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anonymous

  • one year ago

please help me with this question! Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form. f(x) = (x − 8)2 f(x) = (x − 4)2 − 13 f(x) = (x − 4)2 + 3 f(x) = (x − 4)2 + 16

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  1. anonymous
    • one year ago
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    The question is Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

  2. misty1212
    • one year ago
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    HI!!

  3. anonymous
    • one year ago
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    hi!

  4. misty1212
    • one year ago
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    do you have to show your work or can you just do it?

  5. anonymous
    • one year ago
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    you don't have to show your work

  6. misty1212
    • one year ago
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    ok then lets to it quick you have \[f(x)=x^2-8x+3\] right?

  7. anonymous
    • one year ago
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    yes

  8. misty1212
    • one year ago
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    what is half of 8?

  9. anonymous
    • one year ago
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    4

  10. misty1212
    • one year ago
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    so go right to \[f(x)=(x-4)^2+k\]

  11. misty1212
    • one year ago
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    to find \(k\) plug in 4 in to \[f(x)=x^2-8x+3\]

  12. anonymous
    • one year ago
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    so I solve f(x)= (x-4)^2 +k and for k, I plug in 4?

  13. misty1212
    • one year ago
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    yes, let me know what you get and i will check it

  14. anonymous
    • one year ago
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    okay I'll solve it real quick :)

  15. anonymous
    • one year ago
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    I squared the -4 in the parenthesis and got f(x)= (x-16)+4

  16. anonymous
    • one year ago
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    is that all that I have to do?

  17. misty1212
    • one year ago
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    whoa hold on

  18. misty1212
    • one year ago
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    lets back up a second

  19. misty1212
    • one year ago
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    we got half of 8 is 4, so it s going to look like \[f(x)=(x-4)^2+k\]

  20. misty1212
    • one year ago
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    now we need the number \(k\) right?

  21. anonymous
    • one year ago
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    yes and you said to plug 4 in for k right?

  22. misty1212
    • one year ago
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    no

  23. misty1212
    • one year ago
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    to find \(k\) start with \[f(x)=x^2-8x+3\] and find \(f(4)\) \[f(4)=4^2-8\times 4+3\]

  24. misty1212
    • one year ago
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    let me know what you get now

  25. anonymous
    • one year ago
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    oh! okay. I totally switched up what you said :o I'll re-solve it

  26. anonymous
    • one year ago
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    I got -13 on my calculator, but when I solved it on paper I got 35. Sorry if I'm annoying you, I just don't understand algebra 2 at all :/

  27. misty1212
    • one year ago
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    it is ok dear, not problem the correct one is \(-13\) lets do it with pencil and paper

  28. misty1212
    • one year ago
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    \[f(4)=4^2-8\times 4+3\] first square \[16-8\times 4+3\] then multiply \[16-32+3\] then subtract \[-16+3\] then add \[-13\]

  29. misty1212
    • one year ago
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    final answer to your question "vertex form" is \[f(x)=(x-4)^2-13\]

  30. anonymous
    • one year ago
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    Oh! I solved the exponent first and then subtracted. But thank you so much for helping me! :)

  31. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  32. anonymous
    • one year ago
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    Hey @misty1212 . Can you help me with one more question? It also involves the vertex form and standard form

  33. misty1212
    • one year ago
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    sure happy to!!

  34. anonymous
    • one year ago
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    It's this - Rewrite f(x) = –2(x − 3)2 + 2 from vertex form to standard form.

  35. misty1212
    • one year ago
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    ok this one is going the other way around from vertex to standard how is your algebra?

  36. misty1212
    • one year ago
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    because that is what you need \[-2(x-3)^2+2=-2(x-3)(x-3)+2\] you have to multiply out, distribute, combine like terms

  37. anonymous
    • one year ago
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    Should I solve that and then tell you what I got?

  38. misty1212
    • one year ago
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    sure don't forgot to multiply out first \[(x-3)(x-3)\] requires 4 multiplications then multiply each term by \(-2\)

  39. anonymous
    • one year ago
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    So for the first part of the equation, I got -2x+6^2 +2. Is that right so far?

  40. misty1212
    • one year ago
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    hmm no

  41. misty1212
    • one year ago
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    \(-3\times (-3)=+9\) lets go slow

  42. misty1212
    • one year ago
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    \[(x-3)(x-3)\]first we get \[x^2-3x-3x+9\] let me know when that is clear, or if not, ask

  43. anonymous
    • one year ago
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    Yes that makes sense

  44. misty1212
    • one year ago
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    you sure?

  45. misty1212
    • one year ago
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    you see where each term came from?

  46. misty1212
    • one year ago
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    some times math teachers call it "foil"

  47. anonymous
    • one year ago
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    Yeah. The x's get multiplied to each other and that makes x^2, and then you multiply the 3's to the x's and get -3x-3x. Last, you multiply the -3's to each other and you get 9. But how would we put that into standard form?

  48. anonymous
    • one year ago
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    My answers are f(x) = –2x2 − 18 f(x) = –2x2 + 12x + 20 f(x) = –2x2 + 12x − 16 f(x) = 4x2 − 24x + 38

  49. misty1212
    • one year ago
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    ok so we are at \[x^2-3x-3x+9\] which is the same as \[x^2-6x+9\]

  50. misty1212
    • one year ago
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    then \[-2(x^2-6x+9)=-2x^2+12x-18\]

  51. misty1212
    • one year ago
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    then add the \(2\) at the end, get \[x^2+12x-16\]

  52. anonymous
    • one year ago
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    So the answer would be f(x)= -2x^2+12x-16?

  53. misty1212
    • one year ago
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    yes FLVS?

  54. anonymous
    • one year ago
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    yes. Do you do FLVS?

  55. misty1212
    • one year ago
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    lol no dear, i don't

  56. anonymous
    • one year ago
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    oh haha

  57. misty1212
    • one year ago
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    good luck with the algebra \[\color\magenta\heartsuit\]

  58. anonymous
    • one year ago
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    Thank you so much! And thanks for the help again :)

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