Give direct proportion a starting point and you get y = kx + b, lifted off the origin to cross the y-axis at (0, b).
A taxi starts the meter with a base fare the moment you climb in, then adds on by the mile — so the cost is not a pure proportion; it begins at a nonzero amount. Add that starting value b to a direct proportion y = kx and you get the general linear function y = kx + b. Geometrically, b just lifts the whole line up or down: the line keeps the same steepness k, but now it crosses the y-axis at (0, b) instead of the origin. In the last lesson the line was nailed to the origin; here we set it free to slide. This lesson defines the linear function, reads b as the y-intercept, and draws the graph from two clean points.
Picture a taxi ride. The moment you sit down, the meter already reads a base fare of $3; after that it adds $2 for every mile. So if x is the number of miles, the fare is
y = 2x + 3.
Compare this to a pure direct proportion y = 2x, where the cost would start at $0. Here, at x = 0 miles you already owe $3 — the proportion y = 2x has been lifted by 3. The steady rate ($2 per mile) is the same as before; what is new is the starting amount. A fixed start plus a steady rate is exactly the shape of a linear function.
A fixed starting value plus a steady rate of change gives a linear function y = kx + b: the rate k is how fast y climbs per step of x, and b is the value of y when x = 0.
Whenever a rule has the form
y = kx + b (with k ≠ 0, and k, b constants),
we call y a linear function of x. The name fits its picture: the graph is always a straight line. The two constants each have a job:
• k is the slope — it sets the steepness and the direction (whether the line rises or falls), just as in a direct proportion.
• b is the y-intercept — the lift, the height at which the line crosses the y-axis.
When b = 0 the rule folds back to y = kx — so a direct proportion is simply the special linear function that passes through the origin. (And if instead k = 0, you get the flat line y = b, a constant function — a horizontal line, not a tilted one. We mention it, but it is not the focus here: a genuine linear function needs k ≠ 0.)
A linear function is y = kx + b with k ≠ 0. Its graph is a straight line; k is the slope and b is the y-intercept. The case b = 0 is the direct proportion y = kx through the origin.
Why is b the y-intercept? Just set x = 0, the moment the line meets the y-axis:
x = 0 ⇒ y = k·0 + b = b.
So the line always passes through the point (0, b). Changing b slides the whole line straight up (when b > 0) or down (when b < 0), without ever tilting it — the slope k is untouched. Three lines with the same slope but different lifts are parallel:
y = 2x crosses at (0, 0); y = 2x + 3 crosses at (0, 3); y = 2x − 2 crosses at (0, −2).
b is a y-value — a height on the y-axis — not an x-value. The line crosses the y-axis at (0, b): the first coordinate is always 0, and b is the height.
Two points fix a line, so to graph y = kx + b you only need to find two of its points, plot them, and lay a ruler across. The two easiest points are almost always the intercepts — where the line meets the axes:
• the y-intercept (0, b) — read straight off the formula;
• the x-intercept — where the line crosses the x-axis, found by setting y = 0.
Take y = 2x − 4. The y-intercept is (0, −4). For the x-intercept, set y = 0:
0 = 2x − 4 ⇒ 2x = 4 ⇒ x = 2, so the x-intercept is (2, 0).
Plot (0, −4) and (2, 0), draw the line through them, and the graph is done.
If b = 0 the two intercepts coincide — both are the origin (0, 0) — so they can't fix the line on their own. In that case (a direct proportion) just use the origin and one other point, such as (1, k).
The two crossings come from the same two moves, and they are worth memorizing as a pair:
| To find… | set… | and get | the crossing |
|---|---|---|---|
| the y-intercept | x = 0 | y = b | (0, b) |
| the x-intercept | y = 0 | kx + b = 0 ⇒ x = −b∕k | (−b∕k, 0) |
The x-intercept comes from solving the little equation kx + b = 0, which gives x = −bk. Worked example: take y = −3x + 6.
• y-intercept: set x = 0 → y = 6, so (0, 6).
• x-intercept: set y = 0 → 0 = −3x + 6 → 3x = 6 → x = 2, so (2, 0).
The kbslide widget above tracks both intercepts live as you change k and b — watch how sliding b moves the y-intercept straight up the y-axis while the x-intercept −b∕k drifts along the x-axis.
One small change turns a direct proportion into the whole family of straight lines:
• A linear function is y = kx + b with k ≠ 0; its graph is a straight line, k the slope and b the y-intercept.
• b is the lift: the line crosses the y-axis at (0, b). Changing b slides the line up or down without changing its steepness.
• The case b = 0 is the direct proportion y = kx through the origin; lines with the same k are parallel.
• To graph it, use two points — easiest, the two intercepts: set x = 0 → (0, b); set y = 0 → solve kx + b = 0 → (−b∕k, 0).
Next (21.3 Reading the Line) we let the signs of k and b tell the whole story — rise or fall, which quadrants, parallel or crossing. (And the lesson before this one was 21.1 Direct Proportion, the b = 0 case.)
For the linear function y = 3x − 5, name the slope k and the y-intercept b.
Match it to y = kx + b: k = 3 and b = −5.
Where does y = 3x − 5 cross the y-axis?
Set x = 0: y = 3·0 − 5 = −5. The y-intercept is (0, −5) — that's just b.
Where does y = 3x − 5 cross the x-axis?
Set y = 0 and solve 3x − 5 = 0 → 3x = 5 → x = 53. The x-intercept is (5∕3, 0) — that's −b∕k = −(−5)∕3.
Is y = 2x + 3 a direct proportion?
No. A direct proportion y = kx must pass through the origin (b = 0). Here b = 3 ≠ 0, so the line misses the origin — it crosses the y-axis at (0, 3).
How are the graphs of y = 4x and y = 4x − 7 related?
They have the same slope k = 4, so they are parallel. The second is the first slid down 7: it crosses the y-axis at (0, −7) instead of the origin.
Write the linear function with slope −2 and y-intercept 6, then find its x-intercept.
k = −2, b = 6 → y = −2x + 6. For the x-intercept set y = 0: −2x + 6 = 0 → 2x = 6 → x = 3, so (3, 0).
Six questions to lock it in. Tap the answer you think is right.
This lesson takes the single step from a direct proportion y = kx to the general linear function y = kx + b. The whole idea is that b is a vertical shift: it lifts the line off the origin to cross the y-axis at (0, b) without changing the slope. Reading k as steepness and b as the starting height is the foundation for everything that follows in this stage — graphing, recovering a formula, and tying lines to equations and inequalities.
Three misconceptions are worth heading off. (1) Students often call y = kx + b a "direct proportion" — but only the b = 0 case is, because a proportion must pass through the origin. (2) They read b as an x-coordinate; stress that b is a height on the y-axis, so the crossing is (0, b) with first coordinate 0. (3) They think changing b changes the steepness; the kbslide widget shows the ghost line and the lifted line staying perfectly parallel, so only the position moves. Encourage the two-intercept method for drawing: it turns graphing into two tiny calculations (set x = 0, then set y = 0).
Common Core alignment. 8.F.A.3 (y = mx + b defines a linear function whose graph is a straight line), 8.EE.B.6 (derive y = mx + b; interpret slope and intercept), and HSF-IF.B.4 / HSF-IF.C.7a (interpret and graph intercepts of a linear function). We write the slope as k and the y-intercept as b throughout this stage rather than m.