Yahoo Answers: Answers and Comments for How do I find the vertex of this parabola? [Mathematics]
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From Anonymous
enUS
Mon, 12 Dec 2011 10:44:48 +0000
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Yahoo Answers: Answers and Comments for How do I find the vertex of this parabola? [Mathematics]
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From blacker: A parabola of the style yok = (xh)^2 or y=(x...
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Tue, 29 Nov 2016 01:37:49 +0000
A parabola of the style yok = (xh)^2 or y=(xh)^2+ok has (h,ok) because of the fact the vertex. In all your issues, the xcoordinate of the vertex is 0 as x^2=(x0)^2 y=x^2 vertex(0,0) y=x^2+5 vertex (0.5) y=x^23 vertex (0,3) y=x^22 vertex (0,2) to end the sq. x^2+4x, divide the coefficient of x with the aid of 2. (x+2)^2=x^2+4x+4 so, x^2+4x = (x+2)^24 (x^212x) comparable concept. x^212x=(x6)^236

From Steven: y² = 16x can be written
x = (1/16)y²
so ...
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Mon, 12 Dec 2011 10:52:39 +0000
y² = 16x can be written
x = (1/16)y²
so this is a parabola that opens to the right and has a vertex (0,0).
..to get information about this parabola, use what you learned with parabola's in the form
y = ax² + bx + c ...or... y = a(xh)² + k
just "swap the x's and y's ...but be careful writing the points as the x values and y values must still be written in their proper order.
good luck

From Keith Roberts: First note that the function is quadratic in y...
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Mon, 12 Dec 2011 11:07:31 +0000
First note that the function is quadratic in y as opposed to x. To get it into the familiar form, just swap x & y. We can do this because this amounts to just rotating the parabola, and it does not affect the geometry of the parabola, itself. Thus, x = (1/16)y^2 is transformed to y = (1/16)x^2. The focus = 1/(4a) = 4, and the (transformed) xcomponent of the vertex is x = b/(2a) = 0, and therefore, so does y=0; and the transformed vertex is (0,0). The focus, as a point in the transformed plane, in this case is (0,4). Thus, swapping y and x in order to transform back, we have the vertex (0,0), and focus at (4,0).

From ironduke8159: When x = 0, y = 0 , so vertex at (0,0)
y^2 =...
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Mon, 12 Dec 2011 10:57:43 +0000
When x = 0, y = 0 , so vertex at (0,0)
y^2 = 2px so p = 8
Focus is at (p/2 , 0) =(4,0)

From Martin F: (y  k)² = 4p(x  h) is the standard form of t...
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Mon, 12 Dec 2011 10:51:03 +0000
(y  k)² = 4p(x  h) is the standard form of the equation for a horizontal p'bola with vertex at (h,k) and opening to the right.
y² = 4px is the standard form with vertex at origin . . .
p = distance from vertex to focus
4p = 16, solve for p