# Changing standard form to vertex form

• Nov 1st 2008, 11:24 AM
live_laugh_luv27
Changing standard form to vertex form

y = 0.2955x^2 + 0.1811x + 22.382

If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

Thank you for any help!
• Nov 1st 2008, 11:46 AM
Jhevon
Quote:

Originally Posted by live_laugh_luv27

y = 0.2955x^2 + 0.1811x + 22.382

If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

Thank you for any help!

completing the square for $y = ax^2 + bx + c$

$y = a \bigg( x^2 + \frac bax + \frac ca \bigg)$

$\Rightarrow y = a \bigg[ x^2 + \frac bax + \left( \frac b{2a}\right)^2 - \left( \frac b{2a} \right)^2 + \frac ca \bigg]$

$\Rightarrow y = a \bigg[ \left( x + \frac b{2a}\right)^2 + \frac {4ac - b^2}{4a^2} \bigg]$

$\Rightarrow y = a \bigg(x + \frac b{2a} \bigg)^2 + \frac {4ac - b^2}{4a}$
• Nov 1st 2008, 12:13 PM
Shyam
Quote:

Originally Posted by live_laugh_luv27

y = 0.2955x^2 + 0.1811x + 22.382

If possible, please use the -b/2a version instead of completing the square, because that is what I was taught.

Thank you for any help!

$y = a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$

The vertex is $\left(-\frac{b}{2a}, \;\frac{4ac-b^2}{4a}\right)$

here, a = 0.2955, b = 0.1811, c = 22.382

$y = 0.2955\left(x+\frac{0.1811}{2(0.2955)}\right)^2+\f rac{4(0.2955)(22.382)-(0.1811)^2}{4(0.2955)}$

$y = 0.2955\left(x+0.3064\right)^2+22.3542$
• Nov 1st 2008, 12:21 PM
live_laugh_luv27
Quote:

Originally Posted by Shyam
$y = a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$

The vertex is $\left(-\frac{b}{2a}, \;\frac{4ac-b^2}{4a}\right)$

here, a = 0.2955, b = 0.1811, c = 22.382

$y = 0.2955\left(x+\frac{0.1811}{2(0.2955)}\right)^2+\f rac{4(0.2955)(22.382)-(0.1811)^2}{4(0.2955)}$

$y = 0.2955\left(x+0.3064\right)^2+22.3542$

Thank you so much!! (Sun)
• Nov 1st 2008, 12:25 PM
live_laugh_luv27
One question --
One post has k= 2ac – b2/2a, the other has k= 4ac – b2/2a.
Who is correct?
Thanks!