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K.Binks

  • one year ago

How do you graph the focus (0,8) of a parabola?

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  1. dessyj1
    • one year ago
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    Hi.

  2. dessyj1
    • one year ago
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    What do you mean when you say "focus"? Are you talking about the vertex, or the y-intercept?

  3. K.Binks
    • one year ago
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    I don't really know, I had to graph the parabola y=1/32 x^2 and find the directrix and focus. I know the Directrix is 8 and someone told me the focus was (0,8) but if they hadn't, I wouldn't know how to find it or what it is on my own.

  4. K.Binks
    • one year ago
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    they didnt explain, they just told me what it was (0,8)

  5. dessyj1
    • one year ago
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    Yeah, I do not know how to do that. Maybe @triciaal can help?

  6. triciaal
    • one year ago
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    another reason not to accept answers without understanding I was thinking "doesn't this person know that the focus is just a point"

  7. K.Binks
    • one year ago
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    No, I didn't know that, which is why I was asking someone to explain

  8. dessyj1
    • one year ago
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    Sorry about that. She was talking to me.

  9. triciaal
    • one year ago
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    The focus of a parabola is the point at the center of the parabola. all points on the parabola are the same distance from this point

  10. triciaal
    • one year ago
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    |dw:1438625611026:dw|

  11. triciaal
    • one year ago
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    the focus to the vertex is the same distance of the vertex to the directrix

  12. triciaal
    • one year ago
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  13. anonymous
    • one year ago
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    @K.Binks do you understand?

  14. triciaal
    • one year ago
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    y = x^2 has the vertex at (0,0) the minimum point on the parabola when a is positive when a is a fraction as in less than 1 it stretches out the graph the y value is affected if a was a whole number greater than 1 it would make the graph steeper for a given x value the y value would be much greater

  15. triciaal
    • one year ago
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    don't want to overload you are you understanding this so far?

  16. K.Binks
    • one year ago
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    Yes, I get that so far

  17. K.Binks
    • one year ago
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    If ya'll have more explanations, can you message it to me please? This question is taking too long, I gotta move on. Thank you

  18. anonymous
    • one year ago
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    Sorry I am at work so I am in and out. Your equation to me is in the form \( \huge x^2 = 4ay \) Now to find the focus we need to find a. \( \huge y = \frac{1}{3}x^2 \) \( \huge \frac{1}{3}x^2 = 1y \) The Focus \( \huge 4a = 1\) \( \huge \frac{4a}{4} = \frac{1}{4}\) \( \huge a= \frac{1}{4}\) The directrix = x = 1/4 The latus rectum \( \huge x^2 = 1*\frac{1}{4} \) \( \huge \sqrt{x^2} = \sqrt{\frac{1}{4}} \) \( \huge \sqrt{x^2} = \sqrt{\frac{1}{4}} \) \( \huge x = \frac{1}{2} \) So the latus rectum = \(\huge (\frac{1}{2}, \frac{1}{4}) \) and \(\huge (-\frac{1}{2}, \frac{1}{4}) \) Now we have a stretch though. I have to leave to go to the gym but when I go back home I will finish up and explain more. Here is the graph, which I am not done.

  19. anonymous
    • one year ago
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    https://www.desmos.com/calculator/css6cjlswn

  20. anonymous
    • one year ago
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    The directrix = y = -1/4

  21. anonymous
    • one year ago
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    Anyways, from my post we have the focus, directirx and the latus rectum from the equation \( \huge y = x^2\ \) which is \( \huge x^2 = 4ay \) They only different in yours \(\huge y= \frac{1}{3}x^2 \) is that you have a stretch. Well, since we have all our points, all we need to do is stretch the points so Focus = \( \huge (0,\frac{1}{4}) \) Latus Rectum = \( \huge (\frac{1}{2},\frac{1}{4}) \) and \( \huge (-\frac{1}{2},\frac{1}{4})\) Now we just need to stretch them so after the stretch Focus = \( \huge (0,\frac{1}{3}) \) Latus Rectum = \( \huge (1,\frac{1}{3}) \) and \( \huge (-1,\frac{1}{3})\) And we can graph it like the following https://www.desmos.com/calculator/mxjrmz7abx Ok, well that is my take on the subject. A lot to wrap your brain around :-) . Good luck

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