It is given that f(x)=x^2+ax+b, f(1)=0 and f(2)=5.
a) Find the values of a and b.
b) Find the value of f(3)-f(-3).
f.4 mathz...about functions and graphs
2008-10-07 2:08 am
回答 (2)
2008-10-07 2:30 am
a)
f(1)=1^2+a+b
0=1+a+b
a+b=-1
f(2)=2^2+2a+b
5=4+2a+b
2a+b=1
a+b=-1...(1)
2a+b=1...(2)
from(1),a=-1-b...(3)
sub (3) into (2),
2(-1-b)+b=1
-2-2b+b=1
-b=3
b=3
sub b=3 into (1),
a+3=-1
a=-4
so,a=-4,b=3
b)
f(3)=3^2+3a+b
f(3)=9+3(-4)+3
f(3)=9-12+3
f(3)=0
f(-3)=(-3)^2-3a+b
f(-3)=9-3(-4)+3
f(-3)=9+12+3
f(-3)=24
f(3)-f(-3)=0-24
f(3)-f(-3)=-24
f(1)=1^2+a+b
0=1+a+b
a+b=-1
f(2)=2^2+2a+b
5=4+2a+b
2a+b=1
a+b=-1...(1)
2a+b=1...(2)
from(1),a=-1-b...(3)
sub (3) into (2),
2(-1-b)+b=1
-2-2b+b=1
-b=3
b=3
sub b=3 into (1),
a+3=-1
a=-4
so,a=-4,b=3
b)
f(3)=3^2+3a+b
f(3)=9+3(-4)+3
f(3)=9-12+3
f(3)=0
f(-3)=(-3)^2-3a+b
f(-3)=9-3(-4)+3
f(-3)=9+12+3
f(-3)=24
f(3)-f(-3)=0-24
f(3)-f(-3)=-24
2008-10-07 2:21 am
a) f(x)=x^2+ax+b
f(1)=(1)^2+a(1)+b
=a+b+1
f(2)=(2)^2+a(2)+b
=2a+b+4
a+b+1=0
a=-(b+1)
2a+b+4=5
a=(1-b)/2
-(b+1)=(1-b)/2
2b+2=b-1
b=-3
a=2
b)f(3)-f(-3)
(3)^2+a(3)+b-[(-3)^2+a(-3)+b]
=9+3a+b-9+3a-b
=6a
=12
f(1)=(1)^2+a(1)+b
=a+b+1
f(2)=(2)^2+a(2)+b
=2a+b+4
a+b+1=0
a=-(b+1)
2a+b+4=5
a=(1-b)/2
-(b+1)=(1-b)/2
2b+2=b-1
b=-3
a=2
b)f(3)-f(-3)
(3)^2+a(3)+b-[(-3)^2+a(-3)+b]
=9+3a+b-9+3a-b
=6a
=12
收錄日期: 2021-04-13 16:08:07
原文連結 [永久失效]:
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