Question

E9. Să se determine funcţia
f:R → R, f(x) = ax + b dacă:
a) Gf intersectat cu Ox = {A(2, 0)}, Gf intersectat cu Oy={B(Q 4};
b) Gf intersectat cu Ox={A(3a – 1, 0)}.
Gf intersectat cuOy = {B(0, - 4)};
c) Gf intersectat cu 0x = {A (7 + 2a, 0)}. Gf intersectat cu Oy = {B(0, – 3a)}.

Answer 1

Răspuns:

f(x)=ax+b

a) f(2)=0 <=> 2a+b = 0

f(0)=4 <=> a*0+b = 4 => b = 4

2a+b = 0 => 2a+4=0 => 2a = -4 => a = -2

f(x)= -2x+4

Verificare: f(2)=0 => -2*2+4 = -4+4 = 0 (Adevarat)

                f(0)=4 => -2*0+4 = 4 (Adevarat)

b) f(3a-1)=0 => a(3a-1)+b = 0 <=> $3a^2-a+b$ = 0

f(0)= -4 => a*0+b = -4 => b = -4

$3a^2-a-b$ = 0 <=> $3a^2-a-4$ = 0 <=> $3a^2-a-4 <=> 3a^2+3a-4a-4 <=> 3a(a+1)-4(a+1) <=> (a+1)(3a-4)\\=> a1 = -1 ; a2 = 4/3$

f(x) = -x-4 ; f(x)= $\frac{4x}{3}$-4

Verificare: f(0) = -4 => -0-4 = -4 (Adevarat)

                f(0)= $\frac{4*0}{3} - 4 = -4$

c) f(7+2a) = 0 <=> a(7+2a)+b = 0 <=> $7a+2a^2+b = 0$

f(0)= -3a <=> a*0+b = -3a <=> b = -3a

$7a+2a^2-3a = 0 <=> 2a^2+4a = 0 => 2a(a+2)=0 => \\=> a1 = 0 ; a2 = -2$

=> b = -3a <=> b1 = 0 ; b2 = 6

f(x)= 0*x+6 = 6 ; f(x)= -2x+6

Verificare: (cand a = b = 0) : f(0)=0 (adevarat)

cand a = -2; b = 6 => f(0)= 6 => -2*0+6 = 6 (Adevarat)

Explicație pas cu pas: