• Nov 16th 2006, 12:31 PM
jhonwashington
Hello everybody, how are you doing

sorry to bother
I just have two questions, I would really appreciate if you are so kind to answer step by step, the problem says

write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

1- x^2+2x
2- x=3y^2-12y-36

thanks!
• Nov 16th 2006, 12:47 PM
topsquark
Quote:

Originally Posted by jhonwashington
Hello everybody, how are you doing

sorry to bother
I just have two questions, I would really appreciate if you are so kind to answer step by step, the problem says

write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

1- x^2+2x
2- x=3y^2-12y-36

thanks!

I'm presuming the first is $\displaystyle y = x^2 + 2x$?

This is done by "completing the square." Note that:
$\displaystyle (x + a)^2 = x^2 + 2ax + a^2$

We have: $\displaystyle x^2 + 2x$, so by comparing the two we see that $\displaystyle 2a = 2$, so $\displaystyle a = 1$, so $\displaystyle a^2 = 1$.

But remember we can't simply add a 1, we need to add and subtract 1 so that we don't change the equation:

$\displaystyle y = x^2 + 2x$

$\displaystyle y = x^2 + 2x + 1 - 1$

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

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

The other problem is done exactly the same way.

-Dan
• Nov 16th 2006, 12:58 PM
earboth
Quote:

Originally Posted by jhonwashington
...
write the equation in the form y=a(x-h)^2+k or x=a(y-k)^2+h

1- x^2+2x
2- x=3y^2-12y-36...

Hello, Jhonwashington,

to 1): I assume that it reads:
$\displaystyle y = x^2+2x\ \Longleftrightarrow \ y= x^2+2x+\left(\frac{2}{2} \right)^2-\left(\frac{2}{2} \right)^2$. Thus:
$\displaystyle y = x^2 + 2x \ \Longleftrightarrow\ y = (x+1)^2-1$

to 2)
$\displaystyle x=3y^2-12y-36 = 3\left( \left(y^2-4y+\left(\frac{4}{2} \right)^2\right)-\left(\frac{4}{2} \right)^2 \right) -36$ =
$\displaystyle 3 \left(y-2 \right)^2-12 -36 = 3 \left(y-2 \right)^2-48$

EB
• Nov 24th 2006, 11:31 AM
jhonwashington
Thank you so much for your help topsquark, earboth
God bless you.