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

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

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

Completing the Square:

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

Factoring:

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

Where did I go wrong from here?

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

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

\$\displaystyle 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

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

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

Completing the Square:

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

Factoring:

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

Where did I go wrong from here?

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

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

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

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

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

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

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

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

\$\displaystyle = 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 \$\displaystyle x^2\$ is \$\displaystyle 1\$.

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

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

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

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

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

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