Question

Sa se determine functia de gradul al doilea al carei grafic intersecteaza axa Ox in punctele cu abscisele 3 si -2 si isi atinge extremul in 1/2.
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Answer 1

[tex]f(x)=ax^2+bx+c\\ f(3)=0=>9a+3b+c=0(1)\\ f(-2)=0=>4a-2b+c=0(2)\\ f'( \frac{1}{2} )=2a \frac{1}{2} +b=0=>a+b=0=>b=-a\\ (1)-(2)=>5a+5b=0|:5=>a+b=0\\ (1)=>9a-3a+c=0=>6a+c=0=>c=-6a\\ f(x)=ax^2-ax-6a \\ Pentru\ a=1=>f(x)=x^2-x-6\\ Pentru\ a=-1=>f(x)=-x^2+x+6[/tex]