Calcule as coordenadas da vértice na parábola nas funções: F(X)= -X²+6X-5
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Olá,
na função quadrática
, podemos identificar os seus termos..
se as coordenadas do vértice é dada por,
, daí fazemos..
![\text{V}=\left[-\left\{ \dfrac{6}{2\cdot(-1)}\right\},\left\{ \dfrac{6^2-4\cdot(-1)\cdot(-5)}{4\cdot(-1)}\right\}\right]\\\\\\
\text{V}=\left\{ -\left(\dfrac{6}{-2} \right),-\left( \dfrac{36-20}{-4}\right)\right\}\\\\\\
\text{V}=\left\{-(-3),-\left( \dfrac{16}{-4}\right)\right\}\\\\\\
\text{V}=\{3,-(-4)\}\\\\\Large\boxed{V=(3,4)}](https://tex.z-dn.net/?f=%5Ctext%7BV%7D%3D%5Cleft%5B-%5Cleft%5C%7B+%5Cdfrac%7B6%7D%7B2%5Ccdot%28-1%29%7D%5Cright%5C%7D%2C%5Cleft%5C%7B+%5Cdfrac%7B6%5E2-4%5Ccdot%28-1%29%5Ccdot%28-5%29%7D%7B4%5Ccdot%28-1%29%7D%5Cright%5C%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%0A%5Ctext%7BV%7D%3D%5Cleft%5C%7B+-%5Cleft%28%5Cdfrac%7B6%7D%7B-2%7D+%5Cright%29%2C-%5Cleft%28+%5Cdfrac%7B36-20%7D%7B-4%7D%5Cright%29%5Cright%5C%7D%5C%5C%5C%5C%5C%5C%0A%5Ctext%7BV%7D%3D%5Cleft%5C%7B-%28-3%29%2C-%5Cleft%28+%5Cdfrac%7B16%7D%7B-4%7D%5Cright%29%5Cright%5C%7D%5C%5C%5C%5C%5C%5C%0A%5Ctext%7BV%7D%3D%5C%7B3%2C-%28-4%29%5C%7D%5C%5C%5C%5C%5CLarge%5Cboxed%7BV%3D%283%2C4%29%7D "\text{V}=\left[-\left\{ \dfrac{6}{2\cdot(-1)}\right\},\left\{ \dfrac{6^2-4\cdot(-1)\cdot(-5)}{4\cdot(-1)}\right\}\right]\\\\\\
\text{V}=\left\{ -\left(\dfrac{6}{-2} \right),-\left( \dfrac{36-20}{-4}\right)\right\}\\\\\\
\text{V}=\left\{-(-3),-\left( \dfrac{16}{-4}\right)\right\}\\\\\\
\text{V}=\{3,-(-4)\}\\\\\Large\boxed{V=(3,4)}")
na função quadrática
se as coordenadas do vértice é dada por,
AlessandroSevernini:
Obg
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