How do you find the vertex of this parabola $y=4x-x^2$ ?

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Answer 1 · 2015-04-17T18:58:01

Vertex (2,4)

Rewrite the equation as y = - $(x^2 -4x)$
= - $(x^2 -4x +4)$ +4
= - $(x-2)^2$ +4

The vertex is (2,4)

Answer 2 · 2015-04-17T18:59:38

Vertex is $(2, 4)$

Step 1: Complete the square

$y = -[x^2 - 4x] = -[(x - 2)^2 - 4]$

Step 2: Arrange so that you get the form $(x - x_v)^2 = 4a(y - y_v)$

$y = -[(x - 2)^2 - 4] = -(x - 2)^2 + 4$

$=> (x - 2)^2 = 4 - y$

$=> (x - 2)^2 = -(y-4)$

From here you can conclude that the vertex is at $(2, 4)$

How do you find the vertex of this parabola y=4x-x^2y=4x-x^2?