b)
Daca A(2;2)∈ Gf ⇒ f(2)=2 ⇒ (m²-1)2 + m + 1=2 ⇒2m²-2+m+1=2 ⇒ 2m²+m-1-2=0 ⇒
⇒2m²+m-3=0
a=2
b=1 Δ=b²-4ac=1²-4·2·(-3)=1+24=25
c=-3
[tex] \frac{-1+ \sqrt{25} }{4} = \frac{-1+5}{4} = \frac{4}{4}=1[/tex]
[tex] \frac{-1-5}{4} = \frac{-6}{4} = \frac{-3}{2} [/tex]
Si acum inlocuiesti m cu 1 in f(m²-1)x+m+1