The signs of k and b decide it all — rise or fall, which quadrants, parallel or crossing, shifted or flipped.
Once you can draw a line, the next skill is to read one. A linear function y = kx + b hides its whole story in two signs. The sign of k says whether the line rises or falls; the signs of k and b together say which quadrants it travels through. Two lines with the same k never meet — they are parallel — while different slopes cross exactly once. And sliding or flipping a line changes b and the signs in tidy, predictable ways. This lesson turns "able to draw" into "able to read."
Walk along the graph from left to right, the way you read a sentence. If the line goes uphill, then as x grows, y grows too — the function is increasing. If it goes downhill, y shrinks as x grows — the function is decreasing. Which way it tilts is decided entirely by the slope.
For y = kx + b, bump x up by 1 and y changes by exactly k (that is what slope means — the rise over a run of 1). So:
Notice what is not on the list: b. The intercept lifts the line up or down, but it never changes which way the line tilts. Rise or fall is the slope's job alone.
k > 0 ⇒ the line rises (increasing). k < 0 ⇒ the line falls (decreasing). The y-intercept b plays no part in rise or fall.
A line runs forever in both directions, so it does not sit in just one corner of the plane — it sweeps across three of the four quadrants (or exactly two when it passes through the origin, b = 0). Which three? You can reason it out from just two facts: where it crosses the y-axis — the point (0, b) — and which way it tilts — the sign of k.
Take y = x + 1. It crosses the y-axis above O at (0, 1), and since k > 0 it rises to the upper right. Follow it right and it climbs into Quadrant Ⅰ; follow it left and it dips through (−1, 0) into Quadrant Ⅲ, passing across Quadrant Ⅱ on the way. So it visits Ⅰ, Ⅱ, Ⅲ — every quadrant except Ⅳ.
Do the same reasoning for each pair of signs and a clean table of six cases falls out:
| slope | intercept | direction | quadrants visited |
|---|---|---|---|
| k > 0 | b > 0 | rises | Ⅰ, Ⅱ, Ⅲ |
| k > 0 | b < 0 | rises | Ⅰ, Ⅲ, Ⅳ |
| k > 0 | b = 0 | rises | Ⅰ, Ⅲ |
| k < 0 | b > 0 | falls | Ⅰ, Ⅱ, Ⅳ |
| k < 0 | b < 0 | falls | Ⅱ, Ⅲ, Ⅳ |
| k < 0 | b = 0 | falls | Ⅱ, Ⅳ |
Don't memorize the six rows blindly — read them off the picture. Mark the crossing (0, b), tilt the line by the sign of k, and trace where it goes. The table is just the result of that reasoning, done once for each sign pair.
Dial k and b. The readout names the direction and lists every quadrant the line passes through — straight from the two signs.
Put two linear functions on one plane, y = k1x + b1 and y = k2x + b2, and ask: do they ever meet? The slope decides.
For example, y = 2x + 1 and y = 2x − 3 share the slope 2, so they are parallel — one is just the other slid down 4. But y = 2x + 1 and y = −x + 4 have different slopes, so they cross. Setting them equal: 2x + 1 = −x + 4 → 3x = 3 → x = 1, and then y = 2(1) + 1 = 3, so they meet at (1, 3).
Equal slope ⇒ parallel (the lines never meet). Different slopes ⇒ they cross at one point. To find the crossing, set the two expressions equal and solve for x, then read off y.
Set the slopes equal to make the lines parallel, or different to make them cross. The readout also says which line is higher at x = 2.
Three moves come up again and again, and each one changes the formula in a tidy way.
Shift up or down. To slide a line up c units, add c to every output — so y = kx + b becomes y = kx + (b + c). Only b changes; the slope k is untouched, so the new line is parallel to the old one. Sliding down means c is negative.
Reflect across the x-axis. Flipping over the horizontal axis sends every point (x, y) to (x, −y), so every output flips sign: y = kx + b → y = −kx − b. Both signs flip — the slope and the intercept.
Reflect across the y-axis. Flipping over the vertical axis sends (x, y) to (−x, y), so wherever you read x you now read −x: y = kx + b → y = −kx + b. This time only the slope flips; the y-intercept stays put (the y-axis itself doesn't move).
