Create equations in two or more variables for a relationship and graph them on labeled, scaled axes

Learning goal: Create equations in two or more variables to represent relationships between quantities, including linear, simple exponential and simple quadratic relationships and, with technology where it helps, more complicated ones; and graph those equations on coordinate axes with labels and scales chosen to show the relationship clearly.

Why two variables. An equation in one variable answers one question: for what number is this true? An equation in two variables describes a whole relationship: for every value of one quantity, what is the other? A plumber who charges $40 to come out plus $15 per hour has a cost C that depends on the hours h, and the relationship is C = 40 + 15h. That single equation contains every possible bill, and its graph shows all of them at once. When the situation has three quantities that vary, the equation has three variables: distance, rate and time are tied by d = rt.

Creating the equation. The method is the same translation as for one variable. Name each quantity with a letter and a unit: h hours, C dollars. Decide which quantity depends on which; the independent quantity goes on the horizontal axis. Write the relationship from the sentences in the situation: a fixed amount plus a rate times a quantity is linear; a quantity that is multiplied by the same factor each period is exponential; a quantity that is a product of two lengths that both depend on one variable is quadratic. Then state what values of the input make sense, because the graph should be drawn only there: hours are not negative, and a width cannot exceed the whole perimeter.

A fully worked model: a quadratic relationship. A rectangle is to be made from 40 meters of fencing. How does its area depend on its width? Let w be the width in meters. The perimeter is 2w + 2l = 40, so the length is l = 20 − w, and the area is A = w(20 − w) = 20w − w². This is an equation in two variables, A and w. The width must satisfy 0 < w < 20, since both w and 20 − w must be positive; the endpoints w = 0 and w = 20 are shown in the table and graph only as limiting cases (a flat "rectangle" of area 0), drawn as open circles in the diagram and when sketching by hand. Make a table with a sensible step: w = 0, 2, 4, …, 20 gives A = 0, 36, 64, 84, 96, 100, 96, 84, 64, 36, 0. The graph, in the diagram, is a parabola opening downward, symmetric about w = 10, where the area is greatest at 100 square meters. The graph makes visible what the equation states: the same area occurs at two widths (w = 4 and w = 16 both give 64), and the largest area is the square.

Labels and scales. A graph without labels is a curve, not a representation of a relationship. Each axis carries the quantity's name and unit: "width w (m)" and "area A (m²)". Each axis has a scale: equally spaced tick marks with numbers, chosen so the relevant range of the data fits and the shape is visible. For the area graph the horizontal scale runs 0 to 20 in steps of 2 and the vertical scale 0 to 100 in steps of 20. Choosing the scale is a decision: for the bacteria equation P = 50·2ʰ, the values after 0 to 5 hours are 50, 100, 200, 400, 800 and 1600, so a vertical scale to 1600 in steps of 200 is needed; a scale to 100 would show only two points, and a scale to 10,000 would flatten the curve into the axis. The two axes may use different scales, and often must.

Linear and exponential relationships. A tank holding 300 liters is filled at 25 liters per minute: V = 300 + 25t. The graph is a straight line that crosses the vertical axis at 300 and rises by 25 for each minute; two points, (0, 300) and (10, 550), fix it, and a third, (4, 400), checks it. A population of 200 that grows by 50% each year is P = 200(1.5)ᵗ: 200, 300, 450, 675, 1012.5, 1518.75 for t = 0 to 5. The graph is a curve that rises faster and faster, never a straight line, because each year adds more than the year before. Plotting these two on the same axes would be a mistake, since liters and people are different quantities.

More than two variables. A shopper with $20 buys apples at $1.25 each and oranges at $0.80 each, spending all of it: 1.25a + 0.80b = 20. This is a relationship between two quantities, and its graph is a straight segment from (16, 0), sixteen apples and no oranges, to (0, 25), since 1.25 × 16 = 20 and 0.80 × 25 = 20. Only whole-number points on the segment are purchases. With a third fruit at $0.50, the equation 1.25a + 0.80b + 0.50c = 20 has three variables; it still describes the relationship precisely, though its graph is a surface and a table or technology is the practical way to explore it.

Graphing with technology. For a complicated equation, for instance a height that follows h = −4.9t² + 30t + 2 with decimal coefficients, or an exponential with an awkward base, a graphing tool produces the curve quickly and accurately. The tool does not choose the labels, the units or the window; those remain the student's job, and a graph produced by technology is judged by the same standard: named axes with units, a scale that shows the important features, and a domain that matches the situation.

