👤

Sa se determine functia f:IR ->IR , f(x) = mx+n ; m,n apartinand lui IR , al carei grafic intersecteaza axele unui sistem ortogonal de coordonate in punctele A(0;2) si , respectiv, B(2;0) .

Răspuns :

A(0,2)∈graficului functiei f daca f(0)=2, mx0+n=2, n=2

B(2,0)∈graficului functiei f daca f(2)=0, 2m+n=0, n=2, 2m+2=0, m=-1

f(x)=-x+2
[tex]A(0,2) ... apartine...Gf... daca...f(0)=2=\ \textgreater \ m*0+n=2 \\ B(2,9)...apartine...Gf...daca...f(2)=0=\ \textgreater \ m*2+n=0 \\ \left \{ {{m*0+n=2} \atop {m*2+n=0}} \right. =\ \textgreater \ \left \{ {n=2} \atop {2m+n=0}} \right. =\ \textgreater \ 2m=0-2=\ \textgreater \ 2m=-2 \\ =\ \textgreater \ m=-1 \\ =\ \textgreater \ f(x)=-x+2[/tex]