# What is the vertex form of y=-x^2+5x?

Dec 21, 2017

${\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4}$

#### Explanation:

To find the vertex form, you need to complete the square:

$- {x}^{2} + 5 x$
$= {x}^{2} - 5 x$
$= {x}^{2} - 5 x + {\left(\frac{5}{2}\right)}^{2} - {\left(\frac{5}{2}\right)}^{2}$
$= {\left(x - \frac{5}{2}\right)}^{2} - {\left(\frac{5}{2}\right)}^{2}$
$= {\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4}$

Dec 21, 2017

$y = - {\left(x - \frac{5}{2}\right)}^{2} + \frac{25}{4}$

#### Explanation:

Given -

$y = - {x}^{2} + 5 x$

Vertex

$x = \frac{- b}{2 a} = \frac{- 5}{- 1 \times 2} = \frac{5}{2}$

At $x = \frac{5}{2}$;

$y = - {\left(\frac{5}{2}\right)}^{2} + 5 \left(\frac{5}{2}\right) = - \frac{25}{4} + \frac{25}{2} = \frac{- 25 + 50}{4} = \frac{25}{4}$

Vertex $\left(\frac{5}{2} , \frac{25}{4}\right)$

The vertex form of the quadratic equation is -

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

Where -

$a = - 1$ - coefficient of ${x}^{2}$
$h = \frac{5}{2}$ - x - coordinate of the vertex
$k = \frac{25}{4}$ - y - coordinate of the vertex

Substitute these values in the formula

$y = - 1 {\left(x - \frac{5}{2}\right)}^{2} + \frac{25}{4}$

$y = - {\left(x - \frac{5}{2}\right)}^{2} + \frac{25}{4}$