Sa se determine functia de gradul 2 f:R->R, f(x)=ax^2 +bx+c, a≠0, daca graficul contine punctul A(4;5) si are varful V(2;4)

Răspuns:

Explicație pas cu pas:

Graficul contine punctul A(4,5) ->>> f(4)=5 ->>>>>>>16a+4b+c=5 (1)

V(2,4) ->>>>>>f(2)=4 ->>>4a+2b+c=4 (2)

Dar V ( -b/2a , -Δ/4a)

->>>-b/2a=2->>-b=4a->>>b=-4a (3)

Δ=[tex]b^{2}[/tex]-4ac...

Din b=-4a ->>>>> 16a-16a+c=5 ->>>c=5

->>4a+2b+c=4 ->>>4a-8a+c=4 ->>-4a+5=4->-4a=-1=>>a=1/4

b=-4*a ->b=-1

f(X)=1/4*X^2-x+5