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iwanttogotostanford

  • one year ago

HELP PLEASE IM STUCK: Using the completing-the-square method, find the vertex of the function f(x) = –2x^2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point. Maximum at (–3, 5) Minimum at (–3, 5)

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  1. jim_thompson5910
    • one year ago
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    Vertex form is \(\Large y = a(x-h)^2+k\) The vertex of that equation is \(\Large (h,k)\) From \(\Large -2x^2 + 12x + 5\) we see that \(\Large a = -2, b = 12, c = 5\) Plug the values of 'a' and 'b' into the formula \[\Large h = \frac{-b}{2a}\] to find the value of h. Tell me what you get

  2. anonymous
    • one year ago
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    Oh, I thought you already know the answer to this. Hmmm.

  3. iwanttogotostanford
    • one year ago
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    so it would be A? @jim_thompson5910

  4. jim_thompson5910
    • one year ago
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    why maximum?

  5. iwanttogotostanford
    • one year ago
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    because it increases

  6. jim_thompson5910
    • one year ago
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    you're looking at the graph I'm guessing?

  7. iwanttogotostanford
    • one year ago
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    yes

  8. jim_thompson5910
    • one year ago
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    do you know how to determine without the graph? by just looking at the value of 'a'

  9. iwanttogotostanford
    • one year ago
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    no, but can you just determine by looking at the graph?

  10. jim_thompson5910
    • one year ago
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    well let's say we don't have a graphing calculator since 'a' is a negative number, this means that the parabola opens downward like this |dw:1442446419996:dw| a good way to remember this is to think "The value of 'a' is negative. Negative means sad, so we have a sad face graph"

  11. iwanttogotostanford
    • one year ago
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    ok

  12. jim_thompson5910
    • one year ago
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    and of course, at the peak of this "sad face graph" is the vertex. Which in this case, is the max |dw:1442446478716:dw|

  13. jim_thompson5910
    • one year ago
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    do you know how to find h and k without graphing?

  14. iwanttogotostanford
    • one year ago
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    no

  15. jim_thompson5910
    • one year ago
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    use the formula I posted earlier use h = -b/(2a)

  16. jim_thompson5910
    • one year ago
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    plug in a = -2 and b = 12 into that formula what is the value of h?

  17. iwanttogotostanford
    • one year ago
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    the value of H would be -3 right?

  18. jim_thompson5910
    • one year ago
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    yes, let's do another example Let's say we had y = 3x^2-30x+85 what are the values of a,b,c ? what is the value of h?

  19. iwanttogotostanford
    • one year ago
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    a=3x^2 b=-30 c=85 and then h would be 30/18? i think i got it wrong...

  20. jim_thompson5910
    • one year ago
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    a,b,c are just numbers (no variables attached)

  21. jim_thompson5910
    • one year ago
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    a = 3, b = -30, c = 85

  22. jim_thompson5910
    • one year ago
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    To find h, we do this \[\Large h = \frac{-b}{2a}\] \[\Large h = \frac{-(-30)}{2*3}\] \[\Large h = ???\]

  23. iwanttogotostanford
    • one year ago
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    can you give me one second? please just one minute:-)

  24. jim_thompson5910
    • one year ago
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    Take all the time you need. There's no rush.

  25. iwanttogotostanford
    • one year ago
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    ok, so i worked it out and found that it was B

  26. jim_thompson5910
    • one year ago
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    I'm asking for a number

  27. jim_thompson5910
    • one year ago
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    if you plug a = 3 and b = -30 into that h formula, what result do you get?

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