A função quadrática f(x) = - x ² + 6x - 8 possui:
a)Os zeros -2 e -4
b)Os zeros -2 e 4
c)Os zeros 2 e -4
d)Os zeros 2 e 4
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Boa noite!
Solução!
Você pode resolver essa questão substituído os pontos na função e ver quando a função vale zero,ou aplicar a formula de Bhaskara,pois as raízes são os valores que anulam a função.
![f(x)=- x^{2} +6x-8\\\\\\
a=-1\\\\\
b=6\\\\\
c=-8\\\\\\
x= \dfrac{-b\pm \sqrt{b^{2} -4.a.c} }{2.a}\\\\\\\
x= \dfrac{-6\pm \sqrt{6^{2} -4.-1.-8} }{2.(-1)}](https://tex.z-dn.net/?f=f%28x%29%3D-+x%5E%7B2%7D+%2B6x-8%5C%5C%5C%5C%5C%5C%0Aa%3D-1%5C%5C%5C%5C%5C%0Ab%3D6%5C%5C%5C%5C%5C%0Ac%3D-8%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D+-4.a.c%7D+%7D%7B2.a%7D%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B-6%5Cpm+%5Csqrt%7B6%5E%7B2%7D+-4.-1.-8%7D+%7D%7B2.%28-1%29%7D+ "f(x)=- x^{2} +6x-8\\\\\\
a=-1\\\\\
b=6\\\\\
c=-8\\\\\\
x= \dfrac{-b\pm \sqrt{b^{2} -4.a.c} }{2.a}\\\\\\\
x= \dfrac{-6\pm \sqrt{6^{2} -4.-1.-8} }{2.(-1)}")
![x= \dfrac{-6\pm \sqrt{36-32} }{-2}\\\\\\\\\
x= \dfrac{-6\pm \sqrt{4} }{-2}\\\\\\\\\
x= \dfrac{-6\pm2 }{-2}\\\\\\\\\\\\
Raizes!\\\\\\
x_{1}= \dfrac{-6+2}{-2}= \dfrac{-4}{-2}=2\\\\\\\\
x_{2}= \dfrac{-6-2}{-2}= \dfrac{-8}{-2}=4\\\\\\\\\\\\\
S=\{2,4\}\\\\\\\\\\
Resposta:Alternativa~~d](https://tex.z-dn.net/?f=x%3D+%5Cdfrac%7B-6%5Cpm+%5Csqrt%7B36-32%7D+%7D%7B-2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%0Ax%3D+%5Cdfrac%7B-6%5Cpm+%5Csqrt%7B4%7D+%7D%7B-2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C++%0Ax%3D+%5Cdfrac%7B-6%5Cpm2+%7D%7B-2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0ARaizes%21%5C%5C%5C%5C%5C%5C%0A+x_%7B1%7D%3D+%5Cdfrac%7B-6%2B2%7D%7B-2%7D%3D+%5Cdfrac%7B-4%7D%7B-2%7D%3D2%5C%5C%5C%5C%5C%5C%5C%5C%0Ax_%7B2%7D%3D+%5Cdfrac%7B-6-2%7D%7B-2%7D%3D+%5Cdfrac%7B-8%7D%7B-2%7D%3D4%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AS%3D%5C%7B2%2C4%5C%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AResposta%3AAlternativa%7E%7Ed+++++%0A+ "x= \dfrac{-6\pm \sqrt{36-32} }{-2}\\\\\\\\\
x= \dfrac{-6\pm \sqrt{4} }{-2}\\\\\\\\\
x= \dfrac{-6\pm2 }{-2}\\\\\\\\\\\\
Raizes!\\\\\\
x_{1}= \dfrac{-6+2}{-2}= \dfrac{-4}{-2}=2\\\\\\\\
x_{2}= \dfrac{-6-2}{-2}= \dfrac{-8}{-2}=4\\\\\\\\\\\\\
S=\{2,4\}\\\\\\\\\\
Resposta:Alternativa~~d")
Boa tarde!
Bons estudos!
Solução!
Você pode resolver essa questão substituído os pontos na função e ver quando a função vale zero,ou aplicar a formula de Bhaskara,pois as raízes são os valores que anulam a função.
Boa tarde!
Bons estudos!
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