Question

Exercițiul nr.6 . Răspuns:a=0

Answer 1

Varful V al parabolei unei functii de gradul al doilea, la cazul general

$f(x)=ax^2+bx+c$

are coordonatele

$x_V=\cfrac{-b}{2a}$

$y_V=\cfrac{-\Delta}{4a}$

unde $\Delta=b^2-4ac$

Inlocuind cu datele problemei, obtinem:

$x_V=\cfrac{-(-3)}{2}=\boxed{\cfrac{3}{2}}$

$\Delta=(-3)^2-4a=9-4a$

Deci

$y_V=\cfrac{-(9-4a)}{4}=\cfrac{4a-9}{4}=\cfrac{4a}{4}-\cfrac{9}{4}=\boxed{a-\cfrac{9}{4}}$

Punctul V se va afla pe dreapta aceea doar daca coordonatele sale ii respecta ecuatia, adica:

$x_V+2y_V-1=0$

$\cfrac{3}{2}+2\times \left(a-\cfrac{9}{4}\right)-1=0$

$\cfrac{3}{2}+2a-\cfrac{9}{2}-1=0$

$2a-\cfrac{6}{2}-1=0$

$2a-4=0$

$\boxed{a=2}$