Mathematics
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OpenStudy (anonymous):
Write an equation for the parabola with a vertex at (2, -1) and y-intercept 5.
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OpenStudy (saifoo.khan):
\[y = (x - 2)^2 -1\]
OpenStudy (amistre64):
when x=0 that equals 3 tho, not 5
OpenStudy (saifoo.khan):
Messed up.. o_O
OpenStudy (saifoo.khan):
Yeah..
OpenStudy (saifoo.khan):
figured//
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OpenStudy (amistre64):
gotta scale it right?
OpenStudy (amistre64):
5 = a(-2)^2-1
OpenStudy (amistre64):
its a good start tho, get the basic down and fidget with it
OpenStudy (anonymous):
That was what the question asked, I am confused on how to calculate (a)
OpenStudy (amistre64):
a is just a single unknown that can be algebraed into submission
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OpenStudy (amistre64):
add 1 and divide off the (-2)^2
OpenStudy (anonymous):
My algebra teacher said the answer was \[3/2(x-2)^{2}\]
OpenStudy (anonymous):
PLus 1 at the end sorry
OpenStudy (amistre64):
almost, missing the -1 at the end tho
OpenStudy (anonymous):
oh minus?
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OpenStudy (anonymous):
Ok so how did he get 3 halves?
OpenStudy (amistre64):
5 = a(-2)^2 - 1
+1 +1
----------------
6 = a(4)
/4 /4
---------
6/4 = a
OpenStudy (amistre64):
this is basic equation manipulations; you will need to become quite adept at it for further math skills
OpenStudy (anonymous):
You put in the y intercept (5) in for x?
OpenStudy (amistre64):
when x=0, y=5 so we zero out the x and replace y by 5
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OpenStudy (amistre64):
y = a(x-2)^2 - 1
5 = a(-2)^2 - 1
OpenStudy (anonymous):
Is it also possible to substitute h and k as well instead?
OpenStudy (amistre64):
h and k are your vertex elements
OpenStudy (amistre64):
we plugged those in to determine "a"
OpenStudy (anonymous):
oh ok i get it (hopefully) thank you
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OpenStudy (amistre64):
practice helps :)
OpenStudy (saifoo.khan):
Practise makes a man perfect. ;)