Substitute f(2)=- 1.
-(2+b)^2+8=- 1
B= 1 or -4
F (x) =-(x+ 1) 2+8 or f (x) =-(x-4) 2+8.
Write the standard formula as f (x) =-x 2-2x+7 or f (x) =-x 2+8x-8.
2.( 1)=0->2a+b+ 1=0①
F(x)+ 1=0 has a real root->; b-a^2+ 1<； 0②
① Substituting ② to get a 2+2a > =0.
A & gt=0 or a
A > brought into ① respectively; =0，b & lt=- 1; a & lt=-2，b & gt=3
3. If a=0, then f(x)=(b-8)x, no matter what the value of b is, f(x) can only be a straight line or a point, which does not meet the meaning of the question. So a is not equal to 0.
Because a is not equal to 0, f(x) is a quadratic equation, X =-3, and X = 2 is the solution of the equation.
Substituting x =-3 and x = 2 into the original equation, we get:
9a-3(b-6)-a-ab=0，
4a+2(b-8)-a-ab=0。
Solve a and b from the above two formulas.
The analytical formula of y=f(x) can be obtained by substituting into the original equation.
That's easy to understand.
4.( 1)F( 1)- 1 >； = 0 f( 1)& lt； There are two situations when = is a increasing function or a subtraction function, namely, f(0)=3 or f(2)=3, a = 1 radical, 2 or a = 5 radical, 10. Verify (using the symmetry axis formula) that a= 1- radical sign, 2, a. So a= 1- 2 under radical sign or a = 5 under radical sign 10.
6.
Axis of symmetry x =-a/2
7.( 1, positive infinity)
8. The question means: the square of F(X+T)=(X+T)+ 1 = the square of x+2 (1+t) x+(t+1). There is a real number t that makes the square of x +2( 1+T)X+(T+ 1) < =X constant. The square of x+(t+ 1) < =0 is obtained. Because the range of x is [1.m], that is, when X= 1, the square of x+(2t+1) x+(t+1) = 0 is T=- 1 or T=-. =0, get the square of X-X.
9. let f (x) = 3x-5x+A be -2 < m < 0, 1 < n < 3 to get f(-2)>0, f (0) <; 0，f( 1)& lt； 0，f(3)>0
Just solve the equation.