Fares, distances, pricing plans, optimization, fitting data — and the steady rate that makes the graph straight.
The whole stage now pays off. A real situation with a fixed start and a steady rate is a linear function y = kx + b — so building the model, then graphing it, recovering it from data, and comparing it with another line all become practical tools. We will model a cost, read a speed off a distance graph, find the break-even point between two plans, squeeze the best value out of a closed range, and fit a line to measurements so we can predict beyond them. Then we notice the one thing every linear model shares — a constant rate makes a straight graph — and ask what must happen when the rate itself starts to change. That gentle bend is the door into the next stage.
Most everyday change has two ingredients: where it starts and how fast it changes. Pin those down and the linear function writes itself. The fixed starting value becomes the y-intercept b (the output when x = 0), and the change per unit becomes the slope k (how much the output moves for every 1 that the input gains). Assemble them as the linear function y = kx + b we built back in 21.2.
Take a gym: $20 to join, then $5 per visit. The $20 is owed before you visit even once — that is the start, b = 20. Each visit adds $5 — that is the rate, k = 5. So the cost after v visits is cost = 5v + 20. At v = 0 the cost is $20 (the join fee); at v = 4 it is 5·4 + 20 = $40.
A rate can be negative. A water tank holds 8 L and drains 2 L each minute: the start is b = 8, the rate is k = −2 (it loses 2 L a minute), so the volume after t minutes is V = −2t + 8 — a falling line that hits zero at t = 4 min.
A fixed start plus a steady rate is a linear model y = kx + b: the starting value is b and the rate of change is k. A positive rate climbs, a negative rate falls.
When something moves at a steady speed, the distance it has covered grows in lock-step with time. Distance = speed × time, so d = (speed)·t — a direct proportion through the origin (you start at 0 km at 0 h). Here the b is zero and the speed plays the role of k, the slope.
A car holding 50 km/h covers d = 50t. The slope is 50: a rise of 50 km for every 1 h of run. Reading it the other way, the steepness of a distance–time line literally is the speed — a steeper line means a faster trip. At t = 3 h, d = 50·3 = 150 km; at t = 1 h, d = 50 km.
On a distance–time graph, the slope is the speed: rise (km) over run (h) is km per h. Steeper line, faster motion.
Put two plans on one plane, one line each, and the picture answers “which is better?” at a glance. Where the two lines cross — the intersection that solved a system in 21.5 — is the break-even point, the input at which the plans cost exactly the same. On either side of that crossing, the lower line is the cheaper plan.
Use the hero’s two plans. Plan A: 2x + 10 ($10 base, $2 a unit) and Plan B: 4x ($4 a unit, no base). They cost the same when 2x + 10 = 4x → 10 = 2x → x = 5, and there both cost 4·5 = $20 — the break-even point (5, 20). For fewer than 5 units Plan B is the lower line, so B is cheaper; for more than 5 units Plan A is lower, so A is cheaper. The high base fee of A pays off only once you buy enough units.
The break-even point is the intersection of the two lines. On each side of it, read off the lower line for the cheaper plan; switch plans at the crossing.
A line has no bends. It rises (or falls) at the same steady rate the whole way, so over a closed interval [x₁, x₂] its largest and smallest values can only sit at the two endpoints — never in the middle. If k > 0 the line climbs, so the max is at the right end and the min at the left; if k < 0 it falls, and the two swap.
Say a daily profit is y = −2x + 10 (in hundreds of dollars) and the input x is restricted to [1, 4]. Since k = −2 < 0, the line falls: the highest value is at the left end x = 1 → −2·1 + 10 = 8, and the lowest is at the right end x = 4 → −2·4 + 10 = 2. So on [1, 4] the max is 8 (at x = 1) and the min is 2 (at x = 4) — both at the ends, and every other value lies between them. You never have to test the inside.
On a closed interval, a linear function’s maximum and minimum occur at the endpoints. Check only x₁ and x₂ — a straight line has no interior peak or valley.
Measurements rarely come with a formula attached. But when the data points fall (nearly) on a straight line, we can recover the model: pick two reliable points, find k and b with the method of undetermined coefficients from 21.4, and then the line predicts values we never measured.
