude of skier on lift", caption: none)


        
          + #math.equation(block: false, alt: "a equals 100 plus 150 t")[$a = 100 + 150 t$]
+ The slope tells us that the skier's altitude is increasing at a rate of #math.equation(block: false, alt: "150")[$150$] feet per minute, the vertical intercept that the skier began at an altitude of #math.equation(block: false, alt: "200")[$200$] feet.


        
      
      
        The graph shows the distance, #math.equation(block: false, alt: "d")[$d$] (in meters), traveled by a train #math.equation(block: false, alt: "t")[$t$] seconds after it passes an observer.


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


      
      
        The graph shows the amount of garbage, #math.equation(block: false, alt: "G")[$G$] (in tons), that has been deposited at a dump site #math.equation(block: false, alt: "t")[$t$] years after new regulations go into effect.


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


        
          + #math.equation(block: false, alt: "G equals 25 plus 12.5 t")[$G = 25 + 12.5 t$]
+ The slope tells us that the garbage is increasing at a rate of #math.equation(block: false, alt: "12.5")[$12.5$] tons per year, the vertical intercept that the dump already had #math.equation(block: false, alt: "25")[$25$] tons (when the new regulations went into effect).


        
      
      
        The graph shows the number of barrels of oil, #math.equation(block: false, alt: "B")[$B$], that has been pumped at a drill site #math.equation(block: false, alt: "t")[$t$] days after a new drill is installed.


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


      
      
        The graph shows the amount of money, #math.equation(block: false, alt: "M")[$M$] (in dollars), in Tammy’s bank account #math.equation(block: false, alt: "w")[$w$] weeks after she loses all sources of income.


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


        
          + #math.equation(block: false, alt: "M equals 7000 minus 400 w")[$M = 7000 − 400 w$]
+ The slope tells us that Tammy's bank account is diminishing at a rate of \$#math.equation(block: false, alt: "400")[$400$] per week, the vertical intercept that she had \$#math.equation(block: false, alt: "7000")[$7000$] (when she lost all sources of income).


        
      
      
        The graph shows the amount of emergency water, #math.equation(block: false, alt: "W")[$W$] (in liters), remaining in a southern California household #math.equation(block: false, alt: "t")[$t$] days after an earthquake.


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


      
      
        The formula #math.equation(block: false, alt: "F equals the fraction 9 over 5 C plus 32")[$F = display(frac(9, 5)) C + 32$] defines a function that converts the temperature in degrees Celsius to degrees Fahrenheit.


        + What is the Fahrenheit temperature when it is #math.equation(block: false, alt: "10 degrees")[$10^(∘)$] Celsius?
+ What is the Celsius temperature when it is #math.equation(block: false, alt: "minus 4 degrees")[$− 4^(∘)$] Fahrenheit?
+ Choose appropriate WINDOW settings and graph the equation #math.equation(block: false, alt: "y equals the fraction 9 over 5 x plus 32")[$y = display(frac(9, 5)) x + 32$].
+ Find the slope and explain its meaning for this problem.
+ Find the intercepts and explain their meanings for this problem.


        
          + #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]F
+ #math.equation(block: false, alt: "minus 20 degrees")[$− 20^(∘)$]C
+ #figure(figph[GC graph], alt: "GC graph", caption: none)
+ The slope, #math.equation(block: false, alt: "the fraction 9 over 5 equals 1.8")[$frac(9, 5) = 1.8$], tells us that Fahrenheit temperatures increase by #math.equation(block: false, alt: "1.8 degrees")[$1.8^(∘)$] for each increase of #math.equation(block: false, alt: "1 degrees")[$1^(∘)$] Celsius.
+ #math.equation(block: false, alt: "C")[$C$]-intercept #math.equation(block: false, alt: "open parenthesis minus 17 the fraction 7 over 9 , 0 close parenthesis")[$( − 17 frac(7, 9) , 0 )$]: #math.equation(block: false, alt: "minus 17 the fraction 7 over 9 degrees")[$− 17 attach(frac(7, 9), t: ∘)$] C is the same as #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]F; #math.equation(block: false, alt: "F")[$F$]-intercept #math.equation(block: false, alt: "open parenthesis 0 , 32 close parenthesis")[$( 0 , 32 )$]: #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]C is the same as #math.equation(block: false, alt: "32 degrees")[$32^(∘)$]F.


