Let f (x) be a quadratic function, and f (x-1) + F (2x + 1) = 5x ^ 2 + 2x, find f (x)

Let f (x) be a quadratic function, and f (x-1) + F (2x + 1) = 5x ^ 2 + 2x, find f (x)

Let f (x) = ax ^ 2 + BX + C
f(x-1)=a(x-1)^2+b(x-1)+c
f(2x+1)=a(2x+1)^2+b(2x+1)+c
f(x-1)+f(2x+1)=a(x-1)^2+b(x-1)+c+a(2x+1)^2+b(2x+1)+c
=ax^2-2ax+a+4ax^2+4ax+a+3bx+2c
=5ax^2+(2a+3b)x+2a+2c
=5x^2+2x
So 5A = 5
2a+3b=2
2a+2c=0
a=1 b=0 c=-1
f(x)=x^2-1

Let f (x) = 3 + 5x-2x & sup2;
f（x）=3+5x-2x²
F (x) = three quarters X & sup2-2x

f(x)=-2x²+5x+3=-2(x²-5x/2)+3=-2(x²-5x/2+25/16-25/16)+3=-2(x²-5x/2+25/16)+2×25/16+3=-2(x-5/4)²+49/8f(x)=3/4*x²-2x=(3/4)(x²-8x/3)=(3/4)(x²-8x/3+16/9-16/9)=(3/4)(...

The vertex of quadratic function f (x) = 3 + 5x-2x ^ 2 is

(5 / 4,49 / 8) obtained by vertex coordinate formula

Given that the function f (x) = x - (K-2) x + K + 3K = 5 has two zeros A and B, find the value range of a + B
Such as the title

I'm sorry, I'll just talk about the method, but I didn't calculate the result: first of all, change the original formula to X - (K-2) x + K + 3k-5 = 0, then because there are two zeros A and B, so b-4ac > 0, that is, (K-2) - 4 (K + 3k-5) > 0, so that we can calculate the value range of K. (1) furthermore, a + B = (a + b) - 2Ab (2) according to Weida's theorem