Given y = ax & # 178; + BX + C, and when x = 1, y = - 6; when x = 0, y = - 4; when x = - 2, y = 6, find the value of a, B, C,

Given y = ax & # 178; + BX + C, and when x = 1, y = - 6; when x = 0, y = - 4; when x = - 2, y = 6, find the value of a, B, C,

According to the meaning of the title
a+b+c=-6
c=-4
4a-2b+c=6
∴a= 1
b=-3
c=-4

In y = ax & # 178; + BX + C, if x = 1, y = 2. If x = 2, y = 1. If x = 3, y = 2. Find the value of ABC

X = 1, y = 2; X = 2, y = 1; X = 3, y = 2 substituting y = ax & # 178; + BX + Ca + B + C = 2 (1) 4A + 2B + C = 1 (2) 9A + 3B + C = 2 (3) (3) - (1) 8A + 2B = 0, B = - 4A substituting (1) C = 2-b-a = 2 - (- 4A) - a = 2 + 3AB = - 4A, C = 2 + 3A substituting (2) 4a-8a + 2 + 3A = 1A = 1, C = 2 + 3A = 5, B = - 4A = - 4A = 1, B = - 4, C = 5ab

In the equation y = ax & # 178; + BX + C, when x = 1, y = 2; when x = - 1, y = - 2; when x = 2, y = 3, find the value of a, B, C

a=-1/3 b=2 c=1/3

In the equation y = ax & # 178; + BX + C, when x = - 1, y = 5; when x = 1, y = 1; when x = 2, y = 2
The formula has been listed. I just don't know how to make it easier. Then, I feel dizzy when I do it
a-b+c=5
a+b+c=1
4a+2b+c=2

①a－b＋c＝5
②a＋b＋c＝1
③4a＋2b＋c＝2
The results show that - 2b = 4, B = - 2
Using the formula of (1-3), we can get (4), - 3a-3b = 3, - A-B = 1
Substituting B = - 2 into 4 gives - A + 2 = 1, a = 1
Substituting a = 1 and B = - 2 into 1, we get 1 + 2 + C = 5 and C = 2
a＝1,b＝-2,c＝2

In the equation y = ax & # 178; BX + C, when x = - 1, y = 0, when x = 2, y = 3, when x = 5, y = 60, find the values of a, B, C
It's solved by a cubic equation,

When y = ax & # 178; + BX + CX = - 1, y = 0; when x = 2, y = 3; when x = 5, y = 60, so there is A-B + C = 0, ① 4A + 2B + C = 3, ② 25A + 5B + C = 60, ③ - ①, 3A + 3B = 3, that is, a + B = 1. (1) ③ - ①, 24a + 6B = 60, that is, 4A + B = 10. (2) (2) - (1) 3A = 9, so a = 3, substituting a = 3 into (1), 3 + B = 1, so B = - 2, substituting a = 3, B = - 2 into

In the equation y = ax & # 178; + BX + C, when x is 1,2, - 1 respectively, y is - 6, - 11, - 8 respectively, find the value of a + B-C?
(as above) in the equation y = ax & # 178; + BX + C, when x is 1,2, - 1 respectively, y is - 6, - 11, - 8 respectively, find the value of a + B-C?

X = 1, then a + B + C = - 6 (1)
x=2 4a+2b+c=-11 (2)
x=-1 a-b+c=-8 (3)
(1) + (3) get a + C = - 7 (4)
(2)-2*(1) 2a+c=1 (5)
(5)-(4) a=-8 (6)
Substituting (4) C = 1 (7)
Substituting (6) (7) into (1)
-8+b+1=-6
b=1
a+b-c=-8+1+1=-6

In the equation y = ax & # 178; + BX + C: when x = - 1, y = - 6; when x = 2, y = 6; when x = - 4 and x = 1, the values of Y are equal, and the values of a, B and C are obtained

To get the meaning of a question
a-b+c=-6①
4a+2b+c=6②
16a-4b+c=a+b+c③
③ B = 3A, 4
Substitute ① and ② to get
-2a+c=-6⑤
10a+c=6⑥
Solution
a=1
c=-4
Substituting 4, B = 3
∴a=1 b=3 c=-4

If the quadratic trinomial ax & # 178; + BX + C decomposes the factor to get (x + 8) (x-3), find the value of a + b-c

ax²+bx+c=(x+8)(x-3)=x²+5x-24
The comparison shows that:
a=1,b=5,c=-24
∴a+b-c=30

If the quadratic trinomial AX2 + BX + C is a monomial with respect to X ()
A. a≠0，b=0，c=0B. a=0，b≠0，c=0C. a=0，b=0，c≠0D. a=0，b=0，c=0

One time monomial is the one with the number of times 1, so the conditions that meet the meaning of the question should be a = 0, B ≠ 0, C = 0

If the product of the first binomial x + 3 and the second trinomial ax & # 178; + bx-2 does not contain X & # 178; and X terms, try to determine the values of a and B

(x + 3) x (x + 3) and (x (x + 3) at (x (x-x-2) (x (x-x-3) (x (x-3) (x (x-x-3) (x (x-x-2) \\\35\\\\\35\\\\35\\\\\\\\\\\\\\\\\\\\\\\\\\\\x & # 178