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habbster
 3 years ago
PARABOLA
how to find the vertex and points
y^2=9x
habbster
 3 years ago
PARABOLA how to find the vertex and points y^2=9x

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habbster
 3 years ago
Best ResponseYou've already chosen the best response.0done with that x=(1/9)y^2

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3Okay, now you have your quadratic equation in standard form \[ay^{2}+by+c\] Now the vertex (h,k) is given by \[\huge h=\frac{b}{2a}\]\[\huge k=\frac{4acb^{2}}{4a}\] Alternatively, you can just find h, and then evaluate the function at y = h.

habbster
 3 years ago
Best ResponseYou've already chosen the best response.0example please I still dont get it

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[h=0/\frac{ 2}{ 9 }\] i come up with this

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3Yes, indeed, and finally, you can see that h = ?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3Correct. Now you can either compute for k, or just evaluate x when y = 0.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3@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!

habbster
 3 years ago
Best ResponseYou've already chosen the best response.0oh so if the y is the one with a power the vertex is (k,h)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3I 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
 3 years ago
Best ResponseYou've already chosen the best response.3And yes, you're right.

habbster
 3 years ago
Best ResponseYou've already chosen the best response.0how about the points? where the parabola will pass thru?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.3For 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
 3 years ago
Best ResponseYou've already chosen the best response.0if i replace the value of y is equal to 1 then the value of x is 1/9?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3That is correct, and hence, the point (1/9 , 1) is on the parabola as well.

habbster
 3 years ago
Best ResponseYou've already chosen the best response.0ok i get it thank you!! for your time :)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3No problem :) Terence out ...

habbster
 3 years ago
Best ResponseYou've already chosen the best response.0@terenzreignz what is the focus and directx?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.3I'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
 3 years ago
Best ResponseYou've already chosen the best response.3That pvalue 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 xvalue of your vertex, and consider adding and subtracting (9/4) from it. Since your xvalue 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
 3 years ago
Best ResponseYou've already chosen the best response.3To 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
 3 years ago
Best ResponseYou've already chosen the best response.3What 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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