        
      
      
        If the temperature on the ground is #math.equation(block: false, alt: "70 degrees")[$70^(∘)$] Fahrenheit, the formula #math.equation(block: false, alt: "T equals 70 minus the fraction 3 over 820 h")[$T = 70 − display(frac(3, 820)) h$] defines a function that gives the temperature at an altitude of #math.equation(block: false, alt: "h")[$h$] feet.


        + What is the temperature at an altitude of #math.equation(block: false, alt: "4100")[$4100$] feet?
+ At what altitude is the temperature #math.equation(block: false, alt: "34 degrees")[$34^(∘)$] Fahrenheit?
+ Choose appropriate WINDOW settings and graph the equation #math.equation(block: false, alt: "y equals 70 minus the fraction 3 over 820 x")[$y = 70 − display(frac(3, 820)) x$].
+ Find the slope and explain its meaning for this problem.
+ Find the intercepts and explain their meanings for this problem.


      
      
        In England, oven cooking temperatures are often given as Gas Marks rather than degrees Fahrenheit. The table shows the equivalent oven temperatures for various Gas Marks.


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([Gas Mark], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "9")[$9$]]),
  [Degrees (F)], [#math.equation(block: false, alt: "325")[$325$]], [#math.equation(block: false, alt: "375")[$375$]], [#math.equation(block: false, alt: "425")[$425$]], [#math.equation(block: false, alt: "475")[$475$]],
))


        + Plot the data and draw a line through the data points.
+ Calculate the slope of your line. Estimate the #math.equation(block: false, alt: "y")[$y$]-intercept from the graph.
+ Find an equation that gives the temperature in degrees Fahrenheit in terms of the Gas Mark.


        
          + #figure(figph[Fahrenheit vs gas mark], alt: "Fahrenheit vs gas mark", caption: none)
+ #math.equation(block: false, alt: "m equals 25 , b equals 250")[$m = 25 , " " b = 250$]
+ #math.equation(block: false, alt: "y equals 250 plus 25 x")[$y = 250 + 25 x$]


        
      
      
        European shoe sizes are scaled differently than American shoe sizes. The table shows the European equivalents for various American shoe sizes.


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([American shoe size], [#math.equation(block: false, alt: "5.5")[$5.5$]], [#math.equation(block: false, alt: "6.5")[$6.5$]], [#math.equation(block: false, alt: "7.5")[$7.5$]], [#math.equation(block: false, alt: "8.5")[$8.5$]]),
  [European shoe size], [#math.equation(block: false, alt: "37")[$37$]], [#math.equation(block: false, alt: "38")[$38$]], [#math.equation(block: false, alt: "39")[$39$]], [#math.equation(block: false, alt: "40")[$40$]],
))


        + Plot the data and draw a line through the data points.
+ Calculate the slope of your line. Estimate the #math.equation(block: false, alt: "y")[$y$]-intercept from the graph.
+ Find an equation that gives the European shoe size in terms of American shoe size.


      
      
        A spring is suspended from the ceiling. The table shows the length of the spring in centimeters as it is stretched by hanging various weights from it.


        #figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([Weight, kg], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "22")[$22$]]),
  [Length, cm], [#math.equation(block: false, alt: "25.76")[$25.76$]], [#math.equation(block: false, alt: "25.88")[$25.88$]], [#math.equation(block: false, alt: "26.36")[$26.36$]], [#math.equation(block: false, alt: "26.6")[$26.6$]], [#math.equation(block: false, alt: "26.84")[$26.84$]], [#math.equation(block: false, alt: "27.2")[$27.2$]], [#math.equation(block: false, alt: "28.04")[$28.04$]],
))


        + Plot the data on graph paper and draw a straight line through the points. Estimate the #math.equation(block: false, alt: "y")[$y$]-intercept of your graph.
+ Find an equation for the line.
+ If the spring is stretched to #math.equation(block: false, alt: "27.56")[$27.56$] cm, how heavy is the attached weight?


