wolf
  • wolf
y=x^2-2x-15 how do you find the vertex?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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wolf
  • wolf
and the x intercepts are at 5 and -3?
anonymous
  • anonymous
begin by completing the square
anonymous
  • anonymous
to find the 'x' coordinate of the vertex, use x = -b/2a or similarly take the first derivative, set equal to zero and solve for 'x'. To find the 'y' coord. just plug the 'x' value into the original equation.

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anonymous
  • anonymous
Yes, the 'x' intercepts are 5 and -3: (x+3)(x-5) = 0
wolf
  • wolf
and vertex at 1,-16
anonymous
  • anonymous
if you complete the square you'll arrive at \[y=(x-1)^{2}+13\] This is the standard form for a parabola. From this form, we then know that the vertex is at (1, 13)
wolf
  • wolf
oops i messed up haha
anonymous
  • anonymous
oops, me too. Sorry
wolf
  • wolf
-1,-12?
anonymous
  • anonymous
y=(x−1)2+14 so the vertex is at (1, 14)
wolf
  • wolf
obviously the derivative thing isnt working for me ahha
anonymous
  • anonymous
that should work too. Take the first derivative: dy/dx=2x-2 set the first derivative equal to zero: 2x-2=0, x = 1
anonymous
  • anonymous
plug in x= 1 to the original equation to find the y-coordinate of the vertex
anonymous
  • anonymous
I got the coordinates of the vertex to be (1,-16)... After plugging in x=1. I may be wrong.
wolf
  • wolf
but when i plug 1 in i get -16?
wolf
  • wolf
which is what i had the first time before i got confused
anonymous
  • anonymous
y(1)=\[y(1) = 1^{2}-2+15=14\]
wolf
  • wolf
-15
wolf
  • wolf
thats what you were doing wrong jjanelle haha
anonymous
  • anonymous
1-2+15=14
anonymous
  • anonymous
Oh, janelle, you got the -15 mixed up
wolf
  • wolf
hahah thank yall
anonymous
  • anonymous
I can't read, good call.
anonymous
  • anonymous
:)

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