Question

二次函数y=-1/2mx²+3x-4m的图像与x轴相交于a(4,0),点b,与y轴交于点c。 （1）求二次函数的解析式

（2）点p从点a出发以每秒1个单位的速度沿线段ao向o点运动，到达点o后停止运动，过点p作pq∥ac交oc于点q，将四边形pqca沿pq翻折，得到四边形pqc‘a’，设点p的运动时间为t（1）当t为何值时a'恰好落在二次函数的对称轴上（2）设四边形pqc‘a’落在第一象限内的图形面积为s，求s关于t的函数关系式

Answer 1

题目不清楚，假定为y = -(mx²)/2 + 3x - 4m
(1)
过A(4, 0): 0 = -m*4²/2 + 3*4 - 4m, 12m = 12, m = 1
y = -x²/2 + 3x - 4

(2)
y = -x²/2 + 3x - 4 = (-1/2)(x - 2)(x - 4)
B(2, 0), C(0, -4)
二次函数的对称轴 x = (2 + 4)/2 = 3
AC的斜率k = (-4 - 0)/(0 - 4) = 1
t秒时, PA = t, P(4 - t, 0)
PQ的方程: y - 0 = 1[x - (4 - t)], y = x + t - 4 (i)

x = 0, y = t - 4, Q(0, t - 4)
AA'与垂直, 斜率= -1/k = -1
AA'的方程: y - 0 = -1(x - 4), y = 4 - x (ii)

联立(i)(ii), 二者的交点为M(4 - t/2, t/2)
M为AA'的中点, 设A'(a, b):
4 - t/2 = (a + 4)/2, a = 4 - t

t/2 = (b + 0)/2, b = t
A'在二次函数的对称轴上, a = 4 - t = 3, t = 1

A'(4 - t, t)
与上类似，可以求出C'(-t/2, t/2 - 4)
C'A'的斜率为1, 方程: y - t = 1(x -4 + t), y = x + 2t - 4
y = 0, x = 4 - 2t, C'A'与x轴的交点Z(4 - 2t, 0)

(a)0 ≤ t < 2时, C'A'在y轴上的截距为2t - 4 < 0, C'A'与x轴的交点(Z)在线段OB上, PQ与x轴的交点在线段BA上, 四边形PQC'A'落在第一象限内的图形是以PM, ZA'为上下底的直角梯形
PM = √[(4 - t - 4 + t/2)² + (0 - t/2)²] = t/√2
ZA' = √[(4 - t - 4 + 2t)² + (t - 0)²] = (√2)t
高h = MA' = √[(4 - t - 4 + t/2)² + (t - t/2)²] = t/√2
s = (1/2)(PM + ZA')*MA' =(1/2)( t/√2 + √2t)* t/√2 = 3t²/4

(b) 2 ≤ t ≤ 4时, C'A'在y轴上的截距为2t - 4 ≥ 0, C'A'与x轴的交点Z在y轴左侧, 四边形PQC'A'落在第一象限内的图形是个五边形, 可以分为两个直角梯形, 分别在y = x左上和右下方.
C'A'的方程: y = x + 2t - 4
x = 0, y = 2t - 4
C'A'与y轴的交点Y(0, 2t - 4)
AA'的方程: y = 4 - x
与y = x联立, 交点V(2, 2)
两个直角梯形中一个是OVMP, 一个是OVA'Y
OV = √[(2 - 0)² + (2 - 0)²] = 2√2
PM = √[(4 - t - 4 + t/2)² + (0 - t/2)²] = t/√2
YA' = √[(4 - t - 0)² + (t - 2t + 4)²] = (4-t)√2
MV = √[(4 - t/2 - 2)² + (t/2 - 2)²] = (2 - t/2)√2
VA' = √[(4 - t - 2)² + (t - 2)²] = (t - 2)√2
s = (1/2)(PM + OV)MV + (1/2)(OV + YA')MA'
= (16 - t²)/16 + (6-t)(t - 2)
= -17t²/16 + 8t - 11

Answer 2

(1)
过A(4, 0): 0 = -m*4²/2 + 3*4 - 4m, 12m = 12, m = 1
y = -x²/2 + 3x - 4

(2)
y = -x²/2 + 3x - 4 = (-1/2)(x - 2)(x - 4)
B(2, 0), C(0, -4)
二次函数的对称轴 x = (2 + 4)/2 = 3
AC的斜率k = (-4 - 0)/(0 - 4) = 1
t秒时, PA = t, P(4 - t, 0)
PQ的方程: y - 0 = 1[x - (4 - t)], y = x + t - 4 (i)

x = 0, y = t - 4, Q(0, t - 4)
AA'与垂直, 斜率= -1/k = -1
AA'的方程: y - 0 = -1(x - 4), y = 4 - x (ii)

联立(i)(ii), 二者的交点为M(4 - t/2, t/2)
M为AA'的中点, 设A'(a, b):
4 - t/2 = (a + 4)/2, a = 4 - t

t/2 = (b + 0)/2, b = t
A'在二次函数的对称轴上, a = 4 - t = 3, t = 1

A'(4 - t, t)
与上类似，可以求出C'(-t/2, t/2 - 4)
C'A'的斜率为1, 方程: y - t = 1(x -4 + t), y = x + 2t - 4
y = 0, x = 4 - 2t, C'A'与x轴的交点Z(4 - 2t, 0)

(a)0 ≤ t < 2时, C'A'在y轴上的截距为2t - 4 < 0, C'A'与x轴的交点(Z)在线段OB上, PQ与x轴的交点在线段BA上, 四边形PQC'A'落在第一象限内的图形是以PM, ZA'为上下底的直角梯形
PM = √[(4 - t - 4 + t/2)² + (0 - t/2)²] = t/√2
ZA' = √[(4 - t - 4 + 2t)² + (t - 0)²] = (√2)t
高h = MA' = √[(4 - t - 4 + t/2)² + (t - t/2)²] = t/√2
s = (1/2)(PM + ZA')*MA' =(1/2)( t/√2 + √2t)* t/√2 = 3t²/4

(b) 2 ≤ t ≤ 4时, C'A'在y轴上的截距为2t - 4 ≥ 0, C'A'与x轴的交点Z在y轴左侧, 四边形PQC'A'落在第一象限内的图形是个五边形, 可以分为两个直角梯形, 分别在y = x左上和右下方.
C'A'的方程: y = x + 2t - 4
x = 0, y = 2t - 4
C'A'与y轴的交点Y(0, 2t - 4)
AA'的方程: y = 4 - x
与y = x联立, 交点V(2, 2)
两个直角梯形中一个是OVMP, 一个是OVA'Y
OV = √[(2 - 0)² + (2 - 0)²] = 2√2
PM = √[(4 - t - 4 + t/2)² + (0 - t/2)²] = t/√2
YA' = √[(4 - t - 0)² + (t - 2t + 4)²] = (4-t)√2
MV = √[(4 - t/2 - 2)² + (t/2 - 2)²] = (2 - t/2)√2
VA' = √[(4 - t - 2)² + (t - 2)²] = (t - 2)√2
s = (1/2)(PM + OV)MV + (1/2)(OV + YA')MA'
= (16 - t²)/16 + (6-t)(t - 2)
= -17t²/16 + 8t - 11

Answer 3

1。把a的坐标代入函数方程式，即x=4,y=0,可以求出m的值。再回代回去就得到解析式了。。

Answer 4

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