        
          + #figure(figph[spring length vs weight], alt: "spring length vs weight", caption: none)
+ #math.equation(block: false, alt: "y equals 0.12 x plus 25.4")[$y = 0.12 x + 25.4$]
+ #math.equation(block: false, alt: "18")[$18$] kg


        
      
      
        The table shows the amount of ammonium chloride salt, in grams, that can be dissolved in #math.equation(block: false, alt: "100")[$100$] grams of water at different temperatures.


        #figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([Temperature, #math.equation(block: false, alt: "∘")[$∘$]C], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "21")[$21$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "52")[$52$]]),
  [Grams of salt], [#math.equation(block: false, alt: "33")[$33$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "35.5")[$35.5$]], [#math.equation(block: false, alt: "38.5")[$38.5$]], [#math.equation(block: false, alt: "40.5")[$40.5$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "54")[$54$]],
))


        + Plot the data on graph paper and draw a straight line through the points. Estimate the #math.equation(block: false, alt: "y")[$y$]-intercept of your graph.
+ Find an equation for the line.
+ At what temperature will #math.equation(block: false, alt: "46")[$46$] grams of salt dissolve?


      
      In Problems 33–36,


      + Sketch by hand the graph of the line that passes through the given point and has the given slope.
+ Write an equation for the line in point-slope form.
+ Put your equation from part (b) into slope-intercept form.


      
        #math.equation(block: true, alt: "open parenthesis 2 , minus 5 close parenthesis")[$( 2 , − 5 )$]; #math.equation(block: true, alt: "m equals minus 3")[$m = − 3$]


        
          + #figure(figph[line with given point and slope], alt: "line with given point and slope", caption: none)
+ #math.equation(block: false, alt: "y plus 5 equals minus 3 open parenthesis x minus 2 close parenthesis")[$y + 5 = − 3 ( x − 2 )$]
+ #math.equation(block: false, alt: "y equals 1 minus 3 x")[$y = 1 − 3 x$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 6 , minus 1 close parenthesis")[$( − 6 , − 1 )$]; #math.equation(block: true, alt: "m equals 4")[$m = 4$]


      
      
        #math.equation(block: true, alt: "open parenthesis 2 , minus 1 close parenthesis")[$( 2 , − 1 )$]; #math.equation(block: true, alt: "m equals the fraction 5 over 3")[$m = display(frac(5, 3))$]


        
          + #figure(figph[line with given point and slope], alt: "line with given point and slope", caption: none)
+ #math.equation(block: false, alt: "y plus 1 equals the fraction 5 over 3 open parenthesis x minus 2 close parenthesis")[$y + 1 = frac(5, 3) ( x − 2 )$]
+ #math.equation(block: false, alt: "y equals the fraction minus 13 over 3 plus the fraction 5 over 3 x")[$y = frac(− 13, 3) + frac(5, 3) x$]


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


      
      For Problems 37–40,


      + Write an equation in point-slope form for the line that passes through the given point and has the given slope.
+ Put your equation from part (a) into slope-intercept form.
+ Use your graphing calculator to graph the line.


      
        #math.equation(block: true, alt: "open parenthesis minus 6.4 , minus 3.5 close parenthesis")[$( − 6.4 , − 3.5 )$], #math.equation(block: true, alt: "m equals minus 0.25")[$m = − 0.25$]


        
          + #math.equation(block: false, alt: "y plus 3.5 equals minus 0.25 open parenthesis x plus 6.4 close parenthesis")[$y + 3.5 = − 0.25 ( x + 6.4 )$]
+ #math.equation(block: false, alt: "y equals minus 5.1 minus 0.25 x")[$y = − 5.1 − 0.25 x$]
+ #figure(figph[line with given point and slope], alt: "line with given point and slope", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis 7.2 , minus 5.6 close parenthesis")[$( 7.2 , − 5.6 )$], #math.equation(block: true, alt: "m equals 1.6")[$m = 1.6$]


      
      
        #math.equation(block: true, alt: "open parenthesis 80 , minus 250 close parenthesis")[$( 80 , − 250 )$], #math.equation(block: true, alt: "m equals 2.4")[$m = 2.4$]


