How do you write the quadratic in vertex form given #y=-x^2+6x-5#?

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Answer 1 · 2015-04-29T10:41:23

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=-x^2+6x-5#

#-> y + 5 = -x^2 + 6x# (Transposed -5 to the Left Hand Side)

#-> y + 5 = -1(x^2 - 6x)# (Made the coefficient of #x^2# as 1)

Now we subtract #9# from each side to complete the square

#-> y + 5 - 9 = -1(x^2 - 6x + 3^2)#

#-> y - 4 = -1(x-3)^2 #

# -> color(green)(y = -(x-3)^2 + 4# is the Vertex Form

The vertex of the Parabola is# {3 , 4}#