Linear functions and models

Learning objectives

What you should be able to do

Evaluate and interpret function notation; connect equations, tables, graphs, and descriptions; find a rule from a rate or two observations; interpret initial and anchored values; convert units; and use changes, transformations, and unknown inputs without unnecessary algebra.

Input, output, and a constant rate of change

A function assigns exactly one output to each allowed input. In f(x), the letter f is the function’s name and x is the input. The notation does not mean f multiplied by x. If f(x) = 3x + 4, then f(2) asks for an output, while f(x) = 2 asks which input produces the output 2.

A linear function is written f(x) = mx + b.

Its graph is the nonvertical line y = mx + b. The coefficient m is the rate of output change per unit of input; b = f(0) is the output at input zero. A constant function, with m = 0, is a horizontal line and still gives one output for each input. A vertical line is not the graph of y as a function of x because one x value would correspond to many y values.

Information given

What to do with it

A rule and an input

Substitute the input everywhere the variable appears.

A rule and an output

Set the function expression equal to that output and solve.

A rate and one input-output pair

Use f(x) = f (a) + m(x − a).

Two pairs, (a, f (a)) and (b, f (b)), a ≠ b

Compute m = [ f (b) − f (a)]/(b − a), then use either pair.

A table with unequal input gaps

Compare output change divided by input change, not output change alone.

Why the intercept disappears in a difference. For f(x) = mx + b,

f (v) − f (u) = (mv + b) − (mu + b) = m(v − u).

This means a problem asking only about a change may not require b at all. A six-unit input change always produces six times the slope as the output change, regardless of where the interval starts.

The two quantities you must not confuse

Value: f (10) is the output at input 10.

Change: f (10) − f(4) is the difference between two outputs.

Rate: [ f (10) − f(4)]/(10 − 4) is the output change per input unit. These are three different questions, usually with different units.

A table is evidence, not an unlimited rule by itself. Constant rates show that the displayed points lie on a line. When the problem states that the function is linear, two distinct input-output pairs determine its rule for all allowed inputs. Do not assume arbitrary real-world data follow a line beyond the stated model.

Build and interpret linear models

A model is useful only when its variables and units are clear. If C(n) = 4n + 70 represents a cost in dollars for producing n items, 4 means dollars per additional item and 70 is the modeled fixed cost. It is not necessary to produce zero items in practice for the equation to have an algebraic intercept; interpret the intercept only as far as the context supports.

Initial value versus a value at a different starting time. In

H(t) = 72 + 1.5(t − 8),

72 is the value at t = 8, not at t = 0. This anchored form is exactly H(t) = H(8) + m(t − 8): it measures change from a known point. Expanding gives H(t) = 1.5t + 60, so the value at zero is 60.

A zero of the function. A zero is an input at which the output equals zero. In a draining-tank model, a zero may mark the predicted emptying time. It is an input value, not the y-intercept. A model that predicts negative water after that time should normally be restricted before those outputs occur.

Domain and range. The domain consists of allowed inputs; the range consists of outputs produced by those inputs. An abstract linear function is normally defined for all real inputs. A model can be restricted: time might satisfy 0 ≤ t ≤ 30, while the number of completed items may be a nonnegative integer.

Translate units before comparing coefficients

If h is hours and t is minutes, then h = t/60. A rule C(h) = 42h + 15 becomes C(t) = 42(t/60) + 15 = 0.7t + 15. The hourly rate changes to a per-minute rate; the fixed fee does not change. If the output is instead changed from dollars to cents, multiply the entire output by 100, including both the variable charge and the fixed charge.

Direct proportion is a special case. The rule y = kx has intercept zero, constant ratio y/x = k for x ≠ 0, and a graph through the origin. A general linear rule y = mx + b need not have a constant output-to-input ratio when b ≠ 0. Constant change is the test for linearity; constant ratio is the test for direct proportionality.

Fixed increase versus percent increase. Adding a fixed amount per equal time interval produces a linear model. Increasing by a fixed percentage of the current amount generally produces an exponential model, which belongs in Advanced Math. A percentage of an original fixed amount may instead be a constant addition; read what the percentage is taken of.

Use changes and transformations strategically

A change can be enough information. If a linear function rises by 12 when its input rises by 4, its slope is 3. That determines every other output change, but it does not determine an absolute output until at least one input-output pair is known. Know which kind of information the target needs.

