anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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Jhannybean
  • 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
  • Jhannybean
Following so far?
anonymous
  • anonymous
I think so, yes.

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Jhannybean
  • Jhannybean
Any questions before continuing? :o
anonymous
  • 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
  • Jhannybean
RHS = Right Hand Side (of the equation) LHS = Left Hand Side (of the equation ) :)
anonymous
  • anonymous
Ah, okay that's what I thought. Thank you, just making sure.
Jhannybean
  • 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
  • anonymous
Yes, I believe so.
Jhannybean
  • 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
  • anonymous
12, right?
Jhannybean
  • Jhannybean
That's right, so if we substituted 12 in place of b, what would our c-value be?
anonymous
  • anonymous
36?
Jhannybean
  • 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
  • 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
  • Jhannybean
Do you see it now?
anonymous
  • anonymous
Yes, and to simplify the RHS you would just need to add 18 and 36, since there are no variables, right?
Jhannybean
  • Jhannybean
You are correct!
anonymous
  • anonymous
Okay, so it would end up: \[(t+6)^2=54\] Right?
Jhannybean
  • 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
  • Jhannybean
\[y=a(x-h)^2+k \\ ~~~~~~~~~~~~~\downarrow \\ y=(t+6)^2-54\]
anonymous
  • 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
  • Jhannybean
Yep, we are done. \(\boxed{y=(t+6)^2-54}\)
anonymous
  • anonymous
Oh okay, thank you so much! That helped a lot.
Jhannybean
  • 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
  • Jhannybean
No problem! Glad you were able to follow through :D
anonymous
  • anonymous
I understand it a little bit more, too. I didn't understand it at all before cx
Jhannybean
  • 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
  • anonymous
Well thank goodness I found OpenStudy cx
Jhannybean
  • Jhannybean
:) enjoy!
anonymous
  • anonymous
Thanks! c:

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