Question

Determine o vértice da função f(x) = (-5/16)x² + 5

Answer 1

Olá,

Temos a função:

$\tt \: f(x) = \left( - \dfrac{5}{16} \right) {x}^{2} + 5$

Vértices:

$\tt \: x_{v} = \dfrac{ - b}{2a}$

$\tt \: x_{v} = \dfrac{ - 0}{2 \left( - \dfrac{5}{16} \right)} \\ \\ \tt \: x_{v} = 0$

$\tt \: y_{v} = - \dfrac{ \Delta}{4a}$

$\tt \: y_{v} = - \dfrac{ {0}^{2} - 4 \left( - \dfrac{5}{16} \right)(5)}{4\left( - \dfrac{5}{16} \right)} \\ \\ \tt \: y_{v} = \dfrac{ 4 \left( - \dfrac{5}{16} \right)(5)}{4\left( - \dfrac{5}{16} \right)} \\ \\ \tt \: y_{v} = \dfrac{ \cancel{4} \cancel{ \left( - \dfrac{5}{16} \right)}(5)}{ \cancel{4} \cancel{\left( - \dfrac{5}{16} \right)} } \\ \\ \tt \: y_{v} = 5$

Portanto:

$\boxed{\tt \: (x_{v},y_{v} )= (0,5 )}$