        
          + #math.equation(block: false, alt: "y plus 250 equals 2.4 open parenthesis x minus 80 close parenthesis")[$y + 250 = 2.4 ( x − 80 )$]
+ #math.equation(block: false, alt: "y equals minus 442 plus 2.4 x")[$y = − 442 + 2.4 x$]
+ #figure(figph[line with given point and slope], alt: "line with given point and slope", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 150 , 1800 close parenthesis")[$( − 150 , 1800 )$], #math.equation(block: true, alt: "m equals minus 24")[$m = − 24$]


      
      For Problems 41 and 42,


      + Find the slope of the line. (Note that not all the labeled points lie on the line.)
+ Find an equation for the line.


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


        
          + #math.equation(block: false, alt: "m equals the fraction 2 over 3")[$m = display(frac(2, 3))$]
+ #math.equation(block: false, alt: "y equals the fraction minus 1 over 3 plus the fraction 2 over 3 x")[$y = display(frac(− 1, 3)) + display(frac(2, 3)) x$]


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


      
      For Problems 43 and 44, the equation of line #math.equation(block: false, alt: "l sub 1")[$l_(1)$] is #math.equation(block: false, alt: "y equals q plus p x")[$y = q + p x$], and the equation of line #math.equation(block: false, alt: "l sub 2")[$l_(2)$] is #math.equation(block: false, alt: "y equals v plus t x")[$y = v + t x$].


      + Decide whether the coordinates of each labeled point are + a solution of #math.equation(block: false, alt: "y equals q plus p x")[$y = q + p x$],
+ a solution of #math.equation(block: false, alt: "y equals v plus t x")[$y = v + t x$],
+ a solution of both equations, or
+ a solution of neither equation.
+ Find #math.equation(block: false, alt: "p")[$p$], #math.equation(block: false, alt: "q")[$q$], #math.equation(block: false, alt: "t")[$t$], and #math.equation(block: false, alt: "v")[$v$].


      
        #figure(figph[two lines with labeled points], alt: "two lines with labeled points", caption: none)


        
          + #math.equation(block: false, alt: "open parenthesis minus 4 , 4 close parenthesis")[$( − 4 , 4 )$]: neither; #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$]: #math.equation(block: false, alt: "y equals p x plus q")[$y = p x + q$]; #math.equation(block: false, alt: "open parenthesis 3 , 2 close parenthesis")[$( 3 , 2 )$]: both; #math.equation(block: false, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$]: neither; #math.equation(block: false, alt: "open parenthesis 1 , minus 2 close parenthesis")[$( 1 , − 2 )$]: #math.equation(block: false, alt: "y equals t x plus v")[$y = t x + v$]
+ #math.equation(block: false, alt: "p equals the fraction minus 1 over 3")[$p = display(frac(− 1, 3))$], #math.equation(block: false, alt: "q equals 3")[$q = 3$], #math.equation(block: false, alt: "t equals 2")[$t = 2$], #math.equation(block: false, alt: "v equals minus 4")[$v = − 4$]


        
      
      
        #figure(figph[two lines with labeled points], alt: "two lines with labeled points", caption: none)


      
      For Problems 45–50,


      + Estimate the slope and vertical intercept of each line. (Hint: To calculate the slope, find two points on the graph that lie on the intersection of grid lines.)
+ Using your estimates from (a), write an equation for the line.


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


        
          + #math.equation(block: false, alt: "m equals 4 , b equals 40")[$m = 4 , " " b = 40$]
+ #math.equation(block: false, alt: "y equals 40 plus 4 x")[$y = 40 + 4 x$]


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


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


        
          + #math.equation(block: false, alt: "m equals minus 80 , b equals minus 2000")[$m = − 80 , " " b = − 2000$]
+ #math.equation(block: false, alt: "P equals minus 2000 minus 80 t")[$P = − 2000 − 80 t$]


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


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


        
          + #math.equation(block: false, alt: "m equals the fraction 1 over 4 , b equals 0")[$m = display(frac(1, 4)) , " " b = 0$]
+ #math.equation(block: false, alt: "V equals the fraction 1 over 4 d")[$V = display(frac(1, 4)) d$]


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


      
      
        + Write equations for three lines with slope #math.equation(block: false, alt: "m equals the fraction 3 over 4")[$m = display(frac(3, 4))$]. (Many answers are possible.)
+ Graph all three lines in the same window. What do you notice about the lines?


