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Understanding Algebra for Calculus

Erich McAlister

Section 5.1 Linear Functions

Subsection 5.1.1 Introduction to Linear Functions

Linear functions make up perhaps the most important class of functions in mathematics. Indeed, the basic premise of Calculus is to approximate more complicated functions with linear ones. However, despite their importance to higher mathematics, they come up in rather simple real world settings.

Example 5.1.1.

Fired Up Pizzeria charges \(\$ 9.50\) for a 12 inch pizza with no toppings, then an additional \(\$ 1.50\) per topping. Let’s consider the function \(C\) that takes as an input the number of toppings, \(n\text{,}\) and gives the cost of a pizza, \(C(n)\text{,}\) in dollars. Construct a formula for this function and show its graph.
Solution.
We first give a table of values for its outputs for several inputs:
Table 5.1.2.
\(n\) toppings \(0\) \(1\) \(2\) \(3\) \(4\)
\(C(n)\) dollars \(9.50\) \(11.00\) \(12.50\) \(14.00\) \(15.50\)
The critical thing we notice in this table is that each time the input increases by one unit, an additional \(\$ 1.50\) is added to the cost. Hence, the number of times we add \(1.50\) to the initial (\(n=0\)) cost of \(\$ 9.50\) is equal to the input \(n\text{.}\) This process of repeated addition is precisely the definition of adding \(1.5 n\) to \(9.5\text{,}\) hence we have the formula
\begin{equation*} C(n) = 9.5 + 1.5 n. \end{equation*}
Using the table and/or the formula, we may then graph \(C\text{.}\)
Figure 5.1.3. The graph of \(C(n) = 9.5+1.5n\text{.}\)
In this picture we have plotted the graph between integer values of \(n\text{,}\) even though that may not make sense for this function, so that we can see the defining feature of the graph; it is a straight line. Due to the geometric properties of triangles, we see that the ratio of the change in the output divided by the change input is always equal to \(1.5\text{.}\) This value is called the slope of this line.
Using the introductory example given above, we may characterize linear functions as in the following table:

Definition 5.1.4. Linear Function.

A Linear Function is \(\ldots\)
Table 5.1.5.
Linear Function \(\longrightarrow\) A process applied to the input that adds a fixed amount \(m\) to the
output for each additional unit of input.
\(\longrightarrow\) A function of the form \(y = f(x) = mx+b\text{,}\) where \(m\) and \(b\) are constants.
\(\longrightarrow\) A function whose graph is a non-vertical line.
Note that when we say that a function is of the form \(f(x) = mx+b\text{,}\) what we really mean is that it can be put into that form throught some algebraic manipulation. The graphical meaning of the parameters \(m\) and \(b\) are as follows:

Checkpoint 5.1.7.

(a) Find the slope of the line connecting the point \((2,5)\) to the point \((4,10)\text{.}\)
Slope =
(b) Find the slope of the line connecting the point \((4,10)\) to the point \((8,15)\text{.}\)
Slope =
(c) Find the slope of the line connecting the point \((8,15)\) to the point \((16,20)\text{.}\)
Slope =
(d) Could the table below represent a linear function?
  • ?
  • Yes, it could be linear
  • No, it is not linear
    • \(x\) \(y\)
    2 5
    4 10
    8 15
    16 20
    (e) If the table could represent a linear function, give a formula for the linear function. If it is not a linear function, enter NONE.
    \(y =\)

    Checkpoint 5.1.8.

    Match the function with one of the graphs below.
    A B
    C D
    Error: alt_print_q: Unknown displayMode: PTX.
    Note: You can click on the graphs to enlarge the images.

    Example 5.1.9.

