(x-0)^2=4(y-2)
Started by Xtian
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No, that question isn't asking you to get (x-0)^2=4(y-2).
They are asking you to find a equation of the line of symmetry.
express your original equation for Y in terms of X. then you'll get
Y= (X^2)/4 + 2
use that equation to get the line of symmetry. I believe there was a formula for it but I forgot, if anyone knows, please let us know QQ
What i did was pluging in all x values given and found the one that gives the lowest Y value, in this case, x=0
They are asking you to find a equation of the line of symmetry.
express your original equation for Y in terms of X. then you'll get
Y= (X^2)/4 + 2
use that equation to get the line of symmetry. I believe there was a formula for it but I forgot, if anyone knows, please let us know QQ
What i did was pluging in all x values given and found the one that gives the lowest Y value, in this case, x=0
You're kind of right, but I already plugged that equation in:
(x-h)^2=4a(y-k)
therefore:
(x-0)^2=4(y+2)-16
(x-0)^2=4y-8
=4(y-2)
I got that far. But then destroyer concludes:
x-0=0
+0 = +0
------------------
x=0
I think what they might be doing is that since its an equation for a parabola and they know it's facing up, we're only concerned with the x axis for the line of symmetry.
(x-h)^2=4a(y-k)
therefore:
(x-0)^2=4(y+2)-16
(x-0)^2=4y-8
=4(y-2)
I got that far. But then destroyer concludes:
x-0=0
+0 = +0
------------------
x=0
I think what they might be doing is that since its an equation for a parabola and they know it's facing up, we're only concerned with the x axis for the line of symmetry.
The origin is just (0, 2) so the line of symmetry is x = 0. When put in standard form of (y-k) = a(x-h)^2, the origin is (h, k) and the line of symmetry is x = h.
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x^2 = 4y + 8 - 16
x^2 = 4y - 8
x^2 = 4(y - 2)
Remember that x^2 is short for (x-0)^2 so the origin is (h, k) which is (0, 2).
If you put this in y = format then you'd have y = (1/4)x^2 + 2. Normally y = x^2 has origin (0,0). There's no shift in the x-coordinate because it's still just x^2 (as opposed to something like (x-7)^2). There's a shift in the y-coordinate of the origin because now you are adding 2. So when x = 0, y = 2. That's the origin.
cool! so i guess there's no calculation to do here. If I'm given the equation x^2=4(y+2)-16 and I know the equation to find the vertex of a parabola is (x-h)^2=4a(y-k) then I just identify the origin h and k from the equation. SWEET!