        
          + #math.equation(block: false, alt: "y equals the fraction 3 over 4 x")[$y = display(frac(3, 4)) x$], #math.equation(block: false, alt: "y equals 1 plus the fraction 3 over 4 x")[$y = 1 + display(frac(3, 4)) x$], #math.equation(block: false, alt: "y equals minus 2.7 plus the fraction 3 over 4 x")[$y = − 2.7 + display(frac(3, 4)) x$]
+ #figure(figph[3 lines of slope 3/4], alt: "3 lines of slope 3/4", caption: none)

 #linebreak() 
                    The lines are parallel.


        
      
      
        + Write equations for three lines with slope #math.equation(block: false, alt: "m equals 0")[$m = 0$]. (Many answers are possible.)
+ Graph all three lines in the same window. What do you notice about the lines?


      
      In Problems 53–56, choose the correct graph for each equation. The scales on both axes are the same.


      
        + #math.equation(block: false, alt: "y equals the fraction 3 over 4 x plus 2")[$y = display(frac(3, 4)) x + 2$]
+ #math.equation(block: false, alt: "y equals the fraction minus 3 over 4 x plus 2")[$y = display(frac(− 3, 4)) x + 2$]
+ #math.equation(block: false, alt: "y equals the fraction 3 over 4 x minus 2")[$y = display(frac(3, 4)) x − 2$]
+ #math.equation(block: false, alt: "y equals the fraction minus 3 over 4 x minus 2")[$y = display(frac(− 3, 4)) x − 2$]


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


        
          + II
+ III
+ I
+ IV


        
      
      
        + #math.equation(block: false, alt: "m less than 0")[$m < 0$], #math.equation(block: false, alt: "b greater than 0")[$b > 0$]
+ #math.equation(block: false, alt: "m greater than 1")[$m > 1$], #math.equation(block: false, alt: "b less than 0")[$b < 0$]
+ #math.equation(block: false, alt: "0 less than m less than 1")[$0 < m < 1$], #math.equation(block: false, alt: "b less than 0")[$b < 0$]
+ #math.equation(block: false, alt: "m less than minus 1")[$m < − 1$], #math.equation(block: false, alt: "b less than 0")[$b < 0$]


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


      
      
        + #math.equation(block: false, alt: "y equals 1 plus 2 open parenthesis x plus 3 close parenthesis")[$y = 1 + 2 ( x + 3 )$]
+ #math.equation(block: false, alt: "y equals minus 1 plus 2 open parenthesis x minus 3 close parenthesis")[$y = − 1 + 2 ( x − 3 )$]
+ #math.equation(block: false, alt: "y equals minus 1 plus 2 open parenthesis x plus 3 close parenthesis")[$y = − 1 + 2 ( x + 3 )$]
+ #math.equation(block: false, alt: "y equals 1 plus 2 open parenthesis x minus 3 close parenthesis")[$y = 1 + 2 ( x − 3 )$]


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


        
          + III
+ IV
+ II
+ I


        
      
      
        + #math.equation(block: false, alt: "y equals 2 minus the fraction 2 over 3 open parenthesis x minus 3 close parenthesis")[$y = 2 − display(frac(2, 3)) ( x − 3 )$]
+ #math.equation(block: false, alt: "y equals 2 minus the fraction 3 over 2 open parenthesis x plus 3 close parenthesis")[$y = 2 − display(frac(3, 2)) ( x + 3 )$]
+ #math.equation(block: false, alt: "y equals 2 plus the fraction 3 over 2 open parenthesis x minus 3 close parenthesis")[$y = 2 + display(frac(3, 2)) ( x − 3 )$]
+ #math.equation(block: false, alt: "y equals 2 plus the fraction 2 over 3 open parenthesis x plus 3 close parenthesis")[$y = 2 + display(frac(2, 3)) ( x + 3 )$]


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


      
      In Problems 57–60, find the slope of each line and the coordinates of one point on the line. (No calculation is necessary!)


