Punctele sunt pe grafic nu axele !. A(a;0) ∈ Gf, inseamna ca f(a)=0, iar B(0,-3/2) ∈ Gf ⇒ f(0)=-3/2. Avem de rezolvat sistemul format de ecuatiile :[tex] \frac{2a*a+b+2}{a+a}=0,adica,2a^2+b+2=0,si, \frac{2a*0+b+2}{0+a}=- \frac{3}{2},deci,\frac{b+2}{a}=- \frac{3}{2}. [/tex].
[tex] \left \{ {{2a^2+b+2=0} \atop {2b+4=-3a}} \right.,inlocuim,b= -\frac{3a+4}{2},in,prima,2a^2- \frac{3a+4}{2}+2= 0 [/tex], deci,
4[tex] a^{2}-3a=0 [/tex], sau a(4a-3)=0, doua solutii, [tex] a_{1}=0, b_{1}=-2,si, a_{2}= \frac{3}{4}, b_{2}=- \frac{25}{8}. [/tex]
