Question

f:R-R, f(x)=x²-6x+10. Demonstrati ca f(3-x)=f(3+x) ∈ R

Answer 1

f(x) functiede grad 2
a=1
b=-6
x=-b/2a=-(-6)/2=3 axa de simetrie
deci evident f(3-x) =f(3+x)
as simple as that!!!
 cine vrea sa calculeze, let it be my guest!

Altfel
se observa ca f(x) =(x-3)²+1 care este translatarea pe y a lui g(x) =(x-3)² care are axa de simetrie x=3
iar g(x-3)=(x-6)²
si g(3-x)=(6-x)²=(x-6)²=g(x-3)

sau  si mai  altfel , in o singura expresie
f(x) =(x-3)²+1
f(x-3)=(x-6)²+1
f(3-x)=(6-x)²+1=f(x-3)
C.C.T.D.