7日間コース
Master Quadratic Functions (y=ax²) in 7 Days
Understand parabolas, master graphing, domain/range, rate of change, and solve real-world problems involving quadratic functions.
対象: Junior High 3 / Overseas Japanese students
Concepts & Worked Examples
Understand y = ax², its graph (parabola), value behavior, domain/range, rate of change, and applications with moving points and intersections.
定義: yがxの2乗に比例する → y = ax²(a ≠ 0)
見分け方: y/x² がすべてのデータで一定なら y = ax²
例:1辺 x cm の正方形の面積 y cm² → y = x²(a = 1) 例:半径 x の円の面積 y → y = πx²(a = π)
式の求め方: y = ax² で、x = 3 のとき y = −18 なら: −18 = a(3²) = 9a → a = −2 → y = −2x²
注意: y = ax² は比例(y = ax)ではない。xが2倍になるとyは4倍!
グラフは放物線と呼ばれる。
性質:
例:y = x² | x | −3 | −2 | −1 | 0 | 1 | 2 | 3 | | y | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
y = ax² と y = −ax² は、x軸について対称(線対称)。
重要: xの変域に0が含まれるとき、頂点がmin(a > 0)かmax(a < 0)になる。
例:y = x²、−2 ≤ x ≤ 4 x = 0 で y = 0(最小)、x = 4 で y = 16(最大) yの変域:0 ≤ y ≤ 16
例:y = −2x²、−2 ≤ x ≤ 3 x = 0 で y = 0(最大)、x = 3 で y = −18(最小) yの変域:−18 ≤ y ≤ 0
注意: y = x²、1 ≤ x ≤ 4(0を含まない) x = 1 で y = 1、x = 4 で y = 16 yの変域:1 ≤ y ≤ 16(0ではない!)
変化の割合 = yの増加量 / xの増加量
一次関数との違い: y = ax² では変化の割合は一定ではない!
例:y = 2x²、x = 1 → x = 3 y₁ = 2、y₂ = 18 変化の割合 = (18 − 2) / (3 − 1) = 16/2 = 8
例:y = 2x²、x = −3 → x = −1 y₁ = 18、y₂ = 2 変化の割合 = (2 − 18) / (−1 − (−3)) = −16/2 = −8
公式(y = ax²): x = p から x = q への変化の割合 = a(p + q)
確認:a(1 + 3) = 2 × 4 = 8 ✓
「y = ax²、x = 1 から x = 4 の変化の割合が10。aを求めよ。」 a(1 + 4) = 10 → 5a = 10 → a = 2
例:正方形ABCDの1辺10cm。点Pが毎秒2cmでA→Bに移動。x秒後の△APQの面積y。
例:y = (1/2)x² と A(−4, 8)、B(6, 18) を通る直線
直線の傾き = (18 − 8)/(6 − (−4)) = 1 直線:y = x + 12
y軸との交点 C(0, 12)
△OABの面積 = △OAC + △OBC = (1/2)×12×4 + (1/2)×12×6 = 24 + 36 = 60
ボールを落とすと x 秒後に y = 5x² m 落下。
y = ax² where y = −18 when x = 3. Find a, and find y when x = −4.
解答
−18 = a(9) → a = −2 y = −2x² When x = −4: y = −2(16) = −32
Substitute the known (x, y) to find a. Then use the equation for the new x value.
For y = 3x², find the range of y when −2 ≤ x ≤ 1.
解答
At x = 0: y = 0 (minimum, since a > 0 and 0 is in the domain) At x = −2: y = 3(4) = 12 (maximum) At x = 1: y = 3 Range: 0 ≤ y ≤ 12
Since the domain includes x = 0 and a > 0, the vertex gives the minimum. Compare endpoint values for the maximum.
For y = 2x², find the rate of change from x = −3 to x = 1.
解答
Using the shortcut: a(p + q) = 2(−3 + 1) = 2(−2) = −4 Verification: y₁ = 2(9) = 18, y₂ = 2(1) = 2 (2 − 18)/(1 − (−3)) = −16/4 = −4 ✓
The shortcut a(p + q) gives the rate of change directly. Always verify with the definition if unsure.
A ball falls y = 5x² meters in x seconds. Find the average speed between the 2nd and 4th second.
解答
At x = 2: y = 5(4) = 20 m At x = 4: y = 5(16) = 80 m Average speed = (80 − 20)/(4 − 2) = 60/2 = 30 m/s
Average speed = total distance / total time = change in y / change in x = rate of change.
Basic Practice
Practice finding equations, graphing, domain/range, and rate of change.
式の決定、グラフの性質、変域、変化の割合の基礎を練習。
y = ax² and y = 8 when x = 2. Find a.
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答え
a = 2
For y = −x², does the parabola open upward or downward?
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答え
Downward (a < 0)
For y = (1/2)x², find y when x = −6.
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答え
y = 18
For y = −3x², find the range of y when −1 ≤ x ≤ 2.
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答え
−12 ≤ y ≤ 0
For y = x², find the rate of change from x = 2 to x = 5.
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答え
7 (using shortcut: 1 × (2+5) = 7)
Standard Practice
Graphing, domain/range with tricky cases, and rate of change calculations. ★2-3.
Pattern Practice
All y = ax² patterns systematically drilled. ★2-3.
Applied Practice
Moving point problems and graph intersection applications. ★3-4.
Advanced Practice
Parabola-line intersections, area calculations, and complex applications. ★4-5.
Final Test & Bridge to Next
Comprehensive test. Preview of high school quadratic functions (vertex form, discriminant).