Algebra · Geometry · Modeling · Probability
Make the mathematics move.
Change one thing. Watch the pattern. Explain why it works. Four hands-on labs help you connect formulas, visual models, and real reasoning.
Original learning activities for selected middle and high school concepts. For official assessment questions, visit Official practice.
01 / Algebra I and Algebra II
Find the turning point.
A quadratic has the form f(x) = ax² + bx + c, with a ≠ 0. The sign of a controls whether the parabola opens up or down. Its vertex has x-coordinate −b/(2a).
Worked example
For f(x) = x² − 4x + 3, the vertex is (2, −1). Factoring gives (x − 1)(x − 3), so the real zeros are 1 and 3.
The discriminant D = b² − 4ac tells you how many real zeros to expect: two when D > 0, one repeated zero when D = 0, and no real zeros when D < 0.
Try this: keep a = 1 and b = −4. Raise c from 3 to 4, then to 5. How do the real zeros change?
← Return to study pathsQuadratic Explorer
f(x) = x² − 4x + 3
- Vertex
- (2, −1)
- Discriminant
- 4
- Real zeros
- 1 and 3
Use the discriminant to predict how many times the graph meets the x-axis.
02 / Geometry
One scale factor. Two different effects.
Similar figures have equal corresponding angles and proportional corresponding side lengths. Multiply every length by a positive scale factor k: the perimeter becomes k times as large, while the area becomes k² times as large.
Worked example
A 4-by-3 rectangle has perimeter 14 units and area 12 square units. At k = 2, it becomes 8 by 6: perimeter 28 units, area 48 square units.
Why square the scale factor? Area uses two lengths. (4k)(3k) = 12k². The same relationship applies to all similar plane figures.
Try k = ½. Compare the side lengths, perimeter, and area with the original. Which quantities are halved?
← Return to study pathsSimilarity Lab
| Measure | Original | Scaled copy |
|---|---|---|
| Side lengths | 4 × 3 | 8 × 6 |
| Perimeter | 14 units | 28 units |
| Area | 12 square units | 48 square units |
P′ = 2P · A′ = 4A
Compare the area multiplier with the length multiplier.
03 / Algebra II and modeling
Follow a repeated percent change.
For a fixed percent change each step, use A(t) = P(1 + r)t. Here P is the initial amount, r is the rate as a decimal, and t is a nonnegative whole number of steps.
Worked example
Start with 100 and increase by 20% each step. After 3 steps: A(3) = 100(1.2)³ = 172.8. Each increase is 20% of the current amount.
To find when a positive target T is reached, solve t = ln(T/P) ÷ ln(1 + r) when r ≠ 0. A logarithm undoes an exponent. The result may fall between whole-number steps.
Compare +20% with −20%. Their factors are 1.2 and 0.8. One increase followed by one decrease multiplies by 0.96, so it does not return to the starting amount.
← Return to study pathsGrowth & Decay Lab
A(3) = 100(1.2)³ = 172.8
Use a logarithm to work backward
To reach 200, t = ln(2) ÷ ln(1.2) ≈ 3.802. The first whole-number step at or above the target is 4.
Convert the percent to a decimal, then add it to 1. A 0% change gives a constant model.
04 / Probability and statistics
The first draw changes the second.
A bag contains red and blue counters. Draw twice without replacement: the first counter stays out. After drawing red, there is one fewer red counter and one fewer counter overall.
Worked example
With 4 red and 6 blue counters, P(first red) = 4/10. Given a red first draw, P(second red) = 3/9. So P(two red) = (4/10)(3/9) = 2/15.
Conditional probability uses the information that the first draw was red. By contrast, the overall probability that the second counter is red remains R/N. These are different questions.
With replacement, the bag is reset between draws: P(second red | first red) = R/N. The two color outcomes are then independent.
← Return to study pathsWithout Replacement Lab
R = red · B = blue · every counter is equally likely to be drawn.
| Probability | Without replacement | With replacement |
|---|---|---|
| First red | 4/10 = 2/5 | 4/10 = 2/5 |
| Second red, given first red | 3/9 = 1/3 | 4/10 = 2/5 |
| Two red | (4/10)(3/9) = 2/15 | (4/10)² = 4/25 |
Without replacement, P(second red) = 2/5 overall, while P(second red | first red) = 1/3. The events are dependent.
Multiply P(first red) by P(second blue | first red). The number of blue counters stays the same.
Keep the connection.
Pair an exploration with a study path, review a concept, then apply it to official practice.
These are original instructional models, not official assessment items. The studio develops selected concepts and is part of a broader course of study.