如图,在平面直角坐标系中,抛物线y=ax²+bx+c的对称轴为直线x=-3/2,抛物线与x轴的交点为A、B,
1个回答
(1)顶点在对称轴 x= -3/2上
MC的解析式是y= (3/4)x - 2
x = -3/2,y = -9/8 -2 = -25/8
M(-3/2,-25/8)
(2) y = ax²+bx+c = a[x + b/(2a)]²+ c -b^2/(4a)
对称轴为x = -b/(2a) = -3/2,b= 3a (a)
C(0,-2)
-2 = 0 + 0 +c
c = -2 (b)
顶点M纵坐标 c -b^2/(4a) = -25/8 (c)
(a)(b)(c):a = 1/2,b = 3/2
求抛物线的解析式:y = (1/2)x² + (3/2)x - 2
(3) y = (1/2)x² + (3/2)x - 2 = 0
(x+4)(x-1)= 0
A(-4,0),B(1,0)
半径 = (1+4)/2 = 5/2
圆心P(-3/2,0)
直线MC的解析式是y= (3/4)x - 2,3x - 4y - 8 = 0
圆心和直线MC的距离:|3(-3/2) - 4*0 -8|/√(3²+4²) = (25/2)/5 = 5/2,等于半径,直线MC与圆相切
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