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數學問題請解答 麻煩一下,請附詳解!!
Aug 18th 2013, 13:45

1.
y = ax
² + bx - 7 ...... [1]

圖形以 (-3, 2 ) 為頂點的二次函數:
y = a(x + 3)
² + 2 ...... [2]

[2] [1] 表示同一函數:
a(x + 3)
² + 2 = ax² + bx - 7
ax
² + 6ax + (9a + 2) = ax² + bx - 7

比較常數項:
9a + 2 = -7
9a = -9
a = -1

比較 x 項的係數:
6a = b
b = 6(-1)
b = -6

=====
2.
f(x) = ax² + bx + c

f(0) = 17
a(0)² + b(0) + c = 17
c = 17

f(-4) = f(3)
a(-4)² + b(-4) + c = a(3)² + b(3) + c
7a = 7b
a = b ...... [1]

f(3) = 5
ax² + bx + c = 5
a(3)
² + b(3) + c = 5
9a + 3b + c = 5 ...... [2]

[1] c = 17 代入 [2] 中:
9b + 3b + 17 = 5
12b = -12
b = -1

b = -1 代入 [2] 中:
a = -1

a + b + c = (-1) + (-1) + 17 = 15

=====
3.
二次函數: y = -x² + 2x + 8 ...... [1]
x
軸: y = 0 ...... [2]

[1] = [2]
-x
² + 2x + 8 = 0
x
² - 2x - 8 = 0
(x + 2)(x - 4) = 0
x = -2
x = 4
所以 AB兩點座標分別為 (-2, 0) (4, 0)

AB 線段長 = 4 - (-2) = 6

二次函數:
y = -x
² + 2x + 8
y = -(x
² - 2x) + 8
y = -(x
² - 2x + 1) + 1 + 8
y = -(x - 1) + 9
頂點 C 的座標 = (1, 9)

ΔABC 面積 = (1/2) x 6 x 9 = 27

=====
4.
y = f(x) = x² + x - 6

x = 0 (y )
y = (0)
² + (0) - 6
y = -6
A
點的座標為 (0, -6)

y = 0 (x )
x
² + x - 6 = 0
(x + 2)(x - 3) = 0
x = -2
x = 3
C
B 兩點的座標分別為 (-2, 0) (3, 0)

ΔABC 面積 = (1/2) x [3 - (-2)] x 6 =15

=====
5.
AB 的座標分別為 (a, 0) (b, 0)( a > b)
a b 2x² + 12x + c = 0 之兩根。

兩根之和: a + b = -12/2= -6
兩根之積: ab = c/2

AB 線段長:
a - b = 4
(a - b)
² = 16
(a + b)
² - 4ab = 16
(-6)
² - 4(c/2) = 16
c = 10

=====
6.
二次函數:
f(x) = ax
² + bx + c ...... [1]

x = 2 時有極大值 9 的二次函數:
f(x) = a(x - 2)
² + 9 ...... [2]

f(0) = 1 代入 [1] 中:
a(0)
² + b(0) + c = 1
c = 1

f(0) = 1 代入 [2] 中:
a(0 - 2)
² + 9 = 1
4a = -8
a = -2

所以 f(x) = -2x² + bx + 1

f(2) = 9 :
-2(2)² + b(2) + 1 = 9
2b = 16
b = 8

所以 f(x) = -2x² + 8x + 1

f(3) = -2(3)² + 8(3) + 1 = 7

=====
7.
頂點為 (-1, 3)
設二次函數為 f(x) = a(x+ 1)² + 3

f(1) = 15
a(1 + 1)² + 3 = 15
4a = 12
a = 3

f(x) = 3(x + 1)² + 3

f(0) = 3(0 + 1)² + 3 = 6

參考資料: micatkie

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