Compare properties of two functions when each is represented differently: algebraically, graphically, in a table or in words

Learning goal: Compare properties of two functions each represented in a different way, algebraically, graphically, numerically in a table, or by a verbal description, by extracting the same property from each representation and then comparing. The functions are linear, quadratic and exponential, and the properties are intercepts, rate of change, maximum or minimum, growth factor, and values at chosen inputs.

The method. Every representation hides some properties and displays others. An equation displays coefficients; a graph displays shape and intercepts; a table displays values at particular inputs; a description displays the situation. A comparison question therefore has three steps. Name the property being compared. Extract that property from each function using the tools of its own representation. Compare the two extracted values, stating which function has the larger one and by how much. Comparing a number read off a graph with a coefficient in an equation without first converting them to the same property is the error to avoid.

A fully worked quadratic comparison. Function f is given by a graph, drawn in the figure for this lesson. Before reading on, read from the figure its vertex, its vertical intercept and its values at x = 1 and x = 5. The graph is a parabola opening downward with vertex at (2, 7), passing through (0, 3) and (4, 3). Function g is given algebraically: g(x) = −x² + 6x − 1. Which has the larger maximum? From the graph, the maximum of f is the vertex height, 7, at x = 2. From the equation, the axis of symmetry of g is x = −6/(2 · (−1)) = 3 and g(3) = −9 + 18 − 1 = 8, so the maximum of g is 8, at x = 3. Function g has the larger maximum, by 1. Compare intercepts too: f has vertical intercept 3, read from the graph; g has vertical intercept g(0) = −1, read from the constant term. At the input x = 1, f must be read from the graph, where the curve passes through (1, 6); as a check, if the parabola has the form a(x − 2)² + 7 with 3 = 4a + 7, then a = −1 and f(1) = −1 + 7 = 6, while g(1) = −1 + 6 − 1 = 4. The comparison changes with the property: g wins on maximum, f wins on vertical intercept and on the value at x = 1.

Linear comparisons: rate and intercept. A linear function's two properties are its rate of change and its initial value. In an equation y = mx + b they are m and b. In a table, the rate is the change in output divided by the change in input between two rows, and the initial value is the output at input 0, which may have to be found by working backward with the rate. On a graph, the rate is the slope from two clearly readable points and the initial value is the vertical intercept. In a hypothetical description such as a plumber charges 60 dollars to visit plus 45 dollars per hour, the rate is 45 and the initial value is 60. Suppose a second, hypothetical plumber's charges are tabulated: 2 hours, 130 dollars; 4 hours, 210 dollars. The rate is (210 − 130)/(4 − 2) = 40 dollars per hour, and the initial value is 130 − 2 · 40 = 50 dollars. The first plumber has the higher hourly rate, 45 against 40, and the higher visit fee, 60 against 50, so the first is more expensive for every positive number of hours; if the fees had been reversed the answer would depend on the number of hours, and the comparison would need the input where the two costs are equal.

Exponential comparisons: starting value and growth factor. An exponential function's properties are its starting value and its growth factor per unit, equivalently its percent rate of change. In the equation y = 100(1.4)ˣ they are 100 and 1.4, a 40 percent increase per unit. In a table with equally spaced inputs, the ratio of consecutive outputs is the growth factor per step of the table; when the inputs go up by 1, that is the growth factor per unit. Suppose a table gives 80, 120, 180, 270 at x = 0, 1, 2, 3; the ratios are 120/80 = 1.5, 180/120 = 1.5, 270/180 = 1.5, so the table's function is 80(1.5)ˣ. Compared with 100(1.4)ˣ, the equation has the larger starting value, 100 against 80, but the table has the larger growth factor, 1.5 against 1.4, a 50 percent rate against 40 percent. Which is larger at a given input depends on the input: at x = 0, 100 > 80; at x = 3, 100(1.4)³ = 274.4 and 80(1.5)³ = 270, so the equation is still ahead; at x = 4, 384.16 against 405, and the table has overtaken it. When both starting values are positive, the larger growth factor always wins eventually, however small its start, which is the theme of a later lesson.

