Linear Functions

The machine finally gets a name

The phone-bill machine has been with you since F8: $$20$ plus $$3$ per gigabyte. A1 ran it backwards; A4 drew every month of it at once as $y = 3x + 20$. Now put a second plan next to it — $$10$ up front but $$5$ per gigabyte — and watch the old notation buckle: ”$y = 3x + 20$, and also $y = 5x + 10$, and the first $y$ at $4$ is less than the second $y$ at… ” You can’t even ask which plan is cheaper without pointing awkwardly. The fix is to give each machine a name:

$B(g) = 3g + 20 \qquad\qquad L(g) = 5g + 10$

That’s all function notation is: a name for a rule, with a slot for the input. Read $B(g) = 3g + 20$ as ”$B$ of $g$”: the machine is called $B$, you drop a number of gigabytes into the slot, and out comes a bill. $B(4)$ means “the output of $B$ when the input is $4$” — a single number, computed by substituting: $B(4) = 3(4) + 20 = 32$. Now the comparison that was unaskable is one clean line: is $B(4) < L(4)$? ($32$ versus $30$ — the pricier-per-gig plan wins at four gigs.)

Two questions you can ask a machine

Everything in this unit is one of two moves. Evaluating feeds the machine an input:

Solving hands the machine a target output and asks which input produced it. Which month costs exactly $$50$? That’s the equation $B(g) = 50$, i.e. $3g + 20 = 50$ — and A1 already taught you the undo order: subtract the $20$ first ($3g = 30$), then divide by $3$ ($g = 10$). Forward, the machine multiplies then adds; backwards, you subtract then divide. Same machine, two directions.

Predict before you look: with $f(x) = 2x + 1$, what comes out for input $3$? Then flip to solve and hand it target $9$ — watch the undo boxes run in reverse order. Now set $m = 0$, keep $b = 4$, and ask for target $9$: why does a flat machine have no answer for that? (And which single target has every answer?)

The parentheses are a slot, not a product

Here’s the notation’s one nasty trick. Everywhere else in algebra, $a(b)$ means multiply — so your eye wants $f(4)$ to mean $f \times 4$. If that were true, $f(0)$ would be $0$ for every function in the world; but $B(0) = 20$ — an empty month still costs the base fee. And doubling the input would double the output; but $B(8) = 44$, nowhere near $2 \cdot B(4) = 64$. The parentheses after a function name are a mail slot: $f(4)$ is “what $f$ returns for $4$,” one number, and there is no $f$ floating free to multiply or cancel with anything.

Every fact about $f$ is a point

A4 called a two-variable equation a membership test for points. Function notation is the same test, written tighter: saying $f(4) = 32$ and saying “the graph of $y = f(x)$ passes through $(4, 32)$” are the same sentence. Input across, output up. Three special cases do most of the SAT’s work: $f(0)$ is the y-intercept (the start), solving $f(x) = k$ is finding where the line reaches height $k$, and solving $f(x) = 0$ finds the x-intercept — where the graph touches the floor.

Reading a machine off a table

Often you’re handed no rule at all — just values. Say a table shows $f(0) = 5$, $f(2) = 11$, $f(4) = 17$. Is it linear? Check the steps: each time the input climbs by $2$, the output climbs by $6$. Steady steps in, steady steps out — that’s the linear signature. The rate, though, is per one unit of input:

The eye-catching number is the output step, $6$ — but the inputs move by $2$, so the slope is $6 \div 2 = 3$, not $6$. (Tables step by twos and fives precisely to set this trap.) With the rate and the start, the rule reassembles itself: $f(x) = 3x + 5$, and now $f(50)$ is a substitution instead of a fifty-row table.

What the numbers mean out in the world

The SAT’s favorite function question contains no algebra at all: “In $B(g) = 3g + 20$, what does the $20$ represent?” The answer is always the same two sentences, and units decide everything. The slope is a rate, so its meaning needs the word per: $$3$ per gigabyte. The intercept is a value at zero, the reading before anything happens: $$20$ when $g = 0$, the base fee. Swap them and the sentence turns nonsense — a bill can’t start at "$3$" or grow by "$20$" flat. When the slope is negative — a candle $L(h) = 24 - 2h$ — the rate reads as a loss (“burns $2$ cm each hour”), and one more landmark gains a meaning: the x-intercept is the runs-out moment, $L(h) = 0$ at $h = 12$.

Pick the candle. Before touching anything, predict: raising the burn rate moves which end of the line — and does the “gone at…” marker slide left or right? Then slide only the starting length: the tilt never changes, the whole story just starts higher (A4’s $b$-slider, now with a plot). The scooter’s unlock fee does the same thing in dollars.

The one thing to remember

A function is a named machine: $f(\text{input}) = \text{output}$, so every fact $f(a) = b$ is a point $(a, b)$ on its graph. Evaluating substitutes forward; solving runs the machine backwards. And in any real story, the slope is the rate (the “per” number), the y-intercept is the starting value, and the x-intercept is when it runs out.

Reading the notation

From a table of values

• Linear check: equal input steps ⇒ equal output steps.
• Rate: $\text{slope} = \dfrac{\Delta f}{\Delta x}$ — divide by the input step, which is often $2$ or $5$, not $1$.
• Start: $b = f(0)$ (or work backwards to it). Then $f(x) = (\text{rate})\,x + b$.

Interpreting a model $f(x) = mx + b$