anonymous
  • anonymous
vertex for the graph of y = 2x2 + 12x + 10
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AkashdeepDeb
  • AkashdeepDeb
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php CHECK THIS SITE OUT!! :D
anonymous
  • anonymous
write in the form( x-h)^2=4a(y-k) then (h,k) is the vertex.
DebbieG
  • DebbieG
@surjithayer I don't recognize that form?? Can certainly use vertex form which is \(y=a(x-h)^2 + k\) and then (h,k) is the vertex. OR you can just use the fact that the x-coordinate of the vertex is given by: \(x=\dfrac{-b}{2a}\) to get the x-coordinate, substitute that back into the equation for the y-coordinate.

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anonymous
  • anonymous
y=2(x^2+6*x+5) implies y=2[(x+3)^2-4] (y+8)/2=(x+3)^2; a=1/8; iumplies 4*1/8(y+8)=(x+3)^2 vertex is (-3,-8)
DebbieG
  • DebbieG
If you rearrange the vertex form, you can get \(\dfrac{y-k}{a}=(x-h)^2 \) although I've never seen it used that way. :)
anonymous
  • anonymous
2(x^2+6x+5)=y 2(x^2 +6x+(3)^2-(3)^2+5)=y 2((x)^2+6x+9-9+5)=y 2(x+3)^2-8=y 2(x+3)^2=y+8 (x+3)^2=1/2 (y+8) vertex is (-3,-8)
anonymous
  • anonymous
4a=1/2 a=1/8
DebbieG
  • DebbieG
I'm still not following what your "a" is referring to. The standard form for a quadratic is \(ax^2+bx+c=0\) and here, the equation has a=2. Once you had: \(2(x+3)^2-8=y\) above, that is what is usually taught as "vertex form" \[y=a(x-h)^2 + k\] where a is the coefficient on the leading term; and (h,k) is the vertex (as you found). Using this as our understanding of "a", what you are calling a in the rearragement to \(4a(y-k)=(x-h)^2\) would be \(\dfrac{1}{4a}\). Then: \(4\cdot\dfrac{1}{4a}(y-k)=(x-h)^2\) \(\dfrac{1}{a}(y-k)=(x-h)^2\) \((y-k)=a(x-h)^2\) \(y=a(x-h)^2+k\) And when we solve: \(\dfrac{1}{4a}=\dfrac{1}{8}\) you get a=2, which is as we expect, the leading coefficient.
anonymous
  • anonymous
These are all types of parabola with vertex (0,0) and focus at F |dw:1377971271654:dw|
DebbieG
  • DebbieG
Ok, I see... but I would probably opt for a letter other than "a" to designate the x-coordinate of the focus. Lest great confusion be caused amongst beginning algebra students who have \(y=ax^2+bx+c\) \(y=a(x-h)^2 + k\) \(x=\dfrac{-b}{2a}\) endlessly drilled into their heads, with a having a very standard meaning. :)
anonymous
  • anonymous
it is also a standard notation in india.

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