Hang masses on a spring and record its length. At 2 kg it is 16 cm; at 5 kg it is 25 cm. Slope first: k = 25 − 165 − 2 = 93 = 3 cm per kg. Then back-solve the intercept: b = 16 − 3·2 = 10 cm (the spring’s natural length, at 0 kg). So length = 3·(mass) + 10. Predict 4 kg: 3·4 + 10 = 22 cm — without ever hanging a 4 kg weight.
Choose two points that genuinely lie on the trend — a single off point throws the slope off. With real data, eyeball the line of best fit first, then read two clean points on that line.
Step back and the common thread is plain: every linear model changes at a steady rate, and that is exactly why its graph comes out straight. Equal steps in x always produce equal steps in y — $5 per visit, 50 km per hour, 3 cm per kg — so the points march along a single ruled line.
But what if the increase itself keeps changing? Suppose each extra unit adds less than the one before — the rate keeps shrinking. Then the steps in y are no longer equal, the points can no longer line up, and the graph must bend. That is a whole new family. The next stage meets the inverse-proportion function y = k∕x, whose graph is a smooth curve, not a line — because there the rate of change is never constant. The straight line you have mastered here is the calm baseline against which every curve is measured.
A graph is straight because the rate of change is constant. When the rate itself changes, the graph bends — and that is where curves (starting with y = k∕x in the next stage) begin.
A real situation with a fixed start and a steady rate is a linear model y = kx + b, and the whole stage’s machinery applies to it:
A gym charges $20 to join plus $5 per visit. Write the cost c as a function of the number of visits v, and find the cost of 6 visits.
Start b = 20, rate k = 5: c = 5v + 20. At v = 6: c = 5·6 + 20 = $50.
A car travels d = 60t (d in km, t in h). How fast is it going, and how far does it travel in 4 hours?
The slope is the speed: 60 km/h. In 4 h: d = 60·4 = 240 km.
Plan A costs 2x + 10 dollars and Plan B costs 4x dollars for x units. At how many units do they cost the same, and which is cheaper for 8 units?
Set equal: 2x + 10 = 4x → 10 = 2x → x = 5 units (both cost $20). For 8 units (more than 5), Plan A is the lower line: A = 2·8 + 10 = $26 vs B = 4·8 = $32, so A is cheaper.
Find the maximum and minimum of y = −3x + 12 on the interval [0, 3].
A line’s extremes are at the endpoints. Since k = −3 < 0 it falls: max at the left end x = 0 → y = 12; min at the right end x = 3 → −9 + 12 = 3.
A spring is 16 cm long with a 2 kg mass and 25 cm with a 5 kg mass. Find the length model, then predict the length at 4 kg.
Slope first: k = (25 − 16)/(5 − 2) = 9/3 = 3 cm/kg. Then b = 16 − 3·2 = 10 cm. Model: length = 3m + 10. At m = 4: 3·4 + 10 = 22 cm.
Explain why a constant-speed distance–time graph is a straight line.
The rate of change — the speed — is constant, so equal steps in time always give equal steps in distance. Equal steps in x producing equal steps in y is exactly what draws a straight line; the moment the rate varies, the graph would bend.
Six questions to lock it in. Tap the answer you think is right.
This lesson is the payoff of Stage 21: it shows that a linear function y = kx + b is not an abstraction but a tool. The central move is reading a story for its two parameters — the initial value (b) and the rate of change (k) — and then letting graphing do the work: a break-even is an intersection, a “cheapest plan” is the lower line, an optimum on a range is an endpoint, and a prediction is a point on a fitted line. Encourage learners to name the start and the rate out loud before writing any formula.
The misconception to watch is swapping the roles of k and b (writing cost = 20v + 5 for the gym), expecting a linear function’s largest value somewhere in the middle of an interval (it can only be at an endpoint — a line never peaks), and reading the wrong line as “cheaper” once you pass the break-even point. The closing section also plants a seed: a straight graph means a constant rate, so a changing rate forces a curve — the bridge to inverse proportion in Stage 22.
Common Core: F-LE.A.1–2, F-LE.B.5 (build linear models; interpret the rate and initial value as parameters), 8.F.B.4–5 (construct and qualitatively describe linear relationships), and 8.EE.C.8c / A-REI.C (solve real-world problems with a system — the break-even point).