    Because a line is uniquely determined by two points, it is possible to find the formula for a linear function whose graph contains a given pair of points (provided the two points do not lie on a vertical line). For instance, suppose we are given that \((1,3)\) and \((3,4)\) are on the graph of \(f(x) = mx +b\text{.}\) Find the values of \(m\) and \(b\text{,}\) and hence the formula for \(f(x)\text{.}\)
    Solution.
    First we compute the slope:
    \begin{equation*} m = \frac{4-3}{3-1} = \frac{1}{2}. \end{equation*}
    Then we use the facts that \(f\) is of the form \(f(x) = \frac{1}{2}x+b\) and one of the two points is on the graph by substituting and solving for \(b\text{:}\)
    \begin{align*} 3 = \amp\ \frac{1}{2}\cdot 1 + b \\ \updownarrow \amp \\ b = \amp\ \frac{5}{2}. \end{align*}
    Thus \(f(x) = \frac{1}{2}x + \frac{5}{2}\text{.}\) Another way to see how this process works is by noting that \(\frac{1}{2}\) is added to the output for every extra unit of input. By starting at \((1,3)\) and subtracting \(\frac{1}{2}\) once to get back to when \(x=0\text{,}\) we also get \(b=f(0)=\frac{5}{2}\text{.}\)

    Checkpoint 5.1.10.

    Find the equation of the line through the points \(\left(\frac{5}{3}, -2\right)\) and \(\left(1,4\right)\text{.}\) Write your answer in the form \(y = mx + b\text{,}\) using fractions (no decimal numbers).

    Question 5.1.11.

    In the last example, we noted that finding the formula for a linear function would not work if the line was vertical. Explain the following:
    1. You cannot determine the slope of a vertical line from two points on that line.
    2. The equation of a vertical line has the form \(x=a\text{,}\) where \(a\) is a number.
    In practical applications, one can often recognize linear functions by noting a constant rate of change parameter.

    Example 5.1.12.

    Suppose we wish to find a formula for the profit \(P\) of selling \(q\) widgets. This is the revenue we get from selling the widgets minus the cost of producing them. Suppose we sell the widgets for \(\$ 35\) a piece, and it costs \(\$ 1000\) to set up production, plus \(\$ 15\) to make each widget.
    Solution.
    Observe that the units of \(\$ /\text{widget}\) on the selling price and production cost are both rates of change of the form \((\text{function output})/(\text{function input})\text{.}\) This, along with the fact that these rates are constant, indicates the function is likely a linear function. Going to the formula we get
    \begin{equation*} P = (\text{revenue}) - (\text{cost}) = 35q - (1000 + 15q). \end{equation*}
    This simplifies to the linear function
    \begin{equation*} P = 20q - 1000. \end{equation*}
    Here we see that \(\$ 1000\) will be spent no matter how many widgets are sold, and we get a net profit of \(\$ 20\) for each widget sold.

    Checkpoint 5.1.13.

    You want to choose one long distance telephone company from the following options:
    Company A charges $0.37 dollars per minute (no fixed monthly charge).
    Company B charges $13.75 per month plus $0.26 per minute.
    Company C charges a fixed rate of $50 per month.
    Let \(A(x)\text{,}\) \(B(x)\text{,}\) and \(C(x)\) denote the monthly charges of Company A, B, and C respectively for spending \(x\) minutes on long distance calls.
    (a) Find a formula for the monthly cost of using Company A, \(A(x)\text{:}\)
    \(A(x) \ =\)
    (b) Find a formula for the monthly cost of using Company B, \(B(x)\text{:}\)
    \(B(x) \ =\)
    (c) Find a formula for the monthly cost of using Company C, \(C(x)\text{:}\)
    \(C(x) \ =\)
    (d) If you spend 125 minutes talking long distance in a month, which of the three companies will be cheapest? Enter the letter (A, B, or C) here:
    (e) Given the graph below, match the number of each graph to corresponding company it represents:
    Company A
    Company B
    Company C

    Question 5.1.14.

    A ski shop rents skis for a fixed initial payment of \(A\) dollars and a charge of \(r\) dollars per day they are used. If a skier pays a total of \(A + 6r\) dollars, what is the meaning of the number \(6\text{?}\) Explain your answer as completely as possible.