Worked. Start with y = 2x + 1. Shift it down 4: y = 2x + (1 − 4) = y = 2x − 3. Reflect the original across the x-axis instead: flip both signs → y = −2x − 1.
A vertical shift changes only b — the line stays parallel. A reflection changes the slope's sign too, so the line tilts the other way. Don't flip b and forget k, and don't flip k for a plain shift.
Two lines on one graph; at some input x, which one gives the bigger y? Just look up: the line that sits higher at that x has the larger output. No arithmetic needed to see it — though the formulas confirm it exactly.
Worked. Compare y = 2x + 1 and y = x + 4 at x = 4. The first gives 2(4) + 1 = 9; the second gives 4 + 4 = 8. Since 9 > 8, line A sits higher there — exactly as the picture shows.
At a given x, the higher point on the graph is the larger output. Comparing two expressions becomes comparing two heights — the bridge to solving inequalities and systems with graphs (next, in 21.5).
Rise or fall is the slope's job: k > 0 rises (increasing), k < 0 falls (decreasing); b doesn't matter. Which quadrants comes from the crossing (0, b) and the tilt — six tidy cases (k > 0, b > 0 → Ⅰ Ⅱ Ⅲ; k > 0, b < 0 → Ⅰ Ⅲ Ⅳ; k < 0, b > 0 → Ⅰ Ⅱ Ⅳ; k < 0, b < 0 → Ⅱ Ⅲ Ⅳ; and the b = 0 cases hit just two). Parallel vs crossing: equal slope ⇒ parallel, different slope ⇒ one crossing (set equal and solve). Shift changes only b (stays parallel); reflect across x flips both signs (y = −kx − b); reflect across y flips only k (y = −kx + b). Compare heights: at a given x, the higher graph has the larger output.
Does y = −5x + 2 rise or fall as x increases?
It falls. The slope is k = −5 < 0, so the function is decreasing — the intercept b = 2 has no effect on direction.
Which quadrants does the graph of y = x − 3 pass through?
Here k = 1 > 0 (rises) and b = −3 < 0 (crosses the y-axis below O). That's the "k > 0, b < 0" case, so it visits Ⅰ, Ⅲ, Ⅳ — every quadrant but Ⅱ. (Check: it cuts the x-axis at (3, 0).)
Are y = 3x + 1 and y = 3x − 6 parallel or crossing?
Parallel. They share the slope k = 3 but have different intercepts, so they never meet — the second is the first slid down 7.
Find where y = 2x + 1 and y = −x + 4 cross.
Set them equal: 2x + 1 = −x + 4 → 3x = 3 → x = 1. Then y = 2(1) + 1 = 3. They cross at (1, 3). (Different slopes, so exactly one crossing.)
Reflect y = −2x + 5 across the x-axis. What is the new formula?
Flip both signs: y = kx + b ↦ y = −kx − b. So −2x + 5 becomes y = 2x − 5.
At x = 3, which is larger: y = 4x − 2 or y = x + 7?
Compute both: 4(3) − 2 = 10 and 3 + 7 = 10. They're equal! Neither is higher — the two lines actually cross at x = 3, so (3, 10) is their meeting point. A reminder that "which is higher" can switch from side to side of a crossing, and at the crossing itself they tie.
Six questions to lock it in. Tap the answer you think is right.
The big idea of this lesson is reading a graph rather than plotting it. Everything follows from two signs: the slope k controls rise or fall (increasing vs decreasing), and the slope together with the y-intercept b controls which quadrants the line visits. Encourage students to reason from the picture — mark the crossing (0, b), tilt by the sign of k, and trace — rather than memorizing the six-case table cold.
Watch for three common slips. First, thinking b affects rise or fall — it never does; only k tilts the line. Second, assuming two lines always cross — equal slopes are parallel and never meet. Third, flipping the wrong sign under a reflection: across the x-axis both signs flip (y = −kx − b); across the y-axis only the slope flips (y = −kx + b); a plain vertical shift changes only b.
This lesson supports Common Core 8.F.B.4 and F-IF.B.6 (rate of change; increasing and decreasing behavior), 8.EE.C.8a (a solution of a pair of linear equations is the intersection point; parallel lines mean no solution), and F-BF.B.3 (the effect of shifts and reflections on a graph). The "compare heights" idea sets up solving inequalities and systems graphically in 21.5.