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

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Answer 1 · 2015-12-16T05:49:48

$y=2(x-5/4)^2-49/8$

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

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

$y=(2x^2-5x)-3$

$y=2(x^2-5/2x)-3$

In $y=ax^2+bx+c$ , $c$ must make the bracketed polynomial a trinomial. So $c$ is $(b/2)^2$ .
$y=2(x^2-5/2x+((5/2)/2)^2-((5/2)/2)^2)-3$

$y=2(x^2-5/2x+(5/4)^2-(5/4)^2)-3$

$y=2(x^2-5/2x+25/16-25/16)-3$

Multiply $-25/16$ by the vertical stretch factor of $2$ to bring $-25/16$ outside of the brackets.
$y=2(x-5/4)^2-3-((25/16)*2)$

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

$y=2(x-5/4)^2-3-25/8$

$y=2(x-5/4)^2-49/8$

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