Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

habbster

  • 2 years ago

PARABOLA how to find the vertex and points y^2=9x

  • This Question is Open
  1. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Isolate x first.

  2. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    done with that x=(1/9)y^2

  3. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Okay, now you have your quadratic equation in standard form \[ay^{2}+by+c\] Now the vertex (h,k) is given by \[\huge h=-\frac{b}{2a}\]\[\huge k=\frac{4ac-b^{2}}{4a}\] Alternatively, you can just find h, and then evaluate the function at y = h.

  4. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    example please I still dont get it

  5. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  6. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[h=-0/\frac{ 2}{ 9 }\] i come up with this

  7. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Yes, indeed, and finally, you can see that h = ?

  8. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    h=0 ?

  9. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Correct. Now you can either compute for k, or just evaluate x when y = 0.

  10. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x is also 0?

  11. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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!

  12. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    And yes, x = 0

  13. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh so if the y is the one with a power the vertex is (k,h)

  14. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  15. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    And yes, you're right.

  16. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how about the points? where the parabola will pass thru?

  17. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  18. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  19. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if i replace the value of y is equal to 1 then the value of x is 1/9?

  20. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    That is correct, and hence, the point (1/9 , 1) is on the parabola as well.

  21. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i get it thank you!! for your time :)

  22. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    No problem :) Terence out ...

  23. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    w8

  24. habbster
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @terenzreignz what is the focus and directx?

  25. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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...)

  26. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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.

  27. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  28. terenzreignz
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  29. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.