Question

..Sa se determine f:R→R,f(x)=ax la a 2 +bx+c, stiind ca punctele A(1;14) B(2;17),C(1/2;41/4)apartin graficului functiei

Answer 1

f(x)=ax²+bx+c

A(1;14)⇒x=1;y=f(x)=14

f(x)=ax²+bx+c⇔14=a*1²+b*1+c⇔14=a+b+c

B(2;17)⇒x=2;y=17

f(x)=ax²+bx+c⇔17=4a+2b+c

C(1/2;41/4)⇒x=1/2;y=41/4

f(x)=ax²+bx+c⇔41/4=a/4+b/2+c⇔41=a+2b+c

14=a+b+c (1)

17=4a+2b+c (2)

41=a+2b+c (3)

17-41=3a⇔-24=3a⇒a=-8 ( am scazut ec.2 cu 3 )

41-14=b⇔b=27 ( am scazut ec. 3 cu 1 )

c=-5

f(x)=-8x²+27x-5