Question

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Answer 1

a) Gf ∩ Ox= A(x,0)
Avem ca: f(x)=0
               2x-3=0
                 x=3/2 ⇒ A(3/2 , 0)
Gf∩ Oy=B(0,y)
f(0)=y⇒ y=-3⇒ B(0,-3)

b)Notam M(x,f(x))
Avem ca OM =3
    √(x-0)²+(f(x)-0)² =3
     √x²+(2x-3)² =3 |()²
        x²+4x²-12x+9=9
         5x²-12x=0
         x(5x-12) =0 ⇒ x=0 sau x=12/5
Atunci avem punctele M₁(0 ; -3) si M₂(12/5 ; 9/5)