anonymous
  • anonymous
Write an equation for the parabola with a vertex at (2, -1) and y-intercept 5.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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saifoo.khan
  • saifoo.khan
\[y = (x - 2)^2 -1\]
amistre64
  • amistre64
when x=0 that equals 3 tho, not 5
saifoo.khan
  • saifoo.khan
Messed up.. o_O

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saifoo.khan
  • saifoo.khan
Yeah..
saifoo.khan
  • saifoo.khan
figured//
amistre64
  • amistre64
gotta scale it right?
amistre64
  • amistre64
5 = a(-2)^2-1
amistre64
  • amistre64
its a good start tho, get the basic down and fidget with it
anonymous
  • anonymous
That was what the question asked, I am confused on how to calculate (a)
amistre64
  • amistre64
a is just a single unknown that can be algebraed into submission
amistre64
  • amistre64
add 1 and divide off the (-2)^2
anonymous
  • anonymous
My algebra teacher said the answer was \[3/2(x-2)^{2}\]
anonymous
  • anonymous
PLus 1 at the end sorry
amistre64
  • amistre64
almost, missing the -1 at the end tho
anonymous
  • anonymous
oh minus?
anonymous
  • anonymous
Ok so how did he get 3 halves?
amistre64
  • amistre64
5 = a(-2)^2 - 1 +1 +1 ---------------- 6 = a(4) /4 /4 --------- 6/4 = a
amistre64
  • amistre64
this is basic equation manipulations; you will need to become quite adept at it for further math skills
anonymous
  • anonymous
You put in the y intercept (5) in for x?
amistre64
  • amistre64
when x=0, y=5 so we zero out the x and replace y by 5
amistre64
  • amistre64
y = a(x-2)^2 - 1 5 = a(-2)^2 - 1
anonymous
  • anonymous
Is it also possible to substitute h and k as well instead?
amistre64
  • amistre64
h and k are your vertex elements
amistre64
  • amistre64
we plugged those in to determine "a"
anonymous
  • anonymous
oh ok i get it (hopefully) thank you
amistre64
  • amistre64
practice helps :)
saifoo.khan
  • saifoo.khan
Practise makes a man perfect. ;)

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