直角坐標系的幾個題目~~~~幫幫忙

1.
y= -x^2 -6x +2 =-(x^2+6x-2)=-(x+3)^2+11
當x=-3時,有最大值11 2.
f(x)= 2x^2 +bx +c=2(x^2+(b/2)x+(c/2))=2(x+(b/4))^2+(8c-b^2)/8 =2(x+2)^2-5
==>b/4=2 , (8c-b^2)/8=-5==>b=8 ,c=3 4.
f(x)= -2x^2 -8x +7=-(2x^2+8x-7)=-2(x+2)^2+15
當x=2時,有最小值 -17
當x=-2時,有最大值 15 5.
f(x)= x^2 -9x +3=(x-(9/2))^2-(69/4)
當x=2時,有最小值 -11
當x=-1時,有最大值 13
6.
f(x)= 2x^2 -10x +k
==>D=b^2-4ac=100-8k=0
==>k=12.5 7.f(x)= -x^2 +2x +15=0
==>x^2-2x-15=0
==>x=5 ,x=-3f(0)=15=y△ABC之面積 =(1/2)*ll 5 0 l l
l -3 0 l
l 0 15 l
l 5 0 l =60



2011-09-16 12:16:17 補充：
3.

y=3/( x^2 -2x +7 ) =3/[(x-1)^2+6]
當x=1時,有最大值1/2

2. f(x)= 2x^2+bx +c 在x=－2時，有最小值－5，求b，c之值
Sol
f(x)=2(x+2)^2－5
=2(x^2+4x+4)－5
=2x^2+8x+3
b=8，c=3

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