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anonymous

  • one year ago

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.

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  1. Jhannybean
    • one year ago
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    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. \(t^2 + 12t=18\)

  2. Jhannybean
    • one year ago
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    Following so far?

  3. anonymous
    • one year ago
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    I think so, yes.

  4. Jhannybean
    • one year ago
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    Any questions before continuing? :o

  5. anonymous
    • one year ago
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    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.

  6. Jhannybean
    • one year ago
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    RHS = Right Hand Side (of the equation) LHS = Left Hand Side (of the equation ) :)

  7. anonymous
    • one year ago
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    Ah, okay that's what I thought. Thank you, just making sure.

  8. Jhannybean
    • one year ago
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    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?

  9. anonymous
    • one year ago
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    Yes, I believe so.

  10. Jhannybean
    • one year ago
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    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?

  11. anonymous
    • one year ago
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    12, right?

  12. Jhannybean
    • one year ago
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    That's right, so if we substituted 12 in place of b, what would our c-value be?

  13. anonymous
    • one year ago
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    36?

  14. Jhannybean
    • one year ago
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    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, \(ax^2+bx+c\) \(t^2+12t+\color{red}{36} = 18 + \color{red}{36}\)

  15. Jhannybean
    • one year ago
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    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!

  16. Jhannybean
    • one year ago
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    Do you see it now?

  17. anonymous
    • one year ago
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    Yes, and to simplify the RHS you would just need to add 18 and 36, since there are no variables, right?

  18. Jhannybean
    • one year ago
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    You are correct!

  19. anonymous
    • one year ago
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    Okay, so it would end up: \[(t+6)^2=54\] Right?

  20. Jhannybean
    • one year ago
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    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 :)

  21. Jhannybean
    • one year ago
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    \[y=a(x-h)^2+k \\ ~~~~~~~~~~~~~\downarrow \\ y=(t+6)^2-54\]

  22. anonymous
    • one year ago
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    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.

  23. Jhannybean
    • one year ago
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    Yep, we are done. \(\boxed{y=(t+6)^2-54}\)

  24. anonymous
    • one year ago
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    Oh okay, thank you so much! That helped a lot.

  25. Jhannybean
    • one year ago
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    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 :)

  26. Jhannybean
    • one year ago
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    No problem! Glad you were able to follow through :D

  27. anonymous
    • one year ago
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    I understand it a little bit more, too. I didn't understand it at all before cx

  28. Jhannybean
    • one year ago
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    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

  29. anonymous
    • one year ago
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    Well thank goodness I found OpenStudy cx

  30. Jhannybean
    • one year ago
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    :) enjoy!

  31. anonymous
    • one year ago
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    Thanks! c:

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