    Subsection 5.1.2 Forms of Linear Functions

    The form \(y = f(x) = mx + b\) of a linear function is known as slope-intercept form. It is a very useful form. A more general (but non-unique) form is known as point-slope form.

    Checkpoint 5.1.15.

    Write the equation for the line \(8 x + 2 y = 9\) in the form slope-intercept form, and enter it in this form.

    Definition 5.1.16. Point-Slope Form of a Linear Function.

    The linear function of slope \(m\) passing through the point \((x_0,y_0)\) has the form
    \begin{equation*} y = f(x) = y_0 + m(x-x_0). \end{equation*}
    In order to understand point-slope form, all you need to do is ask (and answer) the following two questions:
    • Is \(f(x) = y_0 + m(x-x_0)\) the formula for a linear function with slope \(m\text{?}\) The answer is yes. All one needs to do is simplify and collect like terms to see that \(f(x) = mx + (y_0 - mx_0)\text{.}\) Thus we can say \(b = y_0 - mx_0\text{.}\)
    • Does the graph of \(f(x) = y_0 + m(x-x_0)\) pass through the point \((x_0,y_0)\text{?}\) Again the answer is yes because \(f(x_0) = y_0 + m(x_0-x_0) = y_0\text{.}\)
    The graphical difference between point-slope form and slope-intercept form is that slope-intercept form puts on display the \(y\)-intercept, whereas slope-intercept form displays another point on the graph. Equipped with point-slope form, it is somewhat easier to find a formula for a linear function given two points; all you need to do is compute the slope.

    Checkpoint 5.1.17.

    A line’s equation is given in point-slope form:
    \({y}={5\!\left(x-2\right)+12}\)
    This line’s slope is .
    A point on this line that is apparent from the given equation is .

    Checkpoint 5.1.18.

    Write an equation for the line through the points \((2,3)\) and \((10,5)\) in point-slope form \(y = y_0 + m(x-x_0)\text{.}\)
    \(y =\) +

    Question 5.1.19.

    The slope-intercept form of the formula for a linear function is actually equal to the point-slope form for a specific point on the graph. What point is that?
    Since the graphs of linear functions are lines, it makes sense to ask whether the two lines are parallel or perpendicular. To see whether two lines are parallel, they simply must have the same slope. So see whether two lines are perpendicular, consider the following figure:
    Figure 5.1.20. Perpendicular lines with triangles indicating their slopes.
    From this (with triangles added to illustrate slope), we see that the magnitude of the change in \(y\) for an associated change in \(x\) on one graph is the same as the magnitude of the change in \(x\) for an associated change in \(y\) on the perpendicular graph. Also, the sign of the slope is opposite. Hence the slope of a line perpendicular to a line of slope \(m\) is \(-\dfrac{1}{m}\text{.}\)

    Checkpoint 5.1.21.

    Find the equation of the line though the point \((-3, -2)\) and perpendicular to the line \(y = -{\frac{3}{4}} x + 1\text{.}\) Write your answer in the form \(y = mx + b\text{.}\)

    Checkpoint 5.1.22.

    Write an equation for the line through the point \((-7,-6)\) and parallel to the line \(y = 8 x - 11\) in point-slope form \(y = y_0 + m(x-x_0)\text{.}\)
    \(y =\) +

    Question 5.1.23.

    How can you tell the lines \(y = 3x+1\) and \(y=3x+17\) never intersect, without doing any algebra?

    Question 5.1.24.

    Do the graphs of \(f(x) = 3x+1\) and \(g(x) = 25 + 3(x-8)\) intersect? Why or why not?

    Subsection 5.1.3 Applications of Linear Functions

    Linear functions arise in real-world applications involving one variable that changes at a constant rate relative to another. Once the linear function is found, the problems usually boil down to
    • finding an output for a given input (evaluating),
    • finding the input for a given output (solving a simple equations), or
    • finding when two linear functions are equal (find when graphs intersect).
    Each of these problems is actually relatively straightforward from an algebraic point of view, given that you have formulas for your functions. The main difficulty is turning the words of the real world into mathematics.

