求解一道二次函数题~~~~

2024-11-25 21:18:06
推荐回答(3个)
回答1:

y=ax^2+bx+c=a[x+b/(2a)]^2+c-b^2/4a,图像经过A(0,4),顶点在x轴上,且对称轴在y轴的右侧,
c=4,a>0
对称轴x=-b/(2a),b<0
4-b^2/(4a)=0
b^2=16a
b=-4√a
对称轴x=-b/(2a)=-(-4√a)/(2a)=2/√a
y=ax^2-(4√a)x+4
y=x与二次函数图像自左向右分别交于P(x1,y1),Q(x2,y2),OP:PQ=1:3
OP:OQ=1:4,x1=y1,x2=y2
ax^2-(4√a)x+4=x
ax^2-(1+4√a)x+4=0
x1=y1=[1+4√a-√(1+8√a)]/(2a)
x2=y2=[1+4√a+√(1+8√a)]/(2a)
OP^2=(x1)^2+(y1)^2=2{[1+4√a-√(1+8√a)]/(2a)}^2
OQ^2=(x2)^2+(y2)^2=2{[1+4√a+(1+8√a)]/(2a)}^2
OP^2:OQ^2=1:16
16OP^2=OQ^2
16*2{[1+4√a-√(1+8√a)]/(2a)}^2=2{[1+4√a+(1+8√a)]/(2a)}^2
a=1
y=x^2-4x+4
y=x与二次函数y=x^2-4x+4图像自左向右分别交于P(x1,y1),(x2,y2)两点
x^2-4x+4=x
x1=y1=1,x2=y2=4
OP^2=2,OQ^2=32
OP=√2,OQ=4√2
PQ=3√2
OA=4,过A作AB⊥直线y=x,交直线y=x于B点,则AB=4*sin45°=2√2
△PAQ的面积=PQ*AB/2=3√2*2√2/2=6
AP=√10,PQ=3√2
△APD∽△QAP
PD/AP=AP/PQ
PD=AP^2/PQ=10/(3√2)=5√2/3OD=OP+PD=8√2/3
xD=yD=(8√2/3)*sin45°=8/3
点D的坐标(8/3,8/3)
答:
1、二次函数的解析式:y=x^2-4x+4
2、三角形PAQ的面积=6
3、在线段PQ上存在一点D,使△APD∽△QAP,点D的坐标(8/3,8/3)

回答2:

做过了

回答3:

c=4

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