[tex]\sf y = -2x^2+8x[/tex]
Para achar a altura máxima, basta achar o y do vértice ou derivar igual a 0 e substituir o ponto encontrado.
1º y do vértice :
[tex]\displaystyle \sf y_v = \frac{-\Delta}{4a} \\\\ y_v = \frac{-(8^2-4\cdot (-2)\cdot 0 )}{4\cdot(-2)} \\\\\\ y_v=\frac{-64}{-8} \\\\\\ y_v = 8 \\\\ Da{\'i}} \\\\ \huge\boxed{\sf H_ {max} = 8\ m} \checkmark[/tex]
2º Derivando igual a 0 :
[tex]\sf [-2x^2+8x]'=0 \\\\ -4x + 8 = 0 \\\\ 4x = 8 \\\\ x = 2\\\\ Da{\'i}} \ f(2) \ {\'e} \ o\ m{\'a}ximo : \\\\ f(2) = -2\cdot 2^2+8\cdot 2 \\\\ f(2) = -2\cdot 4 + 16 \\\\ f(2) = -8+16 \\\\ f(2) = 8 \\\\ Portanto : \\\\ \huge\boxed{\sf H_ {max} = 8\ m} \checkmark[/tex]
