Question

Cho f(x) = ax²+bx+c, biết f(4) = f(-4). CMR: f(x) = f(-x)

Answer 1

\(f\left(4\right)=a.4^2+b.4+c=16a+4b+c\)

\(f\left(4\right)=a.\left(-4\right)^2+b.\left(-4\right)+c=16a-4b+c\)

\(f\left(4\right)=f\left(-4\right)\Rightarrow16a+4b+c=16a-4b+c\\ \Rightarrow16a+4b+c-16a+4b-c=0\\ \Rightarrow8b=0\\ \Rightarrow b=0\)

Ta có: \(f\left(x\right)=ax^2+bx+c=ax^2+0x+c=ax^2+c\) (1)

\(f\left(-x\right)=a\left(-x\right)^2+b\left(-x\right)+c=ax^2+0\left(-x\right)+c=ax^2+c\) (2)

Từ (1), (2)\(\Rightarrow f\left(x\right)=f\left(-x\right)\)