Question

Punctul V(1, -4) este varful parabolei asociate functiei f:R->R , f(x) = $x^{2}$ + ax +b. Se cere f(2)

Answer 1

V(-b/2a;-∆/4a)

-b/2a =1

-∆/4a=-4. ∆= b²-4ac

f(x)=x²+ax+b

a=1 b=a. c=b

a=-2×1

∆=a²-4×1×b=a²-4b

{a=-2

{(-a²+4b)/4×1=-4

(4+4b)/4b=-4

1+b=-4

b=-5

b=-5

f(x)=x²-2x-5

f(2)=2²-2×2-5=-5

f(2)=-5