若f(x)=x^2+ax+b,f(-1)=0及f(1)=-4,求a,b的值

f(-1)=(-1)^2+(-1)a+b=0
1-a+b=0
a-b=-1 -----(1)

f(1)=(1)^2+(1)a+b=-4
1+a+b=-4
a+b=-4-1
a+b=-5 ----(2)
(1)-(2):
(a-b)+(a+b)=-1+(-5)
2a=-6
a=-3

From(1)
(-3)-b=-1
-b=2
b=-2

f(-1)=0
f(-1)=-1^2 + a(-1) +b
0= 1-a+b -----------------(1)

f(1)=-4
f(1)= 1^2+ a(1) +b
4= 1+a+b -------------------(2)

(1)-(2)
-4 = -2a
a = 2

Substitude a=2 into (1),
0 = 1-(2)+b
b = 1

f(x) = x^2 + ax + b

f(-1) = 0
1 - a + b = 0 ......(1)

f(1) = -4
1 + a + b = -4 ......(2)

(1) + (2):
2 + 2b = -4
b = -3
a = -2

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