How do you write #y= -2x^2+4x+5# into vertex form?

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Answer 1 · 2015-04-30T05:45:48

The vertex form of a quadratic function is given by
#y = a(x - h)^2 + k#, where #(h, k)# is the vertex of the parabola.

We can use the process of Completing the Square to get this into the Vertex Form.

#y=-2x^2+4x+5#

#-> y - 5 = -2x^2 + 4x# (Transposed 5 to the Left Hand Side)

#-> y - 5 = -2(x^2 - 2x)# (Made the coefficient of #x^2# as 1)

Now we subtract #2# from each side to complete the square

#-> y - 5 - 2 = -2(x^2 - 2x ) - 2#

#-> y - 5 - 2 = -2(x^2 - 2x + 1)#

#-> y - 5 - 2 = -2(x^2 - 2x + 1^2)#

#-> y - 7 = -2(x-1)^2 #

#-> color(green)( y = -2(x-1)^2 + 7# is the Vertex Form

The vertex of the Parabola is# {1 , 7}#