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Vertex of the parabola

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Identify the vertex of the parabola y = (x + 2)2 – 3

    2. Relevant equations



    3. The attempt at a solution
    I would love to have an attempt but I have gone through my book and notes and I'm still confused on how to do this. If i could see an example even so i can work through all of them would be great!
     
  2. jcsd
  3. Nov 18, 2008 #2

    gabbagabbahey

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    First, there must be a typo in your post, because that equation does not describe a parabola, but rather a straight line. (unless you mean [itex]y=(x+2)^2-3[/itex])

    If you are given the equation of a parabola in the form: y=A(x-B)^2+C, you should know that its vertex is at x=B and y=C.... so the vertex of your parabola is ___?
     
  4. Nov 18, 2008 #3
    Sorry yes. I forgot to put that to the power. So the vertex woulc be (B, C)
     
  5. Nov 18, 2008 #4

    gabbagabbahey

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    Yes, but what are the values of B and C for your parabola?
     
  6. Nov 18, 2008 #5
    (2,3)
     
  7. Nov 18, 2008 #6

    gabbagabbahey

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    are you sure?...be careful with your minus signs :wink:
     
  8. Nov 18, 2008 #7
    That's the part I was unsure about. It has been 10 years since I've done any of this and can't seem to remember a single thing. If it -3 in the problem. would that make it a negative 3?
     
  9. Nov 18, 2008 #8

    tiny-tim

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    Hi Kristinanne! :smile:

    If you need to check that you have the right formula,

    remember that the vertex will be where y is a minimum,

    and (without calculus), (x + 2)2 – 3 is a minimum when … ? :wink:
     
  10. Nov 18, 2008 #9
    Yeah I went back through it and realized that they were both negative.
     
  11. Nov 18, 2008 #10
    Without derivation you could find the vertex by putting [tex]x=\frac{-b}{2a}[/tex] in the original function y, so that you got a point [tex](\frac{-b}{2a},y)[/tex] which the vertex passes through. You should also find the y intercept (0,c) and then you could draw curve "upwards" or "downwards" depending of a.
     
  12. Nov 18, 2008 #11

    Mentallic

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    Take another look at tiny-tim's post. This would have to be (in my opinion) the best way to approach the problem if you have trouble remembering the formulas.

    The vertex of the parabola is its maximum or minimum, depending on whether it curves upwards or downwards. If you are given:

    [tex]y=(x+2)^{2}-3[/tex]

    What is (in this case) the minimum value y can have, and its corresponding x value will give you the vertex.

    Hint: [tex]m^{2}\geq 0[/tex], therefore the minimum value m can have so it equals 0 is 0.
     
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