g(x) = 2x − 5. Find x when g(x) = 11.
Two variables are proportional if y/x is always the same constant, k. Every proportional relationship passes through (0, 0).
| 6 apples cost $4. Cost of 12 apples? | 4/6 = c/12 → c = $8 (the ratio stays constant) |
| c = (2/3)n | General equation: cost c for n apples |
Proportional means 'y is always the same multiple of x' — double the input, double the output. The graph is a straight line through the origin, and the equation is y = kx with no added constant. If there's any startup fee or head start, it's not proportional.
| Is the table x: 2, 4, 6 → y: 5, 10, 15 proportional? | Yes — y/x = 2.5 every time, so y = 2.5x. A constant ratio is the fingerprint of proportionality. |
| Is y = 3x + 1 proportional? | No — at x = 0, y = 1, so the line misses the origin. The '+1' breaks the pure-multiple relationship. |
6 apples cost $4. Write and solve a proportion for the cost of 15 apples.
A conversion factor like "5,280 ft / 1 mile" equals 1, so multiplying by it never changes the actual quantity — only the units. Chain several together to convert step by step, canceling units as you go.
| 4.5 mi × (5280 ft / 1 mi) = 23,760 ft | The "mi" units cancel, leaving feet |
Unit conversion is multiplying by clever forms of 1. Write the conversion as a fraction so the unit you want to cancel sits on the opposite side of the fraction bar, and let units cancel like factors. If the units cancel to what you want, the arithmetic is set up right.
| Convert 3.5 hours to seconds. | 3.5 hr × (60 min/1 hr) × (60 s/1 min) = 12,600 s. 'hr' and 'min' cancel diagonally. |
| Convert 90 ft/s to miles per hour. | 90 ft/s × (3600 s/1 hr) × (1 mi/5280 ft) ≈ 61.4 mph. |
Convert 1 mile to feet given 5,280 ft/mile, then to inches given 12 in/ft.
Not every straight line passes through the origin. The general linear form is y = mx + b, where m is the slope (rate of change) and b is the y-intercept. If b ≠ 0, the relationship is not proportional.
| f(−2) = −1, f(1) = 5 | slope = (5−(−1))/(1−(−2)) = 6/3 = 2 |
| y = 2x + 3 | Substitute a point to solve for b: −1 = 2(−2)+b → b = 3 |
Non-proportional linear functions still have a constant rate (slope) but start from somewhere other than zero — think of a race with a head start. y = mx + b: m is the per-step rate, b is where you begin.
| A candle is 12 in tall and burns 0.5 in/hr. Write the equation. | h = 12 − 0.5t. Starting height 12, negative rate because it shrinks. |
| What's the starting value in y = 4x − 9? | −9, the value when x = 0. A negative start is fine — think debt or below-zero temperature. |
Identify the slope and y-intercept of y = (3/2)x − 3.
If an equation isn't already solved for y, isolate y the same way you'd solve any equation — then the slope and intercept are easy to read off.
| 2y − 6x = 12 | Starting equation |
| 2y = 6x + 12 | Add 6x to both sides |
| y = 3x + 6 | Divide both sides by 2 → slope = 3, y-intercept = 6 |
Every line is a story with two numbers: where it starts (y-intercept b) and how it moves (slope m = rise/run). To graph, plot b on the y-axis, then use slope as movement directions: numerator = vertical steps, denominator = horizontal steps. Negative slope means the vertical step goes down.
| Graph y = −&frac32x + 4. | Start at (0, 4). Slope −3/2: down 3, right 2 → next point (2, 1). Repeat. |
| Two points on a line: (0, −2) and (4, 6). Write the equation. | b = −2 straight from the first point. m = (6−(−2))/(4−0) = 2 → y = 2x − 2. |
Rearrange x − 3y = 6 into y = mx + b form.
Find the slope from the two points first, then substitute one point into y = mx + b and solve for b.
| (2, 5) and (5, 17) | slope = (17−5)/(5−2) = 12/3 = 4 |
| 5 = 4(2) + b → b = −3 | Substitute point (2,5) into y = 4x + b |
| y = 4x − 3 | Final equation |
To write a line's equation you need exactly two ingredients: a slope and one point. Get the slope first (from two points, a table, or the story), then plug your known point into y = mx + b to solve for b — or use point-slope form y − y₁ = m(x − x₁) and simplify.
| Write the line through (2, 7) with slope 3. | 7 = 3(2) + b → b = 1 → y = 3x + 1. |
| Write the line through (−1, 4) and (3, −4). | m = (−4 − 4)/(3 − (−1)) = −2. Then 4 = −2(−1) + b → b = 2 → y = −2x + 2. |
Find the equation of the line through (0, 4) and (2, 10).
In a real-world linear model, the slope always tells you how fast the output changes per unit of input, and the y-intercept always tells you the starting amount (the value at input = 0).
| Jannine starts with $450, saves $5/week | s = 5w + 450 — the $5 is the slope (rate), $450 is the intercept (starting amount) |
Linear modeling: find the flat part and the per-unit part in the story. The per-unit rate (per mile, per month, per ticket) is the slope; the one-time amount is the y-intercept. Then the model answers two kinds of questions: plug in x to predict y, or set y equal to a target and solve backwards.
| A plumber charges $60 to show up plus $45/hr. Cost of a 3.5-hr job? | C = 60 + 45t → C(3.5) = 60 + 157.50 = $217.50. |
| Same plumber; the bill was $240. How long was the job? | 240 = 60 + 45t → t = 4 hours. Same model, solved in reverse. |
A model is s = 5w + 450. What is s when w = 10?
