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jimthompson5910 (jim_thompson5910):
do you have an example to work with?
OpenStudy (anonymous):
Completing the square, assuming you're talking about a parabola.
OpenStudy (anonymous):
y = 3x2 + 19x – 40
jimthompson5910 (jim_thompson5910):
ok step 1 is to find the x coordinate of the vertex
jimthompson5910 (jim_thompson5910):
you use this formula
x = -b/(2a)
in this case
a = 3
b = 19
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jimthompson5910 (jim_thompson5910):
tell me what you get
OpenStudy (anonymous):
-19/6
jimthompson5910 (jim_thompson5910):
good
jimthompson5910 (jim_thompson5910):
now that you know the x coordinate of the vertex, you plug this into the original equation to find the y coordinate of the vertex
jimthompson5910 (jim_thompson5910):
y = 3x^2 + 19x – 40
y = 3(-19/6)^2 + 19(-19/6) – 40
....
....
....
y = ???
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OpenStudy (anonymous):
-841/12 ??
jimthompson5910 (jim_thompson5910):
you got it
jimthompson5910 (jim_thompson5910):
so the equation
\[\large y = 3x^2 + 19x - 40\]
turns into
\[\large y = 3\left(x + \frac{19}{6}\right)^2 - \frac{841}{12}\]
after completing the square
jimthompson5910 (jim_thompson5910):
remember that after completing the square and getting an equation into vertex form, you will have this basic form
y = a(x-h)^2 + k
where (h,k) is the vertex
OpenStudy (anonymous):
Thank you so much! Do you think you could do another example?
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jimthompson5910 (jim_thompson5910):
sure
OpenStudy (anonymous):
y = 2x2 - 12x + 22
jimthompson5910 (jim_thompson5910):
a = ??
b = ??
OpenStudy (anonymous):
a= 2 b=-12
jimthompson5910 (jim_thompson5910):
x = -b/(2a)
x = ??
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jimthompson5910 (jim_thompson5910):
to be technical, it should be
h = -b/(2a)
h = ??
OpenStudy (anonymous):
h=3
jimthompson5910 (jim_thompson5910):
this is the x coordinate of the vertex (and the axis of symmetry)
jimthompson5910 (jim_thompson5910):
plug it into the original equation to find the y coordinate of the vertex (ie the value of k)
jimthompson5910 (jim_thompson5910):
throughout this whole process, 'a' stays the same and it is a = 2 in this case
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OpenStudy (anonymous):
-40=y
jimthompson5910 (jim_thompson5910):
no, that's not correct
OpenStudy (anonymous):
4
jimthompson5910 (jim_thompson5910):
y = 2x^2-12x+22
y = 2(3)^2-12(3)+22
y = 4
correct
OpenStudy (anonymous):
so vertex = (3,4)
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jimthompson5910 (jim_thompson5910):
x coordinate of the vertex is 3 ----> h = 3
y coordinate of the vertex is 4 ----> k = 4
jimthompson5910 (jim_thompson5910):
a = 2 (given)
h = 3 (just found this)
k = 4 (just found this)
y = a(x-h)^2 + k ... vertex form
y = 2(x-3)^2 + 4 ... plug in a = 2, h = 3, k = 4
so
2x^2-12x+22
converts to
y = 2(x-3)^2 + 4
which is now in vertex form
OpenStudy (anonymous):
THANK YOU SO MUCH! this helps tremendously! It's actually not that hard!
jimthompson5910 (jim_thompson5910):
nope it's not too bad once you know a handy formula
you could go another route and complete the square, but that gets ugly sometimes
OpenStudy (anonymous):
yeah that one is really confusing....I'll stick with this one:)