A week of MAT 117.

When does the cheaper plan stop being cheaper?

MAT 117 · Linear functions as rates of change · about 690 words

A lot of algebra comes down to one idea: a quantity that changes at a steady rate. A taxi that charges the same amount per mile, a tank that drains the same amount every minute, a weekly deposit into savings. Each of these can be described by a linear function, and once you can write it down, you can answer questions about it quickly.

A linear function has the form y = mx + b. The letter x is the input, often time or quantity. The letter y is the output, often cost, distance or amount. The two numbers that make each situation different are m and b.

The slope, m, is a rate of change. It tells you how much y changes when x goes up by 1. If a gym charges $15 per month, the slope is 15 dollars per month. A slope is always output units per input unit. If the slope is negative, the quantity is shrinking, like a candle that loses 2 centimeters per hour.

The intercept, b, is the starting value. It is the value of y when x is 0. If the gym also charges a $40 sign-up fee, you owe $40 before a single month has passed. So the cost after x months is y = 15x + 40. After 6 months that is 15(6) + 40 = $130.

When a problem gives you two data points instead of a rate, find the slope by dividing the change in output by the change in input: m = (y2 - y1) / (x2 - x1). Then substitute one point into y = mx + b to solve for b.

Worked example 1. A 4-mile taxi ride costs $13 and a 10-mile ride costs $25. Assume the cost is linear. The slope is (25 - 13) / (10 - 4) = 12 / 6 = 2, so the taxi charges $2 per mile. To find b, use the first point: 13 = 2(4) + b, so 13 = 8 + b and b = 5. The function is y = 2x + 5: a $5 base fare plus $2 per mile. Check with the other point: 2(10) + 5 = 25. It works.

Worked example 2. A water tank holds 300 liters and drains at 12 liters per minute. The starting value is 300 and the rate is -12, negative because the amount goes down. So W = -12t + 300, where t is minutes. When is it empty? Set W = 0: 0 = -12t + 300, so 12t = 300 and t = 25. The tank is empty after 25 minutes.

Many real decisions come down to comparing two linear functions: one plan with a low start and a high rate, another with a high start and a low rate. For small inputs the first is cheaper. For large inputs the second wins, because its smaller slope eventually makes up for its bigger head start. Where the two cost the same is the break-even point.

Worked example 3. Streaming Service A costs $8 per month with no fee. Service B costs $5 per month plus a one-time $24 device fee. Plan A is y = 8x and Plan B is y = 5x + 24. Set them equal: 8x = 5x + 24. Subtract 5x from both sides to get 3x = 24, so x = 8. At 8 months, both cost $64. Check each side: at 4 months A costs $32 and B costs $44; at 12 months A costs $96 and B costs $84.

Solving the equation found one number, 8. But the real question, which plan is better, needed another step: figuring out which side of 8 you are on and why. A good answer to a word problem always returns to the words. State the answer in context, with units, and test it with a value on each side.

Real situations are rarely perfectly linear. Phone plans have data caps, gyms raise prices, and tanks drain more slowly as they empty. A linear model is a useful approximation over a limited range, and part of using one well is knowing where it stops being true.

Sources: J. Abramson, College Algebra, OpenStax (2017); J. Stewart, L. Redlin and S. Watson, College Algebra, 7th ed., Cengage (2016).