Question

 

giúp mik vs ạ 

Answer 1

f(0)=2023

=>\(a\cdot0^2+b\cdot0+c=2023\)

=>c=2023

=>\(f\left(x\right)=ax^2+bx+2023\)

f(1)=2027

=>\(a\cdot1^2+b\cdot1+2023=2027\)

=>a+b=4

f(-1)=2025

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+2023=2025\)

=>a-b=2

mà a+b=4

nên \(a=\dfrac{2+4}{2}=3;b=4-3=1\)

Vậy: \(f\left(x\right)=3x^2+x+2023\)

\(f\left(2\right)=3\cdot2^2+2+2023=12+2025=2037\)