# How do you write -2x^2+2x-5 in vertex form and identify the vertex, y intercept and x intercept?

May 25, 2015

The vertex form of a parabola is
$y = m {\left(x - a\right)}^{2} + b$ where the vertex is at $\left(a , b\right)$

Given $- 2 {x}^{2} + 2 x - 5$ (note that I added a "squared" to the first term that was not there in the original question statement)

Extract the $m$
$y = \left(- 2\right) \left({x}^{2} - x\right) - 5$

Complete the square
$y = \left(- 2\right) \left({x}^{2} - x + {\left(\frac{1}{2}\right)}^{2}\right) - 5 - \left(- 2\right) {\left(\frac{1}{2}\right)}^{2}$

$y = \left(- 2\right) {\left(x - \frac{1}{2}\right)}^{2} + \left(- \frac{9}{2}\right)$
which is in vertex form
with the vertex at $\left(\frac{1}{2} , - \frac{9}{2}\right)$

The y-intercept is the value of $y$ when $x = 0$
$y = - 2 {x}^{2} + 2 x - 5$ with $x = 0$ becomes
$y = - 5$

The x-intercepts are the values of $x$ when $y = 0$
$y = - 2 {x}^{2} + 2 x - 5$
or, more easily (using the vertex form),
$y = \left(- 2\right) {\left(x - \frac{1}{2}\right)}^{2} + \left(- \frac{9}{2}\right)$ with $y = 0$ becomes

${\left(x - \frac{1}{2}\right)}^{2} = \frac{9}{2} \cdot \left(- \frac{1}{2}\right)$

$\left(x - \frac{1}{2}\right) = \pm \frac{3}{2} i$ (Note there are no Real solutions for the x-intercept)

$x = \frac{1 \pm 3 i}{2}$