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

Jan 10, 2016

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

#### Explanation:

The vertex form of quadratic function is

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

Where $\left(h , k\right)$ is the vertex.

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

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

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

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

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

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

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

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

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

Jan 10, 2016

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

#### Explanation:

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