Subtract observations before solving for coefficients. From f (u) = mu + b and f (v) = mv + b, subtraction removes b. This works even when the input is an unknown letter. It is often more efficient than treating the observations as two unrelated equations.

New function

Expanded form, if f(x) = mx+b

Useful interpretation

g(x) = f(x) + c

mx + (b + c)

Every output shifts by c; slope is unchanged.

g(x) = f(x − h)

mx + (b − mh)

Right shift by h; input is evaluated before the function.

g(x) = a f(x)

amx + ab

All outputs are scaled by a.

g(x) = f (ax + d) + c

amx + (md + b + c)

Input scaling changes the slope to am; constants must be collected carefully.

These formulas follow by substitution, so memorizing a large transformation chart is unnecessary. To compute f (2x − 3), replace the input in the rule for f with the entire expression 2x − 3. Keep parentheses until distribution is complete.

Combine functions with matching inputs and units. If R(n) is revenue in dollars and C(n) is cost in dollars for the same number n of items, then profit is P(n) = R(n)−C(n). Subtract the entire cost expression. An initial cost reduces profit; it does not disappear because it is not attached to n.

Choose what must be determined

Asked for a value? You generally need a rate and an anchor.

Asked for a difference? The slope may be enough.

Asked for a parameter in a transformed rule? Substitute the given input-output condition and solve.

Asked for a physical prediction? Check both the algebra and the model’s valid input range.

Calculator check. Numeric function evaluations and intersections are useful checks. A graph does not explain what m means in liters per minute, whether the variable counts completed items, or why a constant is an initial value rather than an anchored value. Those interpretations must come from the definitions.

15 worked examples

3.01. Evaluate a function at a negative input

The function f is defined by f(x) = −3x + 8. What is f(−2)?

A) 2 B) −14 C) 14 D) −6

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Recognize the structure. Replace the input variable with −2, using parentheses.

Work it through. Substitute directly:

f(−2) = −3(−2) + 8 = 6 + 8 = 14.

Answer: C, 14.

Check. Moving from input 0 to input −2 is a change of −2. With slope −3, the output changes by (−3)(−2) = 6, from 8 to 14.

Avoid the trap. Function notation is not multiplication. Also, a negative slope does not imply that every output is negative.

3.02. Find the input for a specified output

For f(x) = 2.5x + 7, what value of x satisfies f(x) = 27?

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Recognize the structure. The number 27 is an output, so set the rule equal to 27.

Work it through. Solve 2.5x + 7 = 27 2.5x = 20 x = 8.

Answer: 8.

Check. f(8) = 2.5(8) + 7 = 27.

Avoid the trap. Evaluating f (27) answers a different question. Here the unknown is the input that produces 27, not the output produced by input 27.

3.03. Recover a function from an unevenly spaced table

The table gives values of a linear function f . What is f(7)?

x

1

4

10

f(x)

18

30

54

A) 14 B) 28 C) 42 D) 54

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Recognize the structure. The slope is output change divided by input change. Then use a nearby known input.

Work it through. Between the first two rows, m = (30 − 18)/(4 − 1) = 12/3 = 4. From input 4 to input 7 is an increase of 3, so the output increases by 4(3) = 12. Therefore f(7) = 30 + 12 = 42.

Answer: C, 42.

Check. The final interval also gives rate (54 − 30)/(10 − 4) = 24/6 = 4. The rule is f(x) = 4x + 14.

Avoid the trap. The output changes 12 and 24 are different because the input changes are different. Their rates still agree.

3.04. Give a graph’s slope its contextual meaning

The graph models water volume V, in liters, after t minutes. At what rate is the volume changing?

Water volume V against time t: a descending line passes through (0, 180) and (12, 60), then reaches the time axis at 18. Dashed guides mark t = 12.tV(0,180)(12,60)1218
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Recognize the structure. Use the volume difference over the time difference.

Work it through. The rate is (60 − 180)/(12 − 0) = −120/12 = −10 liters per minute.

Answer: The volume changes at −10 liters per minute: it decreases by 10 liters each minute.

Avoid the trap. The value 180 is the initial volume, not the rate. The value 60 is the volume remaining at 12 minutes, not the amount lost.

