0rタcxS2c= b^2-4ac=0rタcxS1c_______= b^2-4ac<0rタcxS]нcXȡֵ̓(x= -bb^2-4ac ֵ෴̓i..." />

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2017꼉ƒԔW֪Rcw{

rg2017-06-13 14:56:00   ԴonW     [wС ]
6.タcxSc
= b^2-4ac>0rタcxS2c
= b^2-4ac=0rタcxS1c
_______
= b^2-4ac<0rタcxS]нcXȡֵ̓(x= -bb^2-4ac ֵ෴̓iʽӳ2a)
a>0rx= -b/2a̎ȡСֵf(-b/2a)=4ac-b²/4a;{x|x<-b/2a}ǜp
{x|x>-b/2a};タ_;ֵ{y|y4ac-b^2/4a}෴׃
b=0rタČQSyS@rżʽ׃Ξy=ax^2+c(a0)
ֵʽ
7.ֵʽ
ٮx=1r y=a+b+c
ڮx=-1r y=a-b+c
ۮx=2r y=4a+2b+c
ܮx=-2r y=4a-2b+c
κ|
8.xR
ֵ(ʽֻӑՓa0raС0rՈxƔ)[(4ac-b^2)/4a
oF);[toF)
żԣb=0rżb0rż
ԣo
ʽ
y=ax^2+bx+c[һʽ]
a0
a>0tタ_ڳ;a<0tタ_ڳ;
ǘOֵc(-b/2a(4ac-b^2)/4a);
Ȧ=b^2-4ac,
>0DcxSڃc
([-b-̦]/2a0)([-b+̦]/2a0);
=0DcxSһc
(-b/2a0);
<0DcxSoc;
y=a(x-h)^2+k[cʽ]
˕rOֵc(hk)h=-b/2ak=(4ac-b^2)/4a;
y=a(x-x1)(x-x2)[cʽ(pʽ)](a0)
QSX=(X1+X2)/2 a>0 XR(X1+X2)/2rYSXa>0XQ(X1+X2)/2rYSX

˕rx1x2麯cXSăɂcXY뼴ʽ(һcһԪηB
)
cʽY=A(X-X1)(X-X2) ֪ɂxScһcOcʽɽcXֵX1 X2ֵ
26.2 ú^cһԪη
1. タ cxSйccęM ô rֵ0 Ƿ̵һ
2. κĈDcxSλPϵN]йcһcЃɂc@һԪη̸Nr]ЌЃɂȵČЃɂȵČ
26.3 H}cκ
ճaͿʹʡrgЧʵȆ}ЩɚwYκֵСֵ