Compare the rate of change and initial value of two functions given as a table, graph, equation or description

Learning goal: Compare properties of two functions when each is represented in a different way, algebraically by an equation, graphically, numerically in a table of values or by a verbal description, by finding the same property in each form. Decide which function has the greater rate of change, which has the greater initial value and which gives the greater output for a chosen input.

One function, four forms. A single linear function can be shown as an equation such as y = 3x + 5, as a table of values, as a straight-line graph or as a sentence such as "start at 5 and add 3 for each step." A comparison question usually hands you two functions in two different forms and asks about a property: the rate of change, the initial value, or the output at some input. The mistake to avoid is comparing a number in an equation with a number in a table directly. Extract the property from each representation first, in the form's own way, and then compare the extracted numbers.

Rate of change. For a linear function the rate of change is the amount the output changes for each increase of 1 in the input. In an equation y = mx + b it is the coefficient m. In a table, pick two rows and divide the change in y by the change in x. In a graph, pick two points whose coordinates you can read exactly and do the same division. In a verbal description it is the amount added or removed per unit, such as "2 cm each week."

Initial value. The initial value is the output when the input is 0. In an equation y = mx + b it is b. In a table it is the y in the row where x = 0; if there is no such row, work backwards using the rate of change. On a graph it is the height where the line crosses the vertical axis. In a description it is the starting amount, such as "is 8 cm tall when planted."

Four functions to compare. All four are linear functions; a table or graph shows only some of a function's points, so that assumption is stated rather than deduced. Function A is given by a table: x = 0, 2, 4, 6 and y = 5, 11, 17, 23. Function B is given algebraically: y = 4x + 1. Function C is given verbally: "A seedling is 8 cm tall when planted and grows 2 cm each week," with x in weeks and y in centimeters. Function D is given graphically: it is the line drawn in this lesson's graph. Read two points where the line crosses grid intersections exactly; the marked ones are (0,3) and (5,13).

For A, the output rises by 6 when the input rises by 2, so the rate of change is 6 ÷ 2 = 3 per unit. The row x = 0 gives the initial value 5. So A is y = 3x + 5. The common error here is to report the rate as 6, the difference in y, without dividing by the change in x. Since the table steps by 2, not by 1, the change per one unit is half of 6.

For B, read the equation: the rate of change is 4 and the initial value is 1.

For C, the rate of change is 2 cm per week and the initial value is 8 cm, so C is y = 2x + 8.

For D, the rate of change is (13 − 3) ÷ (5 − 0) = 10 ÷ 5 = 2, and the initial value is 3 because the line passes through (0,3). D is y = 2x + 3.

Now the comparisons are direct. The greatest rate of change belongs to B (4 is greater than 3 and 2). The greatest initial value belongs to C (8). C and D have the same rate of change, 2, so their graphs are parallel lines; C is always 5 units above D and they never meet. A has a smaller rate than B but a larger initial value, so the two must cross: solve 3x + 5 = 4x + 1 to get x = 4, where both give 17. Before x = 4, A is larger (at x = 2, A gives 11 and B gives 9). After x = 4, B is larger (at x = 10, A gives 35 and B gives 41).

Output at a chosen input. Comparing two functions at one input means evaluating each. At x = 4: A gives 17, B gives 4 × 4 + 1 = 17, C gives 2 × 4 + 8 = 16 and D gives 2 × 4 + 3 = 11. A greater rate of change does not mean a greater output at every input; B has the greatest rate but is not the greatest at x = 4 alone, since A ties it, and it is smaller than A and C at x = 2.

Reading graphs carefully. A graph shows a rate only relative to its scales. Two lines drawn on different grids cannot be compared by how steep they look; compute the rate from coordinates. Choose points where the line crosses grid intersections so the reading is exact, and state that a read value is approximate if the crossing is not on a grid point.

A fully worked model. Function E is linear, with the table x = 1, 2, 3 and y = 9, 14, 19. Function F is the equation y = 6x − 2. Which has the greater rate of change, which has the greater initial value, and which is greater at x = 10? For E, y rises 5 when x rises 1, so the rate is 5. There is no x = 0 row; stepping back one unit from (1,9) subtracts 5, so the initial value is 4, and E is y = 5x + 4. For F, the rate is 6 and the initial value is −2. F has the greater rate of change (6 > 5); E has the greater initial value (4 > −2). At x = 10, E gives 54 and F gives 58, so F is greater there. They are equal where 5x + 4 = 6x − 2, at x = 6, where both give 34.

Comparing properties that are not rates. Two functions can also be compared on whether they increase or decrease, on where they are equal, or on which is larger over a range of inputs. The same principle applies: translate each representation into the property you need, then compare numbers, never appearances.

Worked table versus equation: A linear function's table gives x = 1, 3, 5 and y = 7, 13, 19; an equation gives y = 2.5x + 4. The table's rate of change is (13 − 7) ÷ (3 − 1) = 6 ÷ 2 = 3, and its initial value is 7 − 3 × 1 = 4, so the table is y = 3x + 4. The equation's rate is 2.5 and its initial value is 4. Both start at 4, but the table's function grows faster, so for every positive input the table's output is larger; at x = 8 the table gives 28 and the equation gives 24.

Worked description versus graph: A description says "a tank holds 20 liters and drains 4 liters each hour." A graph is a line through (0,12) and (3,0). The description has rate −4 and initial value 20. The graph has rate (0 − 12) ÷ (3 − 0) = −4 and initial value 12. The rates are equal, so the lines are parallel; the described tank starts with 8 liters more and empties at t = 5 hours (20 ÷ 4), two hours after the graphed tank empties at t = 3.

Worked graph versus equation, crossing point: A graph passes through (2,9) and (4,15); an equation is y = 2x + 7. The graph's rate is (15 − 9) ÷ (4 − 2) = 3 and its initial value is 9 − 3 × 2 = 3, so the graphed function is y = 3x + 3. The equation's rate is 2 and initial value 7. At x = 2 the equation is larger (11 versus 9); at x = 4 both give 15; at x = 6 the graph is larger (21 versus 19). Reading to the right, the function with the greater rate of change eventually overtakes the other, provided it starts below it; here it starts below and passes it at x = 4.

Representation task: Copy the four functions A, B, C and D from the lesson into one table with columns for the representation given, the rate of change, the initial value and the output at x = 4. Then draw all four on one pair of labeled axes from x = 0 to x = 10 and mark the point where A and B cross. Write two sentences: one comparing rates of change and one explaining why C and D never meet.

Independent performance: Create two linear functions of your own with different rates of change and different initial values. Present the first as a table with no x = 0 row and the second as a verbal description of a real or hypothetical situation. Swap with a partner. For the partner's pair, find each function's rate of change and initial value, state which is greater in each property, evaluate both at x = 3 and at x = 12, and find the input where they are equal. Success criteria: each rate is computed as change in y divided by change in x; the initial value of the table function is found by working back to x = 0; every comparison names the property being compared; the crossing input is checked by substituting into both functions.