anonymous
  • anonymous
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)
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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misty1212
  • misty1212
i swear i just saw this question a second ago
misty1212
  • misty1212
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?
misty1212
  • misty1212
lol i mean on the RIGHT

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anonymous
  • anonymous
no i dont :( @misty1212
misty1212
  • misty1212
ok ready? it is not hard
anonymous
  • anonymous
ready @misty1212
misty1212
  • misty1212
what is half of 2?
misty1212
  • misty1212
...
anonymous
  • anonymous
1 @misty1212
misty1212
  • misty1212
ok and \(1^2=1\) so we go right from \[y+3=x^2-2x\] to \[y+3+1=(x-1)^2\] or \[y+4=(x-1)^2\]
misty1212
  • misty1212
now we need the focus, but first we need the vertex the vertex we see from our eyeballs is \((1,-4)\)
misty1212
  • misty1212
then we compare to the standard form \[4p(y-k)=(x-h)^2\] and see that in this case \(4p=1\)so \(p=\frac{1}{4}\)
misty1212
  • misty1212
so we travel \(\frac{1}{4}\) units up from \((1,-4)\) and get to \((1,-\frac{15}{4})\)
misty1212
  • misty1212
what to check it
anonymous
  • anonymous
yes please thank you @misty1212

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