f(x)=2x^2+ax+b之圖形過點(2,3),且頂點在y=4x-3上,求數對(a,b)?
回答 (3)
F(x) = 2x² + ax + b
F(x)之圖形過點 (2, 3):
3 = 2(2)² + a(2) + b
b = -2a - 5 ...... [1]
F(x) = 2x² + ax + b
F(x) = 2[x² + (a/2)x] + b
F(x) = 2[x² + (a/2)x + (a/4)²] - 2(a/4)² + b
F(x) = 2[x + (a/4)]² - 2(a/4)² + b
F(x) = 2[x + (a/4)]² - (a²/8) + b
頂點 (-a/4, -(a²/8) + b) 在 y = 4x - 3 上:
-(a²/8) + b = 4(-a/4) - 3
b = (a²/8) - a - 3 ...... [2]
[2] = [1]:
(a²/8) - a - 3 = -2a - 5
(a²/8) + a + 2 = 0
a² + 8a + 16 = 0
(a + 4)² = 0
a = -4
代入 [1] 中:
b = -2(-4) - 5
b = 3
(a, b) = (-4, 3) ...... (答案)
Sol
頂點(m,4m-3)
y=2(x-m)^2+(4m-3)
3=2(2-m)^2+4m-3
3=2(m^2-4m+4)+4m-3
2m^2-4m+2=0
(m-1)^2=0
m=1
y=2(x-1)^2+4-3=2x^2-4x+3
(a,b)=(-4,3)
收錄日期: 2021-04-18 00:22:43
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