Question

Sa se determine functia f:R-R, f(x)=ax^2+bx+c, a diferit de 0,daca:
f(2/3)=f(-3/5); f(0)=-6; a+c= 9.

plsss

Answer 1

Inlocuiesti in functie valorile pe care ti le da ipoteza:
[tex]f(\frac{2}{3})=a*(\frac{2}{3})^2+b*\frac{2}{3}+c = \frac{4a}{9}+\frac{2b}{3}+c = \frac{4a+6b+9c}{9} \\ f(-\frac{3}{5})=a*(-\frac{3}{5})^2+b*(-\frac{3}{5})+c=\frac{9a}{25}+\frac{-3b}{5}+c=\frac{9a-15b+25c}{25} \\ f(0)=c; \\ f(\frac{2}{3})=f(-\frac{3}{5})\ \textless \ =\ \textgreater \ \frac{4a+6b+9c}{9}=\frac{9a-15b+25c}{25} \\ f(0)=-6=\ \textgreater \ c=-6 =\ \textgreater \ \frac{4a+6b-54}{9}=\frac{9a-15b-150}{25}|*225 =\ \textgreater \ \\=\ \textgreater \ 25(4a+6b-54)=9(9a-15b-150)=\ \textgreater \ \\100a+150b-1350=81a-135b-1350 =\ \textgreater \ \\ 100a-81a+150b+135b=0=\ \textgreater \ 19a+ 285b=0 \\ [/tex]
Dar noi stim ca a+c=9 si c = -6 => a-6=9 => a =15.
=> 19 * 15+285b=0 => 285+285b=0=>285(1+b)=0=>b=-1

Atunci avem functia f:R->R,f(x)=15x^2-x-6