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terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
Isolate x first.
 one year ago

habbster Group TitleBest ResponseYou've already chosen the best response.0
done with that x=(1/9)y^2
 one year ago

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

habbster Group TitleBest ResponseYou've already chosen the best response.0
example please I still dont get it
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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)
 one year ago

habbster Group TitleBest ResponseYou've already chosen the best response.0
\[h=0/\frac{ 2}{ 9 }\] i come up with this
 one year ago

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

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

habbster Group TitleBest ResponseYou've already chosen the best response.0
x is also 0?
 one year ago

terenzreignz Group TitleBest 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!
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
And yes, x = 0
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
And yes, you're right.
 one year ago

habbster Group TitleBest ResponseYou've already chosen the best response.0
how about the points? where the parabola will pass thru?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

habbster Group TitleBest ResponseYou've already chosen the best response.0
if i replace the value of y is equal to 1 then the value of x is 1/9?
 one year ago

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

habbster Group TitleBest ResponseYou've already chosen the best response.0
ok i get it thank you!! for your time :)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
No problem :) Terence out ...
 one year ago

habbster Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz what is the focus and directx?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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...)
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
That 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.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago
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