    Example 5.1.25.

    As one climbs a mountain, the air temperature decreases by about \(3^{\circ}\)F for every \(1000\) foot increase in elevation. Suppose the temperature is \(60^{\circ}\)F when at an elevation of \(2000\) feet above sea level. Suppose we want the answer to the following questions:
    1. What will the temperature be at an elevation of \(14,439\) feet?
    2. What should the elevation be to be a “crisp” room temperature of \(65^{\circ}\)F?
    Solution.
    Our first task must be to find a formula for temperature as a function of elevation. To achieve this, we first have to name variables (this can be hard at first). Let \(T\) be the temperature, in \(^{\circ}\)F, at an elevation \(h\) feet above sea level. The problem describes a rate of change of \(-3/1000\) in \(^{\circ}\)F/ft; this will be the slope of our linear function. This slope can often be seen using the units as the units of slope must be output units divided by input units. The sign is negative because the temperature decreases at this rate as \(h\) increases. The problem then describes a point on the graph of this function, \((2000,60)\text{.}\) Using point-slope form, we have
    \begin{equation*} T = -\frac{3}{1000}(h-2000) + 60. \end{equation*}
    The answer to part (a) is then obtained by substituting \(14,493\) for \(h\) to get
    \begin{equation*} T = -\frac{3}{1000}(14493-2000) + 60\approx 22.521^{\circ}\mbox{F}. \end{equation*}
    To answer part (b), we need to substitute \(65\) for \(T\text{,}\) and solve for \(h\text{:}\)
    \begin{align*} 65 = \amp\ -\frac{3}{1000}(h-2000) + 60\\ \updownarrow \amp \\ h = \amp\ \frac{1000}{3}\approx 333.33\ \text{ft}. \end{align*}

    Checkpoint 5.1.26.

    A theater manager graphed weekly profits as a function of the number of patrons and found that the relationship was linear. One week profit was $10,444 when 1316 patrons attended. Another week 1526 patrons produced a profit of $12,334.
    (a) Find a formula for weekly profit, \(y\text{,}\) as a function of the number of patrons, \(x\text{.}\) Do not enter any commas in your formula.
    \(y =\)
    (b) How much will profits increase if 1 more patron goes to the theater?
    $ (no comma in your answer)
    (c) What number of patrons is closest to the break-even point (that is, the number of patrons for which there is as close to zero profit as possible)?
    number = patrons (no comma in your answer)
    d) Find a formula for the number of patrons as a function of the profit. Do not enter any commas in your formula.
    \(x = \ \)
    f) If the weekly profit was $15,934, how many patrons attended the theater? number = patrons (no comma in your answer)

    Question 5.1.27.

    Consider the example 5.1.25. Why is the formula for \(T\) as a function of \(h\) given in the last example more useful to a climber whose elevation is near 2000 feet than to one near sea level? What equivalent formula would be more useful to someone near sea level? Explain.

    Question 5.1.28.

    The ideal gas law says that
    \begin{equation*} PV = nR(T-z), \end{equation*}
    where \(P\) is the pressure of a gas, \(V\) is its volume, \(T\) is its temperature in \(^{\circ}\)C, \(n\) is the number of moles of gas, \(R\) is a constant, and \(z\) is the value of absolute zero in \(^{\circ}\)C. If \(n\) and \(V\) are constant, like when the gas is in a sealed container, we have a linear function for \(P\) as a function of \(T\text{:}\)
    \begin{equation*} P = \frac{nR}{V} (T-z). \end{equation*}
    1. What is the slope of the graph of \(P\) (just in symbols, not the number)?
    2. Suppose we measure \(P\) to be \(3.5\) when \(T = 70\text{,}\) and \(3.602\) when \(T = 80\text{.}\) Find the slope of the graph of \(P\) from this data.
    3. Use the data and the value of the slope from part (b) to give an approximate value of \(z\text{.}\) Is this close to the value you may have learned in a science class (or can look up)?
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