# quadratic function general form to standard form

• Mar 9th 2010, 08:11 PM
Anemori
quadratic function general form to standard form
Ok here is what I did. I'm kinda weak from factoring

$f(x)= 2x^2+8x-2$

group:
$f(x)= (2x^2+8x)-2$

Completing the Square:

$f(x)= (2x^2+8x+16-16)-2$
$f(x)= (2x^2+8x+16)-16-2$
$f(x)= (2x^2+8x+16)-18$

Factoring:

$f(x)= 2(x^2+4x+8)-18$

Where did I go wrong from here?

$f(x)= 2(x+2)-18$

bcoz if I FOIL 2(x+2)(x+2) it will come out:

$2(x^2+4x+4)$
• Mar 9th 2010, 08:14 PM
Prove It
Quote:

Originally Posted by Anemori
Ok here is what I did. I'm kinda weak from factoring

$f(x)= 2x^2+8x-2$

group:
$f(x)= (2x^2+8x)-2$

Completing the Square:

$f(x)= (2x^2+8x+16-16)-2$
$f(x)= (2x^2+8x+16)-16-2$
$f(x)= (2x^2+8x+16)-18$

Factoring:

$f(x)= 2(x^2+4x+8)-18$

Where did I go wrong from here?

$f(x)= 2(x+2)-18$

bcoz if I FOIL 2(x+2)(x+2) it will come out:

$2(x^2+4x+4)$

You can't complete the square unless the coefficient of $x^2$ is $1$.

$f(x)= 2x^2+8x-2$

$= 2(x^2 + 4x - 1)$

$= 2(x^2 + 4x + 2^2 - 2^2 - 1)$

$= 2[(x + 2)^2 - 5]$

$= 2(x + 2)^2 - 10$.
• Mar 9th 2010, 08:33 PM
Anemori
Quote:

Originally Posted by Prove It
You can't complete the square unless the coefficient of $x^2$ is $1$.

$f(x)= 2x^2+8x-2$

$= 2(x^2 + 4x - 1)$

$= 2(x^2 + 4x + 2^2 - 2^2 - 1)$

$= 2[(x + 2)^2 - 5]$

$= 2(x + 2)^2 - 10$.

yea i look back my answer and i made a mistake after the grouping.
thanks!