iwanttogotostanford
  • iwanttogotostanford
Using the completing-the-square method, rewrite f(x) = x^2 − 8x + 3 in vertex form.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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iwanttogotostanford
  • iwanttogotostanford
@urbanmorgans @jjwalton04 @Rishi_K
iwanttogotostanford
  • iwanttogotostanford
WOULD IT BE: f(x) = (x − 4)2 + 3??
jjwalton04
  • jjwalton04
do they have the same product?

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More answers

iwanttogotostanford
  • iwanttogotostanford
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
iwanttogotostanford
  • iwanttogotostanford
yes they do @jjwalton04
anonymous
  • anonymous
x23456y−9−12−13−12−9
iwanttogotostanford
  • iwanttogotostanford
so, would it be C???
jjwalton04
  • jjwalton04
is that the only one with the same product?
iwanttogotostanford
  • iwanttogotostanford
not sure...
iwanttogotostanford
  • iwanttogotostanford
yes?
jjwalton04
  • jjwalton04
then your right'
iwanttogotostanford
  • iwanttogotostanford
thanks!
jjwalton04
  • jjwalton04
No prob bob
iwanttogotostanford
  • iwanttogotostanford
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
johnweldon1993
  • johnweldon1993
\[\large (x-4)^2 + 3 = x^2 - 8x + 16 + 3 = x^2 - 8x + 19\] 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 \[\large (x-4)^2 - 13\]
iwanttogotostanford
  • iwanttogotostanford
@johnweldon1993 ok, thank you !!!! i would've gotten that wrong then...
jjwalton04
  • jjwalton04
i think the answer for your second question is 36 though
jjwalton04
  • jjwalton04
@johnweldon1993 am i right?
johnweldon1993
  • johnweldon1993
Correct

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