Question

f(x) = -x^2 + 4x = - (x-2) ^2 +4

Answer 1

$V\'ertice~~~~~V(X_{v},~Y_{v})\\\\\\Onde~X_{v}=-\dfrac{b}{2a}~~e~~Y_{v}=f(X_{v})=-\dfrac{\Delta}{4a}$
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$f(x)=-x^{2}+4x=-x^{2}+4x+0~~~~(a=-1,~~b=4,~~c=0)$

Achando o x do vértice:

$X_{v}=-\dfrac{b}{2a}=-\dfrac{4}{2(-1)}=2$

Achando o y do vértice:

$Y_{v}=f(X_{v})\\\\Y_{v}=f(2)\\\\Y_{v}=-2^{2}+4\cdot2\\\\Y_{v}=-4+8\\\\\boxed{\boxed{Y_{v}=4}}$