Question

determine os zeros ou raizes de cada função abaixo
$y = - {x}^{2} + 4x + 5$
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Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-x^{2}+4x+5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-1{;}~b=4~e~c=5\\\\\Delta=(b)^{2}-4(a)(c)=(4)^{2}-4(-1)(5)=16-(-20)=36\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(4)-\sqrt{36}}{2(-1)}=\frac{-4-6}{-2}=\frac{-10}{-2}=5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(4)+\sqrt{36}}{2(-1)}=\frac{-4+6}{-2}=\frac{2}{-2}=-1\\\\S=\{5,~-1\}$