Linear Functions, Intercepts, Rate of Change & Slope — Lessons 5.1–5.3
Essential Question: What is a linear function?
A race car goes 210 mph. After x hours it has gone y = 210x miles.
| x (hours) | y (miles) | Change in y |
|---|---|---|
| 0 | 0 | — |
| 1 | 210 | +210 |
| 2 | 420 | +210 |
| 3 | 630 | +210 |
| 4 | 840 | +210 |
| 5 | 1050 | +210 |
x changes by 1 every time, and y changes by 210 every time. That makes it linear.
| x | y = x2 | Change |
|---|---|---|
| 1 | 1 | — |
| 2 | 4 | +3 |
| 3 | 9 | +5 |
| 4 | 16 | +7 |
| 5 | 25 | +9 |
| x | y = x3 | Change |
|---|---|---|
| 1 | 1 | — |
| 2 | 8 | +7 |
| 3 | 27 | +19 |
| 4 | 64 | +37 |
| 5 | 125 | +61 |
x changes by 1 each time, but the y changes keep growing. Not linear.
Linear functions change by equal amounts over equal intervals.
f(x) = mx + b m = slope (rate of change), b = y-interceptGiven: x2 − x1 = x4 − x3 (two equal-size intervals), and f(x) = mx + b.
Prove: f(x2) − f(x1) = f(x4) − f(x3)
| Statement | Reason |
|---|---|
| x2 − x1 = x4 − x3 | Given |
| m(x2 − x1) = m(x4 − x3) | Multiplication Property of Equality |
| mx2 − mx1 = mx4 − mx3 | Distributive Property |
| mx2 + b − mx1 − b = mx4 + b − mx3 − b | Add and subtract b (adds 0) |
| mx2 + b − (mx1 + b) = mx4 + b − (mx3 + b) | Distributive Property |
| f(x2) − f(x1) = f(x4) − f(x3) | Definition of f(x) |
Any linear function can be written as a linear equation.
Ax + By = C A, B, C are real numbers; A and B are not both 0.
A solution is any ordered pair (x, y) that makes the equation true. The graph shows all the solutions.
5x + y = 103x + 4y = 8x/2 + y/4 = 3/2 (multiply by 4 → 2x + y = 6)y = 3x + 2 (linear, not standard form)2x2 + y = 6 — x squaredx + y2 = 1 — y squared3xy + y = 4 — x times yy = 5/x — x in denominatory = 10 − 5x| x | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| y | 15 | 10 | 5 | 0 | −5 |
A = −4, B = 1, C = 11. Solve for y: y = 11 + 4x
| x | −4 | −2 | 0 | 2 | 4 |
|---|---|---|---|---|---|
| y | −5 | 3 | 11 | 19 | 27 |
Does the in-between make sense? That decides whether the dots connect.
| Situation | Function | Type | Domain |
|---|---|---|---|
| Hans buys video games, $5 each | f(x) = 5x | Discrete | {0, 1, 2, …} |
| Kristoff pays $6 per umbrella sold | f(x) = 6x | Discrete | {0, 1, 2, …} |
| Hot air balloon at 85 mph | y = 85x | Continuous | x ≥ 0 |
| Gallons of water over time | y = 4x | Continuous | x ≥ 0 |
Particle velocity: −4x + 3y = 15. When does it reach y = 30 km/hr?
Jordan’s profits: $150, $300, $450, $600 in weeks 1–4. Each week adds $150, so f(x) = 150x and f(9) = $1350.
Essential Question: How can you identify and use intercepts in linear relationships?
(a, 0) — y is always 0.(0, b) — x is always 0.| Equation | x-intercept | y-intercept |
|---|---|---|
| −5x + 6y = 60 | −12 | 10 |
| 8x + 7y = 28 | 3.5 | 4 |
| −6x − 8y = 24 | −4 | −3 |
| 2x − 3y = −6 | −3 | 2 |
| −4x − 5y = 40 | −10 | −8 |
| 3⁄5x + 1⁄2y = 30 | 50 | 60 |
C/A y-intercept = C/B
In a real-world problem, each intercept answers a question.
Miners start 90 ft underground and ride an elevator up for 3 minutes.
A negative y-intercept means the starting value is below zero (underground, below sea level, below 0°, owing money).
| Situation | Function | x-intercept | y-intercept |
|---|---|---|---|
| Sandia Peak tram | 4500 − 300x | 15 min to reach the peak | starts 4500 m from the peak |
| Hot air balloon | 750 − 25x | lands after 30 min | starts 750 m up |
| Experiment temperature | 5⁄2x − 70 | reaches 0°C at 28 hr | starts at −70°C |
| Trout in lake | 350 − 14x | gone after 25 yr | 350 trout at start |
| Air temp at sunrise | 3x − 6 | 0°C after 2 hr | −6°C at sunrise |
| Brake pads | 480 − 4x | enough for 120 cars | 480 pads at start |
| Connor’s 10 km race | 10 − 1⁄4x | finishes at 40 min | 10 km to go at start |
| Kim owes $245, pays $35/wk | 245 − 35x | paid off in 7 wk | owes $245 at start |
| Bank account, $8 fee/mo | 360 − 8x | $0 after 45 mo | $360 at start |
Two points make a line. The intercepts are the easiest two to find.
