What is the vertex of #y= -7(2x-1)^2-3#?

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Answer 1 · 2016-01-10T09:11:17

The vertex is #(1/2,-3)#

The vertex form of quadratic function is

#y=a(x-h)^2+k#

Where #(h,k)# is the vertex.

Our problem is
#y=-7(2x-1)^2-3#

Let us try to convert this to the form #y=a(x-h)^2+k#

#y=-7(2(x-1/2))^2 -3 #

#y=-7(2^2)(x-1/2)^2-3#

#y=-7(4)(x-1/2)^2 - 3#

#y=-28(x-1/2)^2 - 3#

Now comparing with #y=a(x-h)^2 +k#

We can see #h=1/2# and #k=-3#

The vertex is #(1/2,-3)#

Answer 2 · 2016-01-10T11:58:24

#Vertex (1/2, -3)#

This is actually the vertex form of y.
x-coordinate of vertex:
(2x - 1) = 0 --> #x = 1/2#
y-coordinate of vertex: y = -3