Question

Fie functia f:R->R , f(x)=(1-m)x+m+2n-4 , unde m si n sunt numere reale.
determinati numerele reale m si n stiind ca A(-1,1) si B(-2,0) apartin graficului functiei f .

va rog ... am nevoie urgent

Answer 1

A(-1,1) e pe grafic atunci f(-1)=1 rezulta (1-m)(-1)+m+2n-4=1
rezulta m-1+m+2n-4-1=0
2m+2n=6
m+n=3
B(-2,0) e pe grafic atunci f(-2)=0 rezulta (1-m)(-2)+m+2n-4=0
rezulta 2m-2+m+2n-4=0
3m+2n-6=0
3m+2n=6

m+n=3 /x2
2m+2n=6
3m+2n=6
atunci m=0
m+n=3
atunci n=3

asadar m=0, n=3

Answer 2

$f:\mathbb R \longrightarrow \mathbb R;~f(\mathrm{x)=x(1-m)+m+2n-4} \\ \\ \mathrm{A(-1;1) \in G}_f\longrightarrow f(\mathrm{-1)=1}=\bold{m-1+m+2n-4} \\ \mathrm{B(-2;0) \in G}_f \longrightarrow f(\mathrm{-2)=0=\bold{2m-2+m+2n-4}} \\ \\ \rm m-1+m+2n-4=1 \\ 2m+2n-5=1 \\ 2m+2n=6 \\ m+n=3 \\ \\ 2m-2+m+2n-4=0 \\ 3m+2n-6=0 \\ 3m+2n=6 \\ \\ m+n=3|\cdot3 \\ 3m+2n=6 \\ 3m+3n=9 \\ ------(-) \\ \not3m+3n-\not3m-2n=9-6 \\ \bold{n=3} \\ m=3-3 \\ \bold{m=0}$