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)#