Descriptions. A verbal description must be translated into the property before comparing. The sentence a car loses 15 percent of its value each year describes exponential decay with factor 0.85; the sentence a car loses 2,000 dollars of value each year describes a linear function with rate −2000. To compare them you need their starting values and a chosen number of years, because a percent loss shrinks in dollars as the car's value falls while a fixed dollar loss does not. Suppose both cars start at 20,000 dollars. After one year the first is worth 17,000 and the second 18,000; after five years the first is worth 20000(0.85)⁵ ≈ 8,874 dollars and the second 10,000 dollars; after ten years the first is worth about 3,937 dollars and the second is worth 0. The property compared, value after n years, has to be computed from each description in its own way.

Reporting a comparison. A complete answer names the property, gives the value found from each representation with a phrase saying how it was found, and states the conclusion: g has the larger maximum, 8 from the vertex of the equation against 7 from the vertex on the graph. A conclusion without the two values is a guess; two values without the method cannot be checked.

Worked linear comparison across a graph and a description: Function A is a line on a graph passing through (0, 4) and (3, 13). Function B is described: start at 10 and add 2 for each unit increase in the input. Rate of A: (13 − 4)/(3 − 0) = 3; rate of B: 2. Initial value of A: 4; of B: 10. A has the greater rate and B the greater initial value. They are equal where 3x + 4 = 2x + 10, at x = 6, where both equal 22; for inputs above 6, A has the larger output, and below 6, B does.

Worked quadratic comparison across a table and an equation: Function f is tabulated with symmetric values: f(0) = 5, f(1) = 8, f(2) = 9, f(3) = 8, f(4) = 5. Function g is g(x) = 2(x − 1)² + 3. The table's outputs rise to 9 at x = 2 and fall symmetrically, so f has a maximum of 9 at x = 2. The vertex form of g shows a minimum of 3 at x = 1 and no maximum, since the coefficient 2 is positive. Comparing minimums instead: f, being a downward parabola, has no minimum; g has minimum 3. Comparing vertical intercepts: f(0) = 5 from the table; g(0) = 2 · 1 + 3 = 5 from the equation; they are equal.

Worked exponential comparison across a table and a description: A table lists an investment at 500, 530, 561.80 dollars at years 0, 1, 2. A description says a second investment starts at 600 dollars and grows 4 percent per year. The table's ratios are 530/500 = 1.06 and 561.80/530 = 1.06, so the first grows 6 percent per year from 500. The second is 600(1.04)ᵗ. The second has the larger starting value; the first has the larger growth rate. After 10 years the first is 500(1.06)¹⁰ ≈ 895.42 dollars and the second is 600(1.04)¹⁰ ≈ 888.15 dollars, so the first has just overtaken the second.

Representation task: Make a two-column comparison chart for f, the graphed parabola with vertex (2, 7) through (0, 3), and g(x) = −x² + 6x − 1. Rows: representation, opens up or down, vertex, maximum value, axis of symmetry, vertical intercept, value at x = 1, value at x = 5. Fill each cell with the value and a short note on how it was found (read from graph, computed from equation), then write a paragraph stating which function is larger for each property.

Independent performance: Create three comparison problems and solve them. In the first, compare two linear functions, one from a table and one from a description, on rate and initial value, and find the input where they are equal. In the second, compare two quadratic functions, one from a graph you draw and one from an equation, on maximum or minimum and on vertical intercept. In the third, compare two exponential functions, one from a table and one from an equation, on starting value and growth factor, and find an input at which the one with the smaller start has overtaken the other. Success criteria: each property is extracted from each representation by the method appropriate to that representation and the method is named; every computed value is correct; and each conclusion states which function has the larger value of the named property and the two values being compared.