Standard form: 3⁄4x + 1⁄2y = 3
x-int: 3⁄4x = 3 → x = 4
y-int: 1⁄2y = 3 → y = 6
Plot (4, 0) and (0, 6). (Same line as 3x + 2y = 12.)
| Equation | Standard form | x-int | y-int |
|---|---|---|---|
| 18y = 12x + 108 | −12x + 18y = 108 | −9 | 6 |
| 3y = −5x − 30 | 5x + 3y = −30 | −6 | −10 |
| −6y = −4x + 24 | 4x − 6y = 24 | 6 | −4 |
| 9y = 3x + 18 | −3x + 9y = 18 | −6 | 2 |
| −3y = 7x − 21 | 7x + 3y = 21 | 3 | 7 |
| 3⁄2x = −4y − 12 | 3⁄2x + 4y = −12 | −8 | −3 |
The sail’s slanted edge is 10x + 4y = 240. The sail is the triangle under the line.
x-int: 240/10 = 24 ft (base)
y-int: 240/4 = 60 ft (height)
For any Ax + By = C (all positive): Area = 1⁄2 · C⁄A · C⁄B = C² / (2AB)
Essential Question: How can you relate rate of change and slope in linear relationships?
rate of change = change in y / change in x| Period | Change in cost / change in years | Rate |
|---|---|---|
| 2003–2004 | (37 − 37) / (4 − 3) | 0 ¢/yr |
| 2004–2006 | (39 − 37) / (6 − 4) | 1 ¢/yr |
| 2006–2008 | (42 − 39) / (8 − 6) | 1.5 ¢/yr |
| 2008–2013 | (46 − 42) / (13 − 8) | 0.8 ¢/yr |
| Segment | Calculation | Slope (km/hr) |
|---|---|---|
| 0.5 → 1 hr | (5 − 3) / (1 − 0.5) | 4 |
| 1 → 2 hr | (7 − 5) / (2 − 1) | 2 |
| 2 → 3 hr | (13 − 7) / (3 − 2) | 6 (steepest) |
For a line, the rate of change is the same between any two points. That constant rate is the slope.
slope = rise / runm = (y2 − y1) / (x2 − x1)
The dashed lines show the run (across) and the rise (up).
| Type | Graph | Why | Example |
|---|---|---|---|
| Positive | Rises left to right | rise and run have the same sign | (5, 3), (10, 8): 5/5 = 1 |
| Negative | Falls left to right | rise and run have opposite signs | (−5, 14), (−1, 2): −12/4 = −3 |
| Zero | Horizontal | rise = 0 → 0/run = 0 | (−5, 6), (8, 6): 0/13 = 0 |
| Undefined | Vertical | run = 0 → can’t divide by 0 | (−4, −17), (−4, −3): 14/0 |
| Points | Work | Slope |
|---|---|---|
| (−2, 3), (1, 0) | (0 − 3) / (1 − (−2)) = −3/3 | −1 |
| (0, −3), (5, 3) | 6/5 | 1.2 |
| (−3, 2), (−2, −3) | −5/1 | −5 |
| (−2, −1), (2, 3) | 4/4 | 1 |
| (−1, 9), (2, −5) | −14/3 | ≈ −4.67 |
| (2, −2), (5, 7) | 9/3 | 3 |
| (−6, 5), (−4, −3) | −8/2 | −4 |
| (−4, 7), (2, −8) | −15/6 | −2.5 |
| (−7, −9), (−1, 5) | 14/6 | 7/3 ≈ 2.33 |
| (12, −7), (2, −2) | 5/(−10) | −0.5 |
| (−3, −10), (−1, −1) | 9/2 | 4.5 |
In a real-world problem, slope is a rate: how much y changes for each 1 unit of x.
Heart rate: (20, 180) and (50, 150)
For every year older, maximum heart rate goes down 1 beat per minute.
| Situation | Points | Slope | Meaning |
|---|---|---|---|
| Plant height | (30, 15), (40, 25) | 1 | grows 1 cm per day |
| °F to °C | (50, 10), (77, 25) | 5/9 | +1 °F = +5⁄9 °C |
| Reservoir | (40, 3000), (60, 4000) | 50 | fills 50 ft³ per hour |
| Money earned | (4, 110), (8, 170) | 15 | earns $15 per hour |
| Peanut butter | (960, 2), (3360, 7) | 1/480 | 1 jar per 480 peanuts |
| Printing | (1000, 310), (3500, 460) | 0.06 | 6¢ per page |
| Bank balance | (3, 860), (8, 585) | −55 | loses $55 per month |
| Crickets | (40, 0), (50, 40) | 4 | 4 more chirps/min per °F |
| kWh | 0 | 200 | 400 | 600 | 1000 | 2000 |
|---|---|---|---|---|---|---|
| Cost ($) | 8 | 8 | 34 | 60 | 112 | 157 |
| Interval (kWh) | 0–200 | 200–400 | 400–600 | 600–1000 | 1000–2000 |
|---|---|---|---|---|---|
| Rate ($/kWh) | 0/200 = 0 | 26/200 = 0.13 | 26/200 = 0.13 | 52/400 = 0.13 | 45/1000 = 0.045 |
Plan: $8 flat for the first 200 kWh, then 13¢ per kWh up to 1000 kWh, then 4.5¢ per kWh after that.
| ISP | 100 → 200 MB | 200 → 400 MB | Constant fee? |
|---|---|---|---|
| A ($54, $74, $94) | 20/100 = $0.20/MB | 20/200 = $0.10/MB | No |
| B ($42, $57, $87) | 15/100 = $0.15/MB | 30/200 = $0.15/MB | Yes |
| C ($60, $72, $96) | 12/100 = $0.12/MB | 24/200 = $0.12/MB | Yes |
For heavy data use, the lowest current rate wins over time. A ($0.10/MB) ends up cheapest and B ($0.15/MB) most expensive. At 1000 MB: A ≈ $154, C ≈ $168, B ≈ $177.