Sometimes the question asks "when" or "how many" — that means setting the model's output to a target value and solving for the input.
| n = 1275 − 75d. When does n = 150? | 1275 − 75d = 150 → −75d = −1125 → d = 15 days |
Harder models just hide the two ingredients better. If the story gives two data points instead of a rate, compute the slope from the points first. Also ask: is the rate positive (filling, earning) or negative (draining, spending)? The sign tells the story's direction.
| A pool had 500 gal at t = 2 min and 800 gal at t = 5 min. Write the model. | m = (800−500)/(5−2) = 100 gal/min. 500 = 100(2) + b → b = 300 → V = 100t + 300. |
| When will that pool hold 1,500 gal? | 1500 = 100t + 300 → t = 12 minutes. |
n = 1275 − 75d. Find n when d = 5.
A horizontal line is y = constant (every point shares the same y). A vertical line is x = constant (every point shares the same x). Vertical lines are not functions.
| Vertical line through (5, −3) | x = 5 — the y-coordinate is irrelevant |
Horizontal and vertical lines are the 'one-ingredient' lines. y = (number) is horizontal: every point has that height, slope 0. x = (number) is vertical: every point has that x, slope undefined — and it's not a function. Memory hook: the variable in the equation tells you which axis the line cuts.
| Write the horizontal line through (−3, 6). | y = 6. Only the height matters; the x-coordinate is irrelevant. |
| What's the slope of x = 2? | Undefined — run is 0, and dividing by zero has no answer. Vertical lines have no slope value. |
State the equation of a vertical line through (−4, 5).
Absolute value gives distance from zero — always non-negative, graphs as a V-shape. Step functions hold one constant output over a whole range of inputs, then jump to a new constant.
| f(x) = |x − 4| + 7, find f(1) | |1−4|+7 = |−3|+7 = 3+7 = 10 |
Absolute value measures distance from zero, so |x| outputs are never negative — that's why the graph is a V that bounces at its vertex. Step functions jump in flat stairs: the output stays constant until the input crosses a breakpoint, then leaps. Both are functions; they just aren't single straight lines.
| Find the vertex of f(x) = |x + 1| − 3. | The inside is zero at x = −1, giving the lowest point (−1, −3). |
| Shipping: $4 for up to 1 lb, $7 for up to 2 lb, $10 for up to 3 lb. Cost for 1.2 lb? | 1.2 lb crosses the 1-lb breakpoint, so it lands on the second stair: $7. |
For f(x) = |x + 3|, find f(0).
Substitute the point's x and y into the equation or inequality. If the result is a true statement, the point lies on (or in the solution set of) the graph.
| Does (2, 10) lie on y = 4x + 2? | 10 = 4(2)+2 = 10 → TRUE → yes, it lies on the graph |
A graph is a set of claims: the point (a, b) claims 'input a produces output b.' To check whether a point satisfies an equation, substitute both coordinates and see if the statement is true. The vertical line test is just checking that no input makes two claims at once.
| Is (3, 5) on the line y = 2x − 1? | 5 = 2(3) − 1 = 5 ✓ yes. Substitution is the test — not eyeballing. |
| Is (−2, 0) on y = x² + 4? | 0 = 4 + 4 = 8? No. The point misses the curve. |
Does (2, 8) lie on x + y ≤ 10?
Solve the inequality for y first (flip the sign if you multiply/divide by a negative), then graph the boundary line — dashed for < or >, solid for ≤ or ≥ — and shade the side that makes the inequality true.
| 3x − 2y ≥ 2 | −2y ≥ −3x + 2 → divide by −2, FLIP → y ≤ (3/2)x − 1 |
A linear inequality shades everything on one side of a boundary line. Solve for y first; then < or ≤ shades below, > or ≥ shades above. Dashed line for strict (<, >), solid for 'or equal.' When in doubt, test a point (0,0 is easiest if it's not on the line).
| Graph y > 2x − 3. | Dashed line y = 2x − 3, shade above. Check (0,0): 0 > −3 ✓, so shade the side containing the origin. |
| Is (1, 5) a solution to y ≤ 3x + 1? | 5 ≤ 3(1) + 1 = 4? No — 5 > 4, so (1,5) is outside the shaded region. |
Graph y ≤ 4. Describe the shaded region in words.
A sequence's input is a term's place in line (1st, 2nd, 3rd...), and its domain is only the natural numbers — never fractions or decimals. It can be defined explicitly (a formula in n) or recursively (each term built from the one before it).
| a(n) = 2n + 1 → 3, 5, 7, 9, 11 | Explicit formula: plug in n = 1, 2, 3... |
| b₁ = 7, bᵢ = bᵢ₋₁ + 4 → 7, 11, 15, 19 | Recursive: start at 7, add 4 to get the next term each time |
An arithmetic sequence adds the same amount each step — it's a linear function wearing sequence clothing. The common difference d is the slope; the explicit formula an = a₁ + d(n − 1) says 'start at the first term, then take (n−1) steps of size d.'
| Sequence: 4, 9, 14, 19, … Find a₁₀. | d = 5, so a₁₀ = 4 + 5(9) = 49. Nine steps after the first term. |
| Which term of 2, 5, 8, … equals 62? | 62 = 2 + 3(n−1) → 60 = 3(n−1) → n = 21. |
These problems pull from earlier units on purpose. Switching between skills feels harder in the moment, but it helps you remember longer and matches how a real exam mixes topics. Try each one on paper first, then reveal.Estos problemas provienen de unidades anteriores a propósito. Cambiar de una destreza a otra se siente más difícil en el momento, pero te ayuda a recordar por más tiempo y se parece a cómo un examen real mezcla los temas. Intenta cada uno en papel primero, luego revela la respuesta.