The misconception to remove. Students often plot the table's points and connect them without asking whether the relationship is really continuous. The area of the rectangle varies continuously with the width, so a smooth curve is right. The number of apples is a whole number, so only the lattice points on the segment are meaningful, and the segment is drawn to show the relationship, not to claim that 3.7 apples can be bought. Stating which applies is part of graphing well.

About the numbers. Every price, fee, amount and business in this lesson's examples, tasks and questions is made up for practice; none describes a real company, product or offer.

Worked linear relationship with intercepts: A shopper spends exactly $20 on apples at $1.25 and oranges at $0.80. Create the equation and graph it. Let a be the number of apples and b the number of oranges: 1.25a + 0.80b = 20. When b = 0, a = 16; when a = 0, b = 25. Axes: "apples a" horizontally, scale 0 to 20 by 2; "oranges b" vertically, scale 0 to 25 by 5. The graph is the segment joining (16, 0) and (0, 25). Check a third point: a = 8 gives 10 + 0.80b = 20, so b = 12.5; the point (8, 12.5) lies on the segment, though it is not a possible purchase because 12.5 oranges is not a whole number.

Worked exponential relationship and its scale: A colony of 50 bacteria doubles every hour. Create the equation, tabulate it and choose scales. P = 50·2ʰ. Table for h = 0 to 5: 50, 100, 200, 400, 800, 1600. Horizontal axis "time h (hours)", 0 to 5 by 1; vertical axis "population P", 0 to 1600 by 200. The plotted points rise slowly at first and then steeply; the smooth curve through them is the graph. The doubling shows on the graph as the height doubling for every step of one to the right, which no straight line does.

Worked quadratic relationship with a restricted domain: A ball is kicked so that its height in meters after t seconds is h = −5t² + 20t, an idealized model that ignores air resistance and rounds half of gravity's acceleration, about 9.8 meters per second per second, to 5. Create the table and graph. For t = 0, 1, 2, 3, 4 the heights are 0, 15, 20, 15, 0. Only 0 ≤ t ≤ 4 makes sense, because before the kick and after landing the ball is on the ground. Axes: "time t (s)", 0 to 4 by 1; "height h (m)", 0 to 20 by 5. The graph is the arch through (0, 0), (1, 15), (2, 20), (3, 15), (4, 0), highest at t = 2.

Worked conversion relationship: Temperature in Fahrenheit depends on Celsius by F = 1.8C + 32. Graph it for C from 0 to 40. Values: C = 0, 10, 20, 30, 40 give F = 32, 50, 68, 86, 104. Axes: "Celsius C (°C)" 0 to 40 by 10; "Fahrenheit F (°F)" 0 to 110 by 10. The graph is a straight line, and the vertical intercept 32 is the freezing point of water in Fahrenheit. The slope 1.8 means each Celsius degree is 1.8 Fahrenheit degrees.

Worked more complicated relationship, graphed with technology: A club rents a hall for a made-up fee of 400 dollars plus 12 dollars of food per member. The cost per member for n members is C = (400 + 12n)/n = 400/n + 12, a rational relationship. The sensible domain is whole numbers n ≥ 1. Values: C(10) = 52, C(20) = 32, C(50) = 20, C(100) = 16. A graphing tool draws the curve quickly, but the window is the student's decision: with n from 0 to 100 and C from 0 to 100 (labels "members n" and "cost per member C (dollars)"), the curve falls steeply and then levels off toward the line C = 12 without reaching it, because the hall-fee share 400/n keeps shrinking while the 12-dollar food cost never goes away. The first few values, such as C(1) = 412, fall above this window, a deliberate choice that keeps the leveling-off visible; a window ending at n = 10 would show only the steep part and hide the behavior the question is about.

Representation task: For the fencing situation, A = w(20 − w), draw the axes with the labels "width w (m)" and "area A (m²)", the horizontal scale 0 to 20 in steps of 2 and the vertical scale 0 to 100 in steps of 20. Plot the eleven points from the table in the lesson, using open circles for the two limiting endpoints (0, 0) and (20, 0), and draw the smooth curve through them. Mark the greatest area with its coordinates and mark the two widths that give an area of 64. Write beneath the graph one sentence explaining why the curve stops at w = 0 and w = 20.

Independent performance: Choose three relationships from everyday situations, one linear, one exponential and one quadratic, and for each: define the variables with units, create the equation, state the sensible domain, build a table of at least six points, choose and justify the scale of each axis, and draw the graph with labeled axes. For the quadratic, identify the highest or lowest point and say what it means in the situation. Then create one equation with three variables for a situation of your choice, and explain in two sentences why a table rather than a single graph is the practical way to show it. Success criteria: every axis has a name, a unit and evenly spaced numbered ticks; every table value satisfies the equation; the domain matches the situation; and the graphs are drawn only over that domain.