Olá,
Temos a função:
[tex] \tt \: f(x) = \left( - \dfrac{5}{16} \right) {x}^{2} + 5 \\ [/tex]
Vértices:
[tex] \tt \: x_{v} = \dfrac{ - b}{2a} \\ [/tex]
[tex] \tt \: x_{v} = \dfrac{ - 0}{2 \left( - \dfrac{5}{16} \right)} \\ \\ \tt \: x_{v} = 0 \\[/tex]
[tex] \tt \: y_{v} = - \dfrac{ \Delta}{4a} \\[/tex]
[tex] \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 \\ \\ [/tex]
Portanto:
[tex] \boxed{\tt \: (x_{v},y_{v} )= (0,5 )} \\ [/tex]
