Question

Faça o estudo do sinal das funções

A) f(x)= -x ao quadrado-x +6

B) f(x)= 3x ao quadrado- 5x +2

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-x^{2}-1x+6=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-1{;}~b=-1~e~c=6\\\\\Delta=(b)^{2}-4(a)(c)=(-1)^{2}-4(-1)(6)=1-(-24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-1)-\sqrt{25}}{2(-1)}=\frac{1-5}{-2}=\frac{-4}{-2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-1)+\sqrt{25}}{2(-1)}=\frac{1+5}{-2}=\frac{6}{-2}=-3\\\\S=\{2,~-3\}$

- - - - - - (-3) + + + + + (2) - - - - - -

b)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~3x^{2}-5x+2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=3{;}~b=-5~e~c=2\\\\\Delta=(b)^{2}-4(a)(c)=(-5)^{2}-4(3)(2)=25-(24)=1\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-5)-\sqrt{1}}{2(3)}=\frac{5-1}{6}=\frac{4}{6}=\frac{2}{3}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-5)+\sqrt{1}}{2(3)}=\frac{5+1}{6}=\frac{6}{6}=1\\\\S=\{\frac{2}{3},~1\}$

+ + + + + (2/3)  - - - - - - (1) + + + + +