Răspuns :
A(m;-1) ∈Gf ⇒f(m)=-1
f(m)=m²-3m+1
m²-3m+1=-1
m²-3m+1+1=0
m²-3m+2=0
m²-m-2m+2=0
m(m-1)-2(m-1)=0
(m-1)(m-2)=0
m-1=0 ∨ m-2=0
m=1 ∨ m=2
f(m)=m²-3m+1
m²-3m+1=-1
m²-3m+1+1=0
m²-3m+2=0
m²-m-2m+2=0
m(m-1)-2(m-1)=0
(m-1)(m-2)=0
m-1=0 ∨ m-2=0
m=1 ∨ m=2
f(m)=m² -3m + 1 = -1
m² - 3m + 2 = 0
m²-m-2m+2=0
m(m-1)-2(m-1)=0
(m-1)(m-2)=0
m-1=0
m=1
m-2=0
m=2
A(1, -1) si A(2,-1) apartin graficului
m² - 3m + 2 = 0
m²-m-2m+2=0
m(m-1)-2(m-1)=0
(m-1)(m-2)=0
m-1=0
m=1
m-2=0
m=2
A(1, -1) si A(2,-1) apartin graficului