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

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