Varful V al parabolei unei functii de gradul al doilea, la cazul general
[tex]f(x)=ax^2+bx+c[/tex]
are coordonatele
[tex]x_V=\cfrac{-b}{2a}[/tex]
[tex]y_V=\cfrac{-\Delta}{4a}[/tex]
unde [tex]\Delta=b^2-4ac[/tex]
Inlocuind cu datele problemei, obtinem:
[tex]x_V=\cfrac{-(-3)}{2}=\boxed{\cfrac{3}{2}}[/tex]
[tex]\Delta=(-3)^2-4a=9-4a[/tex]
Deci
[tex]y_V=\cfrac{-(9-4a)}{4}=\cfrac{4a-9}{4}=\cfrac{4a}{4}-\cfrac{9}{4}=\boxed{a-\cfrac{9}{4}}[/tex]
Punctul V se va afla pe dreapta aceea doar daca coordonatele sale ii respecta ecuatia, adica:
[tex]x_V+2y_V-1=0[/tex]
[tex]\cfrac{3}{2}+2\times \left(a-\cfrac{9}{4}\right)-1=0[/tex]
[tex]\cfrac{3}{2}+2a-\cfrac{9}{2}-1=0[/tex]
[tex]2a-\cfrac{6}{2}-1=0[/tex]
[tex]2a-4=0[/tex]
[tex]\boxed{a=2}[/tex]
