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anonymous

  • one year ago

what is the focus of the parabola given by the equation y=x^2-2x-3? a. (4,-15/4) b. (1,4) c. (1,-4) d. (1,15/4) e. (1,-15/4)

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  1. misty1212
    • one year ago
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    i swear i just saw this question a second ago

  2. misty1212
    • one year ago
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    trick is to write it as \[y+3=x^2-2x\] then complete the square on the left do you know how to do that?

  3. misty1212
    • one year ago
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    lol i mean on the RIGHT

  4. anonymous
    • one year ago
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    no i dont :( @misty1212

  5. misty1212
    • one year ago
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    ok ready? it is not hard

  6. anonymous
    • one year ago
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    ready @misty1212

  7. misty1212
    • one year ago
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    what is half of 2?

  8. misty1212
    • one year ago
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    ...

  9. anonymous
    • one year ago
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    1 @misty1212

  10. misty1212
    • one year ago
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    ok and \(1^2=1\) so we go right from \[y+3=x^2-2x\] to \[y+3+1=(x-1)^2\] or \[y+4=(x-1)^2\]

  11. misty1212
    • one year ago
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    now we need the focus, but first we need the vertex the vertex we see from our eyeballs is \((1,-4)\)

  12. misty1212
    • one year ago
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    then we compare to the standard form \[4p(y-k)=(x-h)^2\] and see that in this case \(4p=1\)so \(p=\frac{1}{4}\)

  13. misty1212
    • one year ago
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    so we travel \(\frac{1}{4}\) units up from \((1,-4)\) and get to \((1,-\frac{15}{4})\)

  14. misty1212
    • one year ago
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    what to check it

  15. anonymous
    • one year ago
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    yes please thank you @misty1212

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