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

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Answer 1 · 2016-01-10T09:11:17

The vertex is $(\frac{1}{2}, - 3)$ $(1/2,-3)$

The vertex form of quadratic function is

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

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

Our problem is
$y = - 7 {(2 x - 1)}^{2} - 3$ $y=-7(2x-1)^2-3$

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

$y = - 7 {(2 (x - \frac{1}{2}))}^{2} - 3$ $y=-7(2(x-1/2))^2 -3$

$y = - 7 (2^{2}) {(x - \frac{1}{2})}^{2} - 3$ $y=-7(2^2)(x-1/2)^2-3$

$y = - 7 (4) {(x - \frac{1}{2})}^{2} - 3$ $y=-7(4)(x-1/2)^2 - 3$

$y = - 28 {(x - \frac{1}{2})}^{2} - 3$ $y=-28(x-1/2)^2 - 3$

Now comparing with $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2 +k$

We can see $h = \frac{1}{2}$ $h=1/2$ and $k = - 3$ $k=-3$

The vertex is $(\frac{1}{2}, - 3)$ $(1/2,-3)$

Answer 2 · 2016-01-10T11:58:24

$V e r t e x (\frac{1}{2}, - 3)$ $Vertex (1/2, -3)$

This is actually the vertex form of y.
x-coordinate of vertex:
(2x - 1) = 0 --> $x = \frac{1}{2}$ $x = 1/2$
y-coordinate of vertex: y = -3

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