3.05. Write a model and interpret a function value

A savings jar initially contains $50. Exactly $8 is added at the end of each week, with no withdrawals. Let S(w) be the amount after w completed weeks. Write S(w) and find S(6).

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Recognize the structure. The initial amount is the intercept; the weekly addition is the slope.

Work it through. After w completed weeks, the additions total 8w dollars. Thus

S(w) = 50 + 8w, S(6) = 50 + 8(6) = 98.

Answer: S(w) = 50 + 8w; S(6) = 98, meaning $98 after 6 completed weeks.

Key takeaway. The relevant domain is nonnegative whole numbers of completed weeks. The line between those inputs is an algebraic extension, not a claim that deposits happen continuously.

Avoid the trap. The phrase “initially contains” refers to S(0), not to the amount after the first week.

3.06. Separate a per-item cost from a fixed cost

A linear function C(n) gives the total cost, in dollars, of producing n items. If C(40) = 230 and C(90) = 430, what fixed cost does the model imply?

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Recognize the structure. The cost increase reveals the per-item rate; then one observation reveals the fixed part.

Work it through. The rate is m = 430 − 23090 − 40 = 20050 = 4.

Write C(n) = 4n + b. Using C(40) = 230 gives 230 = 160 + b, so b = 70.

Answer: $70.

Check. For 90 items, 4(90) + 70 = 430. Both observations fit.

Avoid the trap. 230/40 is an average cost per item at one production level. It includes the fixed cost and is not the model’s slope.

3.07. Solve within a model’s physical domain

A tank initially holds 540 liters of water and loses 18 liters per minute at a constant rate. After how many minutes will 180 liters remain?

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Recognize the structure. Write remaining volume as initial volume minus the amount lost.

Work it through. The model is V(t) = 540 − 18t. Set the desired volume equal to 180:

540 − 18t = 180 18t = 360 t = 20.

Answer: 20 minutes.

Check. The tank is modeled as empty at 540/18 = 30 minutes. The answer lies in the physical interval 0 ≤ t ≤ 30.

Avoid the trap. Dividing 180 by 18 gives the time needed to lose the remaining amount, not the time already elapsed. The amount lost is 540 − 180 = 360.

3.08. Read an anchored value correctly

A plant’s height H(t), in centimeters, is modeled for 0 ≤ t ≤ 20 by H(t) = 72 + 1.5(t − 8), where t is weeks after observation began. What height does the model predict at t = 0?

A) 8 B) 60 C) 72 D) 84

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Recognize the structure. The value 72 is attached to t = 8, because that makes t − 8 = 0.

Work it through. Substitute t = 0:

H(0) = 72 + 1.5(0 − 8) = 72 − 12 = 60.

Answer: B, 60 centimeters.

Another route. Expand to H(t) = 1.5t + 60 and read its intercept.

Avoid the trap. A constant written first is not automatically the value at input zero. Read the structure around the variable before naming an initial value.

3.09. Change the input unit without changing the fixed fee

A service costs C(h) = 42h + 15 dollars for h hours. Which rule gives the same cost for t minutes?

A) C(t) = 42t + 15 B) C(t) = 2520t + 15 C) C(t) = 0.7t + 0.25 D) C(t) = 0.7t + 15

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Recognize the structure. Minutes must be divided by 60 to obtain the input measured in hours.

Work it through. Since h = t/60,

C(t) = 42 (t60) + 15 = 0.7t + 15.

Answer: D, C(t) = 0.7t + 15.

Check. At t = 60 minutes, the new rule gives 42 + 15 = 57, matching the original rule at one hour.

Avoid the trap. Only the time-dependent part is converted. The $15 fixed fee remains dollars; it is not a rate per hour.

3.10. Distinguish linear from directly proportional

A linear function has values f(2) = 9, f(4) = 15, and f(8) = 27. Is f(x) directly proportional to x? Explain using its rule.

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Recognize the structure. Direct proportionality requires a zero intercept, not merely a constant slope.

Work it through. The slope is (15 − 9)/(4 − 2) = 3, also matching (27 − 15)/(8 − 4) = 3. Using f(2) = 9 in f(x) = 3x + b gives b = 3. Thus f(x) = 3x + 3.

Answer: No. The rule is f(x) = 3x + 3, with nonzero intercept 3.

