A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Open

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

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

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

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

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

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

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

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

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

terenzreignz
 one year 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
 one year 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
 one year 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
 one year 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.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.