Can someone help me, please?
The function f(t) = t2 + 12t − 18 represents a parabola.
Part A: Rewrite the function in vertex form by completing the square. Show your work.

- anonymous

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- schrodinger

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- Jhannybean

So first lets set this equation to 0. \(t^2+12t-18=0\)
Now keep the variables on the LHS but move the constants to the RHS. \(t^2 + 12t=18\)

- Jhannybean

Following so far?

- anonymous

I think so, yes.

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## More answers

- Jhannybean

Any questions before continuing? :o

- anonymous

I'm not too sure what the RHS and the LHS stand for- I think I have an idea but I want to be sure.

- Jhannybean

RHS = Right Hand Side (of the equation)
LHS = Left Hand Side (of the equation )
:)

- anonymous

Ah, okay that's what I thought. Thank you, just making sure.

- Jhannybean

Alright, continuing..
We're going to be expanding the LHS by completing the square, and to be consistent, whatever we do to the LHS of the equation, we must do to the RHS of the equation as well.
Now in expanding the LHS, we need to create a quadratic function such that it fits the form \(ax^2+bx+c\). Right now we only have \(ax^2+bx\) but we're missing the \(c\).
Following?

- anonymous

Yes, I believe so.

- Jhannybean

Alright good good.
To expand the LHS, we need to find a \(c\) value. To do that we follow the formula: \(c= \left(\dfrac{b}{2}\right)^2\)
Now what's our value of b?

- anonymous

12, right?

- Jhannybean

That's right, so if we substituted 12 in place of b, what would our c-value be?

- anonymous

36?

- Jhannybean

That's correct :)
We can now complete our quadratic by adding \(\color{red}{36}\) to both sides of our equation. We have now fit the quadratic form, \(ax^2+bx+c\)
\(t^2+12t+\color{red}{36} = 18 + \color{red}{36}\)

- Jhannybean

Now we simplify the LHS of our equation. \[(t+6)^2=18+36\] now we just simplify the right side, move it back over, and we're done!

- Jhannybean

Do you see it now?

- anonymous

Yes, and to simplify the RHS you would just need to add 18 and 36, since there are no variables, right?

- Jhannybean

You are correct!

- anonymous

Okay, so it would end up:
\[(t+6)^2=54\]
Right?

- Jhannybean

Yep, and to fit the vertex form, \(y=a(x-h)^2 +k\) , we just have to move the 54 over by subtracting it from both sides :)

- Jhannybean

\[y=a(x-h)^2+k \\ ~~~~~~~~~~~~~\downarrow \\ y=(t+6)^2-54\]

- anonymous

Is that the finished product or are we not finished?
Because if we aren't finished, I'm not sure what to do after that.

- Jhannybean

Yep, we are done. \(\boxed{y=(t+6)^2-54}\)

- anonymous

Oh okay, thank you so much! That helped a lot.

- Jhannybean

Your question is asking you to rewrite the function, \(y=t^2+12t-18\) in vertex form, \(y=(t+6)^2-54\) by completing the square :)

- Jhannybean

No problem! Glad you were able to follow through :D

- anonymous

I understand it a little bit more, too. I didn't understand it at all before cx

- Jhannybean

Great! That's the goal of using OpenStudy - to answer homework questions and learn more about howto solve problems similar to the ones you are stuck on :D

- anonymous

Well thank goodness I found OpenStudy cx

- Jhannybean

:) enjoy!

- anonymous

Thanks! c:

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