Linear Functions & Graphs

A1.C.1Slope & slope-intercept01/05

Slope and slope-intercept form.

Find the line through these two points

(x_{1}, y_{1}) = (1, 2) and (x_{2}, y_{2}) = (4, 11)

Step01of 05

Slope m measures steepness — how much y changes per unit of x. The formula:

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{Δ y}{Δ x}

"Rise over run." Pick the points in any order — just be consistent on top and bottom.

A1.C.1 — Slope and slope-intercept form.

Find the line through these two points

(x_{1}, y_{1}) = (1, 2) and (x_{2}, y_{2}) = (4, 11)

Step 01 of 05

Slope m measures steepness — how much y changes per unit of x. The formula:

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{Δ y}{Δ x}

"Rise over run." Pick the points in any order — just be consistent on top and bottom.

Step 02 of 05

Plug in our two points.

m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3

Slope is $3$ : every step right by 1, the line rises by 3.

Step 03 of 05

Slope-intercept form bundles slope and y-intercept into one equation:

y = m x + b

m is the slope (we have it: 3). b is where the line crosses the y-axis. Solve for b by plugging in either known point.

Step 04 of 05

Substitute $m = 3$ and the point $(1, 2)$ .

22 b = 3 (1) + b = 3 + b = - 1

Step 05 of 05

Write the final equation.

y = 3 x - 1

Quick check on the second point $(4, 11)$ : $3 (4) - 1 = 11 ✓$ .

Key insight

Two points → one slope → one equation. Once m and b are nailed down, slope-intercept form is the cleanest way to communicate a line.

A1.C.2 — Standard form and point-slope form.

The line we'll rewrite three ways

slope m = 3, passes through (1, 2)

Step 01 of 05

The same line has THREE common written forms. Each one shines for a different job.

Slope-intercept	→ $y = m x + b$ — graphing
Point-slope	→ $y - y_{1} = m (x - x_{1})$ — building a line from a point + slope
Standard	→ $A x + B y = C$ — finding intercepts, mixing variables

Step 02 of 05

Point-slope is fastest if you already have a point and a slope. Plug in directly — no solving needed:

y - 2 = 3 (x - 1)

That's a complete equation. Leave it alone if all you needed was an equation that fits.

Step 03 of 05

Convert to slope-intercept by distributing and isolating y:

y - 2 y - 2 y = 3 (x - 1) = 3 x - 3 = 3 x - 1

Step 04 of 05

Convert to standard form $A x + B y = C$ . Move the x-term to the left and clear fractions if needed.

y - 3 x + y 3 x - y = 3 x - 1 = - 1 = 1

Convention: make $A$ positive and the coefficients integers.

Step 05 of 05

Why standard form is useful. The intercepts pop right out by setting the other variable to $0$ :

x -intercept: y = 0 \Rightarrow y -intercept: x = 0 \Rightarrow 3 x = 1 \Rightarrow x = \frac{1}{3} - y = 1 \Rightarrow y = - 1

Key insight

All three forms describe the same line. Pick the form that matches the question — graphing? slope-intercept. Building from a point? point-slope. Finding intercepts? standard.

A1.C.3 — Parallel and perpendicular lines.

The reference line

y = 2 x + 5

Step 01 of 05

The relationship between two lines is decided by their slopes alone.

Parallel	→same slope $m_{1} = m_{2}$ , never meet
Perpendicular	→slopes are negative reciprocals: $m_{1} \cdot m_{2} = - 1$
Same line	→same slope AND same y-intercept
Intersecting (oblique)	→different slopes

Our reference line has $m = 2$ . Anything parallel uses $2$ ; anything perpendicular uses $- \frac{1}{2}$ .

Step 02 of 05

Find the line PARALLEL to the reference, through $(4, 1)$ . Slope must be $2$ . Use point-slope:

y - 1 = 2 (x - 4)

Convert to slope-intercept:

y = 2 x - 7

Same slope as the reference ( $2$ ), different intercept ( $- 7$ vs. $5$ ) — exactly what "parallel but distinct" means.

Step 03 of 05

Find the line PERPENDICULAR to the reference, through $(4, 1)$ . Slope must be the negative reciprocal of $2$ :

m_{⊥} = - \frac{1}{2}

Flip $2$ to $\frac{1}{2}$ (reciprocal), negate it.

Step 04 of 05

Plug into point-slope and simplify.

y - 1 y - 1 y = - \frac{1}{2} (x - 4) = - \frac{1}{2} x + 2 = - \frac{1}{2} x + 3

Verify: $(2) (- \frac{1}{2}) = - 1$ $✓$ — the slopes multiply to $- 1$ .

Step 05 of 05

Special cases worth memorizing.

Horizontal line	→ $y = c$ , slope $0$
Vertical line	→ $x = c$ , slope undefined
Perp. to horizontal	→vertical (slope undefined)
Perp. to vertical	→horizontal (slope 0)

Key insight

The whole skill compresses to one rule: read slope, then either keep it (parallel) or flip-and-negate it (perpendicular). The point only sets the intercept.

A1.C.4 — Linear inequalities in two variables.

The inequality we'll graph

y > 2 x - 1

Step 01 of 05

An equation $y = 2 x - 1$ draws a single line. An inequality $y > 2 x - 1$ shades a HALF-PLANE — every point on one side satisfies it.

Step 1 — graph the boundary as if it were an equation. Treat the inequality as $y = 2 x - 1$ for the moment.

Step 02 of 05

Solid or dashed boundary?

$\leq$ or $\geq$	→SOLID — boundary points satisfy
$<$ or $>$	→DASHED — boundary points don't

Our inequality is strict ( $>$ ) — draw a dashed line.

Step 03 of 05

Step 2 — pick a test point not on the line. $(0, 0)$ is the easiest if the line doesn't pass through it.

test (0, 0) : 0 > 2 (0) - 1 \Rightarrow 0 > - 1 ✓

True. So $(0, 0)$ is in the solution region — shade the side CONTAINING the origin.

Step 04 of 05

Step 3 — interpret what shading means. Every point in the shaded half-plane makes the inequality true. Points on a dashed boundary do NOT count; points on a solid boundary do.

Quick spot-check on $(2, 5)$ (above the line): $5 > 2 (2) - 1 = 3 ✓$ . And $(2, 0)$ (below): $0 > 3$ is false — correctly excluded.

Step 05 of 05

Common pitfalls.

Multiplying both sides by negative	→FLIP the inequality sign
Solving for y first	→simpler — then "above" or "below" reads off the sign
$y > \dots$ means	→shade ABOVE the line
$y < \dots$ means	→shade BELOW the line

Key insight

Three decisions = full graph: solid vs. dashed (sign type), which side to shade (test point), and that's it. The boundary line itself is the same math you already know from slope-intercept.

Free diagnostic

Test your understanding: 10 questions, ~10 min.

Pulled live from our bank for A1.C · scored instantly with worked solutions · free, no sign-up.

Start diagnostic →Print worksheets

A1.C.1 – A1.C.4 · CCSS F.IF.B.4 / A.CED.A.2 / G.GPE.B.5 / A.REI.D.12·Full curriculum →