      
        #math.equation(block: true, alt: "y plus 1 equals 2 open parenthesis x minus 6 close parenthesis")[$y + 1 = 2 ( x − 6 )$]


        
          #math.equation(block: true, alt: "m equals 2")[$m = 2$]; #math.equation(block: true, alt: "open parenthesis 6 , minus 1 close parenthesis")[$( 6 , − 1 )$]


        
      
      
        #math.equation(block: true, alt: "2 open parenthesis y minus 8 close parenthesis equals 5 open parenthesis x plus 2 close parenthesis")[$2 ( y − 8 ) = 5 ( x + 2 )$]


      
      
        #math.equation(block: true, alt: "y equals 3 minus the fraction 4 over 3 open parenthesis x plus 5 close parenthesis")[$y = 3 − display(frac(4, 3)) ( x + 5 )$]


        
          #math.equation(block: true, alt: "m equals the fraction minus 4 over 3")[$m = display(frac(− 4, 3))$]; #math.equation(block: true, alt: "open parenthesis minus 5 , 3 close parenthesis")[$( − 5 , 3 )$]


        
      
      
        #math.equation(block: true, alt: "7 x equals minus 3 y")[$7 x = − 3 y$]


      
      
        + Draw a set of coordinate axes with a square grid (i.e., with units the same size in both directions). Sketch four lines through the point #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$] with the following slopes:
                #math.equation(block: true, alt: "m equals 3 , m equals minus 3 , m equals the fraction 1 over 3 , m equals the fraction minus 1 over 3")[$m = 3 , " " " " " " m = − 3 , " " " " " " m = frac(1, 3) , " " " " " " m = frac(− 1, 3)$]
+ What do you notice about these lines?


        
          Look for perpendicular lines.


        
        
          + #figure(figph[four lines through the same point], alt: "four lines through the same point", caption: none)
+ The lines with slope #math.equation(block: false, alt: "3")[$3$] and #math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$] are perpendicular to each other, and the lines with slope #math.equation(block: false, alt: "minus 3")[$− 3$] and #math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$] are perpendicular to each other.


        
      
      
        + Draw a set of coordinate axes with a square grid (see Problem 61). Sketch four lines through the point #math.equation(block: false, alt: "open parenthesis 0 , minus 3 close parenthesis")[$( 0 , − 3 )$] with the following slopes:
                    #math.equation(block: true, alt: "m equals the fraction 2 over 5 , m equals the fraction minus 2 over 5 , m equals the fraction 5 over 2 , m equals the fraction minus 5 over 2")[$m = frac(2, 5) , " " " " " " m = frac(− 2, 5) , " " " " " " m = frac(5, 2) , " " " " " " m = frac(− 5, 2)$]
+ What do you notice about these lines?


      
      
        The boiling point of water changes with altitude and is approximated by the formula


        #math.equation(block: true, alt: "B equals f open parenthesis h close parenthesis equals 212 minus 0.0018 h")[$B = f ( h ) = 212 − 0.0018 h$]
        where #math.equation(block: false, alt: "B")[$B$] is in degrees and #math.equation(block: false, alt: "h")[$h$] is in feet. State the slope and vertical intercept of the graph, including units, and explain their meaning in this context.


        
          #math.equation(block: false, alt: "m equals minus 0.0018")[$m = − 0.0018$] degree/foot, so the boiling point drops with altitude at a rate of #math.equation(block: false, alt: "0.0018")[$0.0018$] degree per foot. #math.equation(block: false, alt: "b equals 212")[$b = 212$], so the boiling point is #math.equation(block: false, alt: "212 degrees")[$212^(∘)$] at sea level (where the elevation #math.equation(block: false, alt: "h equals 0")[$h = 0$]).


        
      
      
        The height of a woman in centimeters is related to the length of her femur (in centimeters) by the formula


        #math.equation(block: true, alt: "H equals f open parenthesis x close parenthesis equals 2.47 x plus 54.10")[$H = f ( x ) = 2.47 x + 54.10$]
        State the slope and the vertical intercept of the graph, including units, and explain their meaning in this context.