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anonymous

  • 5 years ago

can anyone explain in great detail the steps to figure out a parabola

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  1. anonymous
    • 5 years ago
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    do you have alot of questions just about the directrix and focus, or do you need a whole bigger picture? I can help, but I just don't want to go off and confuse you or anything.

  2. anonymous
    • 5 years ago
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    the only thing i dont get is how u even get started and what do you figure out first what formula u use i know nothing at all so you cant confuse me just please explain in great detail exactly what i need 2 do

  3. anonymous
    • 5 years ago
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    do you know how to factor polynomials (thy're like the uq's you've been working with)?

  4. anonymous
    • 5 years ago
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    yea i think i remember a little

  5. anonymous
    • 5 years ago
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    okay, so parabolas are polynomial eq's that have a square in y or x, along with a few other expressions. The simplest forms are y = x^2 (which is a a u shape) and x=y^2 (which is a sideways u). More commonly they are in the form y=ax^2+bx+c, or x=ay^2+by+c, where a,b,and c are different constants. Does that make sense so far?

  6. anonymous
    • 5 years ago
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    i been taught 2 use (x-h)^2=4p(y-k) or the reverse is that the same as what u r saying

  7. anonymous
    • 5 years ago
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    whenever a,b and c are added in they more or less change the shape and position of both forms, not really the u, but how wide it gets and where it is positioned. What you mentioned is another way of desribing the equation, but a little more difficult to get a grasp on. I think it helps to get a hold on the y= or x = form, and also the general nature of equation transformations, before looking at the form (x-h)^2=4p(y-k), because without knowing the general form you are looking for overall, it is difficult to be given a quadratic form and be able to get it to (x-h).... and all of that, also without knowing how equations are transformed, it is haredr to realize what all of those constants are saying, and also to get it give you info you need about it. But, they are similar, just different ways of saying it. Anything with a square in it and another variable not square is some sort of parabola. (h,k) would be the origin, and (x, y+p) is th focus, but without being able to manipulate it to give you that, it's harder to figure. (1/4p) is what is subbed for a constanct

  8. anonymous
    • 5 years ago
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    ok so can u teach me a easy way to knock out dis hw im getting tired

  9. anonymous
    • 5 years ago
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    man, my comp is messing up, and to be honest, I would have to have diagrams to explain best figuring out all of the parabola stuff as clear as you seem to need it. It's not as cut and dry as taking an equation and manipulating it, because sometimes the technique that may have worked on one equation won't work on another and there are some other skills that are needed in factoring, completing the square, and just a general understaning of the behavior or quadratic eq's and how they show up. Geometic considerations that can be used to find the focus directrix and vertex. Anytime you can factor the quadratic so y= (x+a)(x+b) (those can be pluss or minus a or b )you have your zeros where the parabola crosses the x-axis. Sometimes you may have a quadratic you can't factor and have to use another equation to pull them out. Also, the vertex is half way in between the focus and directrix. The nature of the parabola is such that, the distance from the focus to any point on the parabola is equal to the distance from that point to the directrix. Before you can even tackle alot of that you have to be able to look at the equation and get it to a place where you can graph it by points or know how to do the manipulations. But that's really the best I can do. I would suggest studying the chapter it is covered in your text book and follwing the examples, and then moving on to the focus and directrix studd because you can hardly ever get the equation to show up with all the info you need for that stuff in my experience. You have to use the geomtery and points to pull it all out. Which is a decent feat to do, and you have to be on top of your fundamentals and general understanding. There is no qucik and easy way to explain it all. It would take a whole hour of class time to get it all across, and even then some previous mention of stuff about it while covering certain algebra and trig tactics, I'm sorry I couldn't help more.

  10. anonymous
    • 5 years ago
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    ok ill just give up

  11. anonymous
    • 5 years ago
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    hold on if i give u the answer to the question can u help me figure out how they got the answer

  12. anonymous
    • 5 years ago
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    not if it's dealing with a focus, or directrix and all of that, because I would need a diagram of the parabola and lines connecting those points I mentioned, to get you to understand where it is all coming from.

  13. anonymous
    • 5 years ago
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    i have the pics of the parabolas 2 cuz all the problems and graphs r in the back

  14. anonymous
    • 5 years ago
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    what I'm saying though is I would have to have drawn them, and you be able to see me walking you through it all. I could see the equation, graph it and figure all that stuff out, but explaining to you how would be impossible without the visual and you seeing me point the stuff out. I tell you what, give me the equation and we'll get you somewhere with it.

  15. anonymous
    • 5 years ago
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    y=1/2x^2

  16. anonymous
    • 5 years ago
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    so in this case, your vertex is (0,0) and your points in the - x and +x directions are symetric. (when x= -1and 1, y=(1/2), x=-2 and 2, y=2, x=-4 and 4, y=8. So you have six points to plot and make your parabola. I would choose the point (2,2) to use as your point used to figure out your focus and all of that. Because the origin is at 0 your focuse point = (0, p) and your directrix is the line y=-p

  17. anonymous
    • 5 years ago
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    do you know the distance formula between two points?

  18. anonymous
    • 5 years ago
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    ok so the book says the focus is (0,1/2) directrix is y=-1/2 and vertex is (0,0)

  19. anonymous
    • 5 years ago
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    that makes sense lookig at it, but knowing how to figure it is the deal, the a constant doesn't always give it so cut and dry. And I've always done it throught the distance formula and solving for p.

  20. anonymous
    • 5 years ago
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    dats wat we was doing in class i just dont get it but if u cant help its ok

  21. anonymous
    • 5 years ago
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    do you know the distance formula?

  22. anonymous
    • 5 years ago
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    nope i only know the formula we been using for dis stuff

  23. anonymous
    • 5 years ago
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    \[d=\sqrt{(x _{2}-x _{1})^2+(y _{2}-y _{1})^2}\] the disespecially in places where it is actually useful and not ance between pt #1 is (0,p) =(x1,y1) and (2,2)=(x2,y2) has to be = to y+p, so d = y+p. You need to kind of memorize that distance formula because it is going to come up alot for really useful things outside of this. In this case it will allow you to solve for p and find the focus and directrix. It works for distances in real life also.

  24. anonymous
    • 5 years ago
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    it comes from the pythagorean theorem relating triangle legths. This is another important thing you ought to kind of know by the time you are relating important points on a quadratic polynomial.

  25. anonymous
    • 5 years ago
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    diagrams help to explan it better too.

  26. anonymous
    • 5 years ago
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    sorry about the typos, this site sucks sometimes along with my typing disregard from the all the way to between in the remark under the distance formula. I didn't ype it in like that, but this thing jumps around on me for some reasonreason and I didn't notice it. These really seriously are all things you should know before getting into the topic you are on, otherwise you won't fully understand the manipulations and all of that. Take care and good luck.

  27. anonymous
    • 5 years ago
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    ok im done dis is way 2bhard math aint dis difficult im a visual learner anyways

  28. anonymous
    • 5 years ago
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    but thank u

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