habbster
  • habbster
PARABOLA how to find the vertex and points y^2=9x
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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terenzreignz
  • terenzreignz
Isolate x first.
habbster
  • habbster
done with that x=(1/9)y^2
terenzreignz
  • terenzreignz
Okay, now you have your quadratic equation in standard form \[ay^{2}+by+c\] Now the vertex (h,k) is given by \[\huge h=-\frac{b}{2a}\]\[\huge k=\frac{4ac-b^{2}}{4a}\] Alternatively, you can just find h, and then evaluate the function at y = h.

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habbster
  • habbster
example please I still dont get it
terenzreignz
  • terenzreignz
Well, I'm used to them being functions of x, so let's use y = f(x) for now, ok? It's practically the same for y, anyway... so suppose y = x² + 4x + 8 then for this case your a = 1 b = 4 c = 8 Plugging them in your formula gives you h = -2 k = 4 So the vertex of this parabola is at (-2 , 4)
habbster
  • habbster
\[h=-0/\frac{ 2}{ 9 }\] i come up with this
terenzreignz
  • terenzreignz
Yes, indeed, and finally, you can see that h = ?
habbster
  • habbster
h=0 ?
terenzreignz
  • terenzreignz
Correct. Now you can either compute for k, or just evaluate x when y = 0.
habbster
  • habbster
x is also 0?
terenzreignz
  • terenzreignz
@habbster SOMETHING REALLY IMPORTANT When you have x², then after computing for h and k, your vertex is at (h,k) BUT when you have a y², then after computing for h and k, your vertex is at (k,h) DON'T FORGET!
terenzreignz
  • terenzreignz
And yes, x = 0
habbster
  • habbster
oh so if the y is the one with a power the vertex is (k,h)
terenzreignz
  • terenzreignz
I had to give you the more difficult, yet general formula to prepare you for more complicated parabolas in the future. However, now, you must remember that whenever your quadratic equation is of the form y = kx² or x = ky² in other words, b = 0 and c = 0 Then your vertex is always, ALWAYS at (0,0)
terenzreignz
  • terenzreignz
And yes, you're right.
habbster
  • habbster
how about the points? where the parabola will pass thru?
terenzreignz
  • terenzreignz
Well, just give a random value for y, say m, and solve for x(m). (which is x evaluated when y = m) in that case, the point (x(m), m) is a point on the parabola. You can choose any real number value for y.
terenzreignz
  • terenzreignz
For example, I like the number 6. Then x = (1/9)y² x = (1/9)(6)² x = (1/9)(36) x = 4 Hence, the point (4,6) is on the parabola.
habbster
  • habbster
if i replace the value of y is equal to 1 then the value of x is 1/9?
terenzreignz
  • terenzreignz
That is correct, and hence, the point (1/9 , 1) is on the parabola as well.
habbster
  • habbster
ok i get it thank you!! for your time :)
terenzreignz
  • terenzreignz
No problem :) Terence out ...
habbster
  • habbster
w8
habbster
  • habbster
@terenzreignz what is the focus and directx?
terenzreignz
  • terenzreignz
I'm really sorry I left too soon... Another form for a parabola is this equation: \[\huge (x - x_{o})^{2} = 4p(y - y_{o})\]or alternatively \[\huge (y - y_{o})^{2} = 4p(x - x_{o})\] First express the equation in this form... (this can usually be done by completing the square, etc...)
terenzreignz
  • terenzreignz
That p-value there is key. Anyway, your equation, written like this, takes the form \[\huge (y - 0)^{2} = 4\left( \frac{9}{4} \right)(x - 0)\]or just simply\[\huge y^{2}=4\left( \frac{9}{4} \right)x\] You can see that the value for p is (9/4). Then since this is a horizontal parabola (you have a y²), then your focus and directrix are both (9/4) units away from your vertex along the horizontal. While that may not sound too appealing, it simply means you take the x-value of your vertex, and consider adding and subtracting (9/4) from it. Since your x-value is 0, then adding (9/4) gives you (9/4) and subtracting (9/4) gives you -(9/4). So, copy the y value, you get two points: \[\large \left( \frac{9}{4},0 \right) \ and \ \left( -\frac{9}{4},0 \right)\]. One of these is your focus.
terenzreignz
  • terenzreignz
To find out which one, it is necessary to find out in which direction your parabola opens. This particular parabola opens to the right, as the coefficient of y² is positive. This means that your focus is to the RIGHT of your vertex, hence, \[\left( \frac{9}{4},0 \right)\]is your focus.
terenzreignz
  • terenzreignz
What about the directrix? It is the line parallel to the y axis passing through the other point we got earlier. So you can see that it is the line \[x = -\frac{9}{4}\]and now you're done.

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