Check. The ratios 9/2, 15/4, and 27/8 are different, though the rates of change are equal.

Key takeaway. A proportional relationship is linear, but a linear relationship does not have to be proportional. This explanatory example targets the distinction, not a separate SAT response format.

3.11. Find a change without finding the intercept

The function f is linear and f(7) − f(2) = 20. What is f (12) − f(9)?

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Recognize the structure. Every output difference equals slope times input difference.

Work it through. The given interval has length 7 − 2 = 5, so its slope is m = 20/5 = 4. The requested interval has length 12 − 9 = 3, hence

f (12) − f(9) = 4(3) = 12.

Answer: 12.

Key takeaway. The intercept is unknown and unnecessary. All functions of the form f(x) = 4x + b give the same requested difference.

Avoid the trap. You cannot determine f (12) alone from this information. The question is solvable precisely because it asks for a difference.

3.12. Subtract an entire cost model to build profit

A business has revenue R(n) = 18n dollars and cost C(n) = 6n + 420 dollars when it produces and sells n items. How many items must it sell to make a profit of $180?

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Recognize the structure. Profit equals revenue minus total cost, including the fixed expense.

Work it through. The profit model is

P(n) = 18n − (6n + 420) = 12n − 420.

Set P(n) = 180: 12n − 420 = 180, so 12n = 600 and n = 50.

Answer: 50 items.

Check. Revenue is $900 and cost is $720, leaving $180. The required count is a nonnegative integer.

Avoid the trap. Subtracting only the variable cost would ignore the fixed $420. Writing 18n − 6n + 420 changes its sign incorrectly.

3.13. Recover a function, then transform its input

The linear function f satisfies f(1) = 7 and f(5) = 19. If g(x) = f (2x − 3) + 4, what is g(x)?

A) 6x − 1 B) 6x + 5 C) 3x + 5 D) 6x + 17

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Recognize the structure. First determine the rule for f; then substitute the entire new input.

Work it through. The slope is (19 − 7)/(5 − 1) = 3. Since f(1) = 7, the intercept is 4, so f(x) = 3x + 4. Therefore g(x) = [3(2x − 3) + 4] + 4 = 6x − 9 + 8 = 6x − 1.

Answer: A, g(x) = 6x − 1.

Check. At x = 2, the inner input is 1. The definition gives g(2) = f(1) + 4 = 11, matching 6(2) − 1 = 11.

Avoid the trap. The outer +4 is added after evaluating f. The inner −3 is multiplied by the slope during evaluation.

3.14. Use differences when an input is unknown

A linear function is defined by f(x) = mx + b. For a constant a, f (a) = b + 18 and f (a + 4) = b + 30. What is a?

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Recognize the structure. Subtract the two function values to remove b and reveal the slope.

Work it through. The input increases by 4 and the output increases by (b + 30) − (b + 18) = 12, so 4m = 12 and m = 3. Next use the first condition:

ma + b = b + 18 ma = 18.

Thus 3a = 18, giving a = 6.

Answer: 6.

Check. For any b, f(6) = b + 18 and f (10) = b + 30 when the slope is 3.

Avoid the trap. The unknown intercept is not an obstacle. Both observations include it in a way that cancels. Do not assume b = 0.

3.15. A shifted function must meet a new value condition

A linear function f satisfies f(2) = 5 and f(7) = 20. The function g is defined by g(x) = f(x) + c. If g(4) = 2g(1), what is c?

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Recognize the structure. Determine the two needed values of f, then apply the condition on g.

Work it through. The slope is (20 − 5)/(7 − 2) = 3. From 5 = 3(2) + b, b = −1, so f(x) = 3x − 1. Consequently f(4) = 11 and f(1) = 2. The given condition becomes

11 + c = 2(2 + c) = 4 + 2c c = 7.

Answer: 7.

Check. Now g(x) = 3x + 6, so g(4) = 18 and 2g(1) = 2(9) = 18.

Avoid the trap. The factor 2 multiplies the entire value g(1), including the added constant c. It is not 2 f(1) + c.

Mastery check and error prevention

Attempt the exit check below, then reveal each answer and explanation.

Topic practice

Before leaving this section

Check that you can distinguish an input from an output, a value from a change, an initial value from an anchored value, and a constant rate from a constant ratio. State a contextual slope with both units and direction.