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Parabolas, Canonical form, help please!

  1. Nov 8, 2011 #1
    Okay I wanna be clear, this is not an homework question.

    But I have a hard time understanding the concept of what I'm gonna show you. The teacher goes way to fast for me, honestly, and when we ask a question it's just like he doesn't wanna take the time to make us understand, kinda makes me feel stupid.

    So I'm writing me question with the hopes that you can point me a tutorial (hopefully video) or if you are good in the subject and can help me a little. I just want to grip the concept so I can have fun doing those because I'd understand...

    [PLAIN]http://img252.imageshack.us/img252/396/interrogations.jpg [Broken]

    I beleive there an equation that's like y = a (x - h)² + b
    You know, I think I understand 80% of that factorization thing, but I have no clue about the parabola.

    Things I'd like to know and remember:

    - How to know if the parabola is pointing and how much is it "open" (I hope you understand what I mean).
    - How to place the parabola at the good place (the - h thing lost me).
    - The canonical form, what's the big deal ? Why can't it be simpler :|

    I'm sorry if this is a stupid question but when I look at my notes it's all just whatever (the teacher goes so fast, I spend the time writing wh he writes on the chalkboard but I can't follow)

    thank you
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 8, 2011 #2

    LCKurtz

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    Yes, y = a(x-h)2+ k is a standard form for a parabola opening up or down. You can read important properties from this equation:

    Vertex is at (h,k)
    Opens up or down if a > 0 or a < 0, respectively
    The larger |a| is the quicker y gets large as you vary x, making the parabola appear "skinnier".

    Since these properties are so easily read from the equation, that is why the standard form is used. To get an equation in standard form if it isn't already, you complete the square. Here's a step by step example.

    y = 3x2+12x+11

    Factor the 3 out of the x terms:
    y = 3(x2 + 4x) + 11
    Complete the square inside the parentheses by adding and subtracting 4
    y = 3((x2 + 4x+ 4) - 4) + 11
    =3((x + 2)2 -4) + 11 = 3(x + 2)2 - 12 + 11 = 3(x + 2)2 - 1

    Now you have the standard form and you can read the vertex as (-2,-1) and see it opens up.
     
  4. Nov 8, 2011 #3

    Deveno

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    there are some general things to know about functions:

    f(x-h) translates (shifts, or moves) the function f(x) h units to the right.

    (if h < 0, it is more convenient to use -h, so f(x+h) translates f(x) h units to the left).

    f(x) + k translates (shifts, or moves) the function f(x) k units up.

    (again, this assumes k > 0. if k < 0, we have:

    f(x) - k moves f(x) k units down).

    so

    y = f(x-h) + k, is the same as:

    y' = f(x'), where:

    y' = y-k
    x' = x-h

    or, equivalently:

    y = y'+k
    x = x'+h

    so y = f(x-h) + k "looks just like" y = f(x), except the origin is now at (h,k).

    so really, whenever we have:

    y = a(x-h)2 + k,

    we have the parabola y = ax2 moved to have its vertex at (h,k) (a change of coordinates).

    figuring out what "h" is, is where one needs to "complete the square". LCKurtz's post is an eloquent demonstration of how to do this.

    every other "qualitative" (behavioral) aspect of a parabola, is controlled by a. if a > 0, it "opens up", if a < 0, it "opens down". the size of a determines how "fast it grows", large |a| parabolas grow fast, and look "skinny", small |a| parabolas grow slow, and look "flat".
     
  5. Nov 8, 2011 #4
    Thank you sooooo much for your explanations

    I need a few precisions if you don't mind !

    I lost you here! The 4 you added and sunstracted ? Why can you do this ?
    Is it -2 because of the "-h" in y = a (x-h)² + k ?

    I understand now! If a was say 1/100 it would almost apear flat if we zoom in ?
    If I understand, say a is 1000 the parabola would look like a straight line going up ? (actually 2 lines very close).

    Well I think you answered my question just above heheh

    One thing I'm not sure is, where do I use my zeros in the equation ?
    I have a couple of zeros, like -5, -3 and o.o, -30 in my example above.

    I see that -5 and -3 are for zero Y.
    But I don't understand the o.o (origin ordonate). I suppose the -30 is Y for X = 0.

    Do I plug it in y= a(x-h)²+k
    -30 = a (0 -h)² + k ?

    I would then resolve the other equation for my h and k, right ? :)
     
  6. Nov 8, 2011 #5
    also, can a = 0 ?
     
  7. Nov 8, 2011 #6

    LCKurtz

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    You need a 4 to complete the square and you have to subtract it back out or you have changed the equation.
    Yes
    Yes, assuming you don't change the scale of the axes.
    When x = h, the squared term is the least it can get so that is the x coordinate of the vertex and k is the y coordinate. Of course, the parabola might not even go through the origin (0,0). You get the x and y intercepts by setting y or x = 0 as usual.
     
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