Linear functions: fixed fees and constant rates of change

Tariff A charges 80 currency units for setup and 20 per hour. Tariff B charges 20 for setup and 35 per hour. A cheaper start need not mean a cheaper long rental. A linear function separates the initial fee from the constant cost of each additional hour.

Use equations and functions. These hypothetical tariffs cover one service, any duration $0\le t\le8$ hours, proportional fractional-hour charges and no other fees or rounding.

Two coefficients, two different roles

With numerical $t$ in hours and $C$ in currency units:

C_A(t)=80+20t,\qquad C_B(t)=20+35t.

These have the form $y=kx+b$ , a linear function in the school convention used here. $b$ is the value at input zero, here the setup fee for a booked but unused rental. $k$ is output change per input unit: 20 currency units/hour for A, 35 for B. The unit of $b$ is currency. Add the rate multiplied by time to the fee, not time alone.

For two distinct inputs:

k=\frac{y_2-y_1}{x_2-x_1}.

For A, moving from 1 to 3 hours changes cost from 100 to 140, so $k=(140-100)/(3-1)=20$ . Half an hour adds 10. This constant rate produces a straight line, or a segment on a restricted domain. Its screen angle also depends on axis scales, so units and increments are more reliable than visual steepness. Linear functions, OpenStax §4.1.

Where the tariffs agree

Equal cost means $80+20t=20+35t$ . Thus $60=15t$ , so $t=4$ hours and both costs are 160. The graphs intersect at $(4,160)$ .

Two tariffs on shared axes

Solid A: 80+20t; dashed B: 20+35t. Equal prices at t=4 h and C=160 currency units. Only the allowed 0–8 hours are shown; uniform scales.

The difference is $C_A-C_B=60-15t$ . Before 4 hours it is positive: A costs more. After 4 it is negative: A costs less. At 2 hours, costs are 120 and 90; at 6, 200 and 230. This conclusion applies only to the permitted 0–8 hours and stated fees.

Axis intercepts and the slope’s sign

Set $t=0$ for the vertical-axis intercept: A gives $(0,80)$ . For the horizontal-axis intercept require $C_A=0$ , formally giving $t=-4$ . That point belongs to the extended line, not our tariff function’s graph, because negative time is excluded.

If $k>0$ , the function increases; if $k<0$ , it decreases; if $k=0$ , it is constant. A negative coefficient could describe a stock of material being consumed steadily. It does not force the output itself to be negative.

When the relationship is direct proportionality

With $b=0$ , we get $y=kx$ . For nonzero $k$ , this is direct proportionality: doubling input doubles output, and $y/x=k$ is constant for $x\ne0$ . The fee-free rule $D(t)=20t$ has this property. Tariff A does not: $C_A(1)=100$ and $C_A(2)=120$ , not 200.

A line equation $ax+by=c$ with $b\ne0$ can be written $y=-(a/b)x+c/b$ . If $b=0$ and $a\ne0$ , it gives the vertical line $x=c/a$ : one $x$ corresponds to many $y$ values, so this is not a graph of $y$ as a function of $x$ .

When reading a tariff, first identify the fixed fee and the unit of change. The intersection then becomes a meaningful comparison threshold.