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.
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/22 22:43:37
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.
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.
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
0=4a-2b+c
2=-b/2a
|12|=4a+2b+c
b=3 , a=-3/4 , c=-9或b=-3,a=3/4,c=9
y=(-3/4)x^2+3x-9或y=(3/4)x^2-3x+9
y=ax^2+bx+c时式子:4a^2+2b+c=0,36a+6b+c=0,4a+2b+c=12.联立1和2得出-32a-8b=o联立2和3得出32a+4b=-12联立得a=-3/4 b=3后代入1中求c=9
所以y=-3/4x^2+3x+9 应该吧