Using the completing-the-square method, rewrite f(x) = x^2 − 8x + 3 in vertex form.
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OpenStudy (iwanttogotostanford):
@urbanmorgans @jjwalton04 @Rishi_K
OpenStudy (iwanttogotostanford):
WOULD IT BE: f(x) = (x − 4)2 + 3??
OpenStudy (jjwalton04):
do they have the same product?
OpenStudy (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
OpenStudy (iwanttogotostanford):
yes they do @jjwalton04
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OpenStudy (anonymous):
x23456y−9−12−13−12−9
OpenStudy (iwanttogotostanford):
so, would it be C???
OpenStudy (jjwalton04):
is that the only one with the same product?
OpenStudy (iwanttogotostanford):
not sure...
OpenStudy (iwanttogotostanford):
yes?
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OpenStudy (jjwalton04):
then your right'
OpenStudy (iwanttogotostanford):
thanks!
OpenStudy (jjwalton04):
No prob bob
OpenStudy (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
OpenStudy (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
\[\large (x-4)^2 - 13\]
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OpenStudy (iwanttogotostanford):
@johnweldon1993 ok, thank you !!!! i would've gotten that wrong then...
OpenStudy (jjwalton04):
i think the answer for your second question is 36 though