Resposta:
[tex]v( - \dfrac{b}{2a} \: , \: - \dfrac{ \Delta}{4a} )[/tex]
a)
[tex]y = 2 {x}^{2} - 10x + 8 \\ a = 2 \: \: b = - 10 \: \: c = 8 \\ xv = - \dfrac{b}{2a} \\ \\ xv = \dfrac{ - ( - 10)}{2(2)} \\ \\ xv = \dfrac{10 \div 2}{4 \div 2} \\ \\ xv = \dfrac{5}{2} \\ \\ yv = f(xv) \\ \\ f( \dfrac{5}{2}) = 2( \dfrac{5}{2} ) {}^{2} - 10( \dfrac{5}{2} ) + 8 \\ \\ yv = 2( \dfrac{25}{4} ) - 25 + 8 \\ \\ yv = \dfrac{25}{2} - 17 \\ \\ yv = \dfrac{25 - 34}{2} \\ \\ yv = - \dfrac{ 9}{2} \\ \\ \blue{ v( \dfrac{5}{2} \: , \: - \dfrac{9}{2} )}[/tex]
b)
[tex]y = - {x}^{2} + 5\\ \\ a = - 1 \:b = 0\: c = 5 \\ \\ xv = \dfrac{-0}{2(-1)}\\ \\ xv = 0\\ \\ \Delta = {b}^2 - 4ac \\ \\ yv = - \dfrac{\Delta}{4a}\\ \\ yv = -\dfrac{{b}^2 - 4ac}{4a}\\ \\ yv = - \dfrac{({0}^2 - 4(-1)(5))}{4(-1)}\\ \\yv = \dfrac{-20}{-4}\\ \\ yv = 5\\ \\\blue{ v(0, 5)}[/tex]
Bons Estudos!
