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iwanttogotostanford

  • one year ago

Using the completing-the-square method, rewrite f(x) = x^2 − 8x + 3 in vertex form.

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  1. iwanttogotostanford
    • one year ago
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    @urbanmorgans @jjwalton04 @Rishi_K

  2. iwanttogotostanford
    • one year ago
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    WOULD IT BE: f(x) = (x − 4)2 + 3??

  3. jjwalton04
    • one year ago
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    do they have the same product?

  4. iwanttogotostanford
    • one year ago
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    these are my choices: f(x) = (x − 8)2 f(x) = (x − 4)2 − 13 f(x) = (x − 4)2 + 3 f(x) = (x − 4)2 + 16

  5. iwanttogotostanford
    • one year ago
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    yes they do @jjwalton04

  6. anonymous
    • one year ago
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    x23456y−9−12−13−12−9

  7. iwanttogotostanford
    • one year ago
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    so, would it be C???

  8. jjwalton04
    • one year ago
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    is that the only one with the same product?

  9. iwanttogotostanford
    • one year ago
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    not sure...

  10. iwanttogotostanford
    • one year ago
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    yes?

  11. jjwalton04
    • one year ago
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    then your right'

  12. iwanttogotostanford
    • one year ago
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    thanks!

  13. jjwalton04
    • one year ago
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    No prob bob

  14. iwanttogotostanford
    • one year ago
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    is this answer right> Which of the following values "completes the square," or creates a perfect square trinomial, for x2 − 12x + ___? –6 12 36 –36 i think it is C

  15. johnweldon1993
    • one year ago
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    \[\large (x-4)^2 + 3 = x^2 - 8x + 16 + 3 = x^2 - 8x + 19\] Doesn't look the same to me \[\large x^2 - 8x + 3\] Complete the square...subtract 3 from both sides of the equation \[\large x^2 - 8x = -3\] Take half the coefficient of the 'x'...and square it...add that to both sides \[\large x^2 - 8x + 16 = -3 + 16\] Rewrite as sum of squares \[\large (x-4)^2 = 13\] Finally subtract 13 from both sides to get vertex form \[\large (x-4)^2 - 13\]

  16. iwanttogotostanford
    • one year ago
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    @johnweldon1993 ok, thank you !!!! i would've gotten that wrong then...

  17. jjwalton04
    • one year ago
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    i think the answer for your second question is 36 though

  18. jjwalton04
    • one year ago
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    @johnweldon1993 am i right?

  19. johnweldon1993
    • one year ago
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    Correct

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