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anonymous

  • 5 years ago

Y=-16t^2+48t+15. Write in vertex form,I don't know why, but my algebra teacher said to divide all the numbers by 16, EXCEPT 15. Any help?

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  1. anonymous
    • 5 years ago
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    you have to bring the 15 over to the other side first, THEN factor out the -16

  2. anonymous
    • 5 years ago
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    oh you dont divide?

  3. anonymous
    • 5 years ago
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    y - 15 = -16(t^2 - 3t) then...

  4. anonymous
    • 5 years ago
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    you have to make that binomial in the parentheses a perfect square trinomial, to do this we take .5 the b term and square it, so .5 of -3 is 1.5, then that squared is our new c term which is 2.25

  5. anonymous
    • 5 years ago
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    i know that but i was just confused on why you don't divide?

  6. anonymous
    • 5 years ago
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    then we take the new c term multiplied by the number we factored out (-16) and put that number on the other side (do not do reverse operations here, if it is negative, you subtract not add). So 2.25 x -16 is -36. So the new y - 15 is now y - 15 - 36 or y - 51

  7. anonymous
    • 5 years ago
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    you dont divide in this, you just factor out the 'a' term, it is like division, but we do not get rid of it

  8. anonymous
    • 5 years ago
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    ok i get it thank you!

  9. anonymous
    • 5 years ago
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    You're welcome, if you need any more help just let me know.

  10. anonymous
    • 5 years ago
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    ok i understand that you have to multiply by -16 and why, but won't that affect the perfect trinomial? or will it turn into: \[(-16)t ^{2}-3t+2.25=-51\]

  11. amistre64
    • 5 years ago
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    you dont need to factor out all the numbers, just the t values will do

  12. amistre64
    • 5 years ago
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    \[y=-16t^2+48t+15\] \[y=(-16t^2+48t)+15\] \[y=-16(t^2-3t)+15\]

  13. amistre64
    • 5 years ago
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    the rest of it is completing the square on your t parts

  14. anonymous
    • 5 years ago
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    did i doit right then?

  15. amistre64
    • 5 years ago
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    hmm, (3/2)^2 = 9/4 sooo..... \[y=-16(t^2-3t+\frac{9}{4})-(-16)(\frac{9}{4})+15\] \[y=-16(t-\frac{3}{2})^2+36+15\] \[y=-16(t-\frac{3}{2})^2+51\]

  16. amistre64
    • 5 years ago
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    yeah, it looks like you did it good

  17. anonymous
    • 5 years ago
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    yay, ok so how would i turn this in to vertex form?

  18. amistre64
    • 5 years ago
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    the final result IS vertex form

  19. anonymous
    • 5 years ago
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    but i thought you would just get a "plus or minus this" answer

  20. amistre64
    • 5 years ago
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    thats not vertex, that +- stuff is solving for the roots

  21. amistre64
    • 5 years ago
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    vertex form is the equation using the vertex as a part of the equation

  22. anonymous
    • 5 years ago
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    ik, i need to find h and k now

  23. amistre64
    • 5 years ago
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    \[y=a(x-h)^2+k\] \[y=-16(x-\frac{3}{2})+51\]

  24. amistre64
    • 5 years ago
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    forgot to put my ^2 on there :/

  25. anonymous
    • 5 years ago
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    yes i have this so far, so now would i square root everything?

  26. anonymous
    • 5 years ago
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    WAIT IM DONE?

  27. amistre64
    • 5 years ago
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    .... its always best that when you get to the end, to stop lol

  28. anonymous
    • 5 years ago
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    oh i got negative 51...

  29. anonymous
    • 5 years ago
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    don't you subtract 15, which turns it negative, and then you take half of -3 and square it which is 2.25. 2.25 times -16 is -36. Don't you then add 36 to negative 15 and get 21?

  30. anonymous
    • 5 years ago
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    help...

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