Pentru orice functie de gradul doi notata la modul general
[tex]f(x)=ax^2+bx+c[/tex]
varful va avea coordonatele [tex](\frac{-b}{2a}, \frac{-\Delta}{4a})[/tex], unde [tex]\Delta=b^2-4ac[/tex].
Pentru [tex]f(x)=x^2-4x-m[/tex] avem:
[tex]a=1 \\b=-4\\ c=-m[/tex]
[tex]\Delta=(-4)^2-4\times1\times(-m)=16+4m[/tex]
Daca [tex]V(2, -1)[/tex], atunci:
[tex]2=\frac{-(-4)}{2\times 1}[/tex] ceea ce este adevarat
si
[tex]-1=\frac{-(16+4m)}{4}[/tex]
[tex]\implies -(16+4m)=-1\times4[/tex]
[tex]\implies -16-4m=-4[/tex]
[tex]\implies 4m=-16+4[/tex]
[tex]\implies 4m=-12[/tex]
[tex]\implies m=-3[/tex]
