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抛物线y=ax^2+bx+c与x轴交与A(-2,0),对称轴是直线x=2,顶点C到x轴的距离是12,求此抛物线的解析式.

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抛物线y=ax^2+bx+c与x轴交与A(-2,0),对称轴是直线x=2,顶点C到x轴的距离是12,求此抛物线的解析式.
y = ax² + bx + c
= a(x² + bx/a) + c
= a[x² + bx/a + (b/2a)²] - a*(b/2a)² + c
= a(x + b/2a)² - b²/4a + c
= a(x + b/2a)² + (4ac - b²)/4a
顶点是[- b/2a,(4ac - b²)/4a]
对称轴x = 2 - b/2a = 2,4a = - b
(4ac - b²)/4a = 12
(- bc - b²)/(- b) = b + c = 12 ...*
当x = - 2,y = 0
4a - 2b + c = 0
c - 3b = 0 ...*
解*得:b = 3,c = 9
a = - 3/4
f(x) = (- 3/4)x² + 3x + 9