# What is the vertex form of y= 2x^2 - 5x – 3 ?

Dec 16, 2015

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} - \frac{49}{8}$

#### Explanation:

To find the vertex form of the equation, we have to complete the square:

$y = 2 {x}^{2} - 5 x - 3$

$y = \left(2 {x}^{2} - 5 x\right) - 3$

$y = 2 \left({x}^{2} - \frac{5}{2} x\right) - 3$

In $y = a {x}^{2} + b x + c$, $c$ must make the bracketed polynomial a trinomial. So $c$ is ${\left(\frac{b}{2}\right)}^{2}$.
$y = 2 \left({x}^{2} - \frac{5}{2} x + {\left(\frac{\frac{5}{2}}{2}\right)}^{2} - {\left(\frac{\frac{5}{2}}{2}\right)}^{2}\right) - 3$

$y = 2 \left({x}^{2} - \frac{5}{2} x + {\left(\frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2}\right) - 3$

$y = 2 \left({x}^{2} - \frac{5}{2} x + \frac{25}{16} - \frac{25}{16}\right) - 3$

Multiply $- \frac{25}{16}$ by the vertical stretch factor of $2$ to bring $- \frac{25}{16}$ outside of the brackets.
$y = 2 {\left(x - \frac{5}{4}\right)}^{2} - 3 - \left(\left(\frac{25}{16}\right) \cdot 2\right)$

y=2(x-5/4)^2-3- ((25/color(red)cancelcolor(black)16^8)*color(red)cancelcolor(black)2)

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} - 3 - \frac{25}{8}$

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} - \frac{49}{8}$