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How to determin the equation of quadratic function?

  1. Jul 25, 2009 #1
    Hi guys whats up. I just have a small question i wish to ask. If you are given vertex(-2,3) and it passes through point(2,5) how do you determine its equation? Its been around 6 months since i last did some math so i am having a hard time remembering. I am not asking you guys to solve it but just to show me how to tackle it =). thanks.
     
  2. jcsd
  3. Jul 25, 2009 #2

    mathman

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    I presume you are asking for a curve of the form y=ax2+bx+c.
    The vertex at (-2,3) gives you an equation y=a(x+2)2+3.
    Plug in the point value gives you 5=16a+3 or a=1/8.
     
  4. Jul 26, 2009 #3
    hi, thanks for the reply but when a friend helped me out this is how we did it.

    we used f(x) = a(x+2) + 3

    then we sub'ed in our point (2,5) to get

    5 = a(2+2)^2 + 3
    5 = a(4)^2 + 3
    5 = 16a + 3
    2 = 16a
    a = 8

    after that we just put a back in to the equation to get

    f(x) = 8(x+2)^2 + 3

    is this correct?

    Also another question i wanted to ask is how would i get the quadratic equation if you are given an X coordinate of the vertex in 2, a = -1 and passes through point (3,4)? This is what i am thinking but i dont know if this is correct.

    f(x) = -1(x -2 )2 + 0
    4 = -1(3-2)2 + 0

    and then i woudl go on from there?
     
    Last edited: Jul 26, 2009
  5. Jul 26, 2009 #4
    Yes.

    [tex](1/8)x^2 + (1/2)x + 7/2 = 1/8 (x+2)^2+3[/tex]
    so it's the same answer mathman gave you.

    You can check it in several ways (this is always a good exercise if you aren't completely sure of your answer). We need to check that (-2,3) and (2,5) satisfy your equation so we insert x=-2 and x=2 to check the values of y:
    [tex]1/8*(-2)^2 + 1/2 * (-2) + 7/2 = 3[/tex]
    [tex]1/8*(2)^2 + 1/2 * 2 + 7/2 = 5[/tex]
    Thus you graph passes through both points. You also need to verify that it has a minimum or maximum at x = -2. This can be checked in a number of ways. In general for an equation of form:
    [tex]ax^2 + bx + c = y[/tex]
    the minimum or maximum is at x = -2a/b. If you don't know this formula, but know a little calculus you can determine local maximum and minimum points by differentiating and setting the derivative equal to 0. This gives us:
    [tex]1/4x + 1/2 = 0[/tex]
    in your case which has the solution x = -2. Since we know that a parabola has exactly one minimum or maximum point and it's differentiable at all points we know that this is it so the vertex is at x = -2. Otherwise you could try to complete the square to get:
    [tex]y = 1/8(x+2)^2 + 3[/tex]
    which shows us that [tex]y\geq 3[/tex] because the square of a real number is non-negative. Hence the vertex is at y= 3 which verifies our answer of (-2,3) for the vertex.

    EDIT: Apparently you edited your post while I replied and changed the correct answer to an incorrect answer. In the first formula you omitted ^2, but I expect this is a typo as you remember it in later formulas. After 2 = 16a you should get 1 = 8a which gives a= 1/8 as mathman pointed out, but you somehow got a=8 which is incorrect (16*8 is not 2). Apart from that you should read mathman's reply as his solution is exactly the same.
     
  6. Jul 26, 2009 #5
    ya that ^2 was a mistake, i forgot to put it there. Thanks for the help though, your explanation makes things more clearer. Its been a while so i kinda forgot how to do this but your explanation helped me out =).

    Also since i changed it to the wrong equation i am guessing i am doing this one wrong as well?

    you are given an X coordinate of the vertex in 2, a = -1 and passes through point (3,4)

    would the equation be this: y = - x² + 9x - 14?

    thanks, help would be appreciated.
     
    Last edited: Jul 26, 2009
  7. Jul 26, 2009 #6
    You did everything right except for the fact that you at some point got the equation 2= 16a and concluded that a =8. This step is not correct. You should have solved for a and gotten a = 1/8 and inserted that and then you would have gotten the correct answer.

    No. The vertex is at the x-coordinate -9/(2(-1)) = 9/2 which is not 2. In general the vertex has an x-coordinate of -b/2a where b and a are the coefficients of x and x^2.

    You are told that the x-coordinate of the vertex is 2 so you know it's of the form:
    f(x) = a(x-2)^2 + c
    where c is the y-coordinate of the vertex. You are told that a=-1 (I'm assuming that this is what was meant by a) so inserting this we get:
    f(x) = -(x-2)^2 + c
    Now inserting the point we get:
    4 = -(3-2)^2 + c
    4 = -(1)^2 + c
    4 = -1 + c
    c = 5
    Thus our function is:
    f(x) = -(x-2)^2 + 5
    Our in expanded form:
    f(x) = -x^2 + 4x + 1
     
  8. Jul 26, 2009 #7
    Ok thanks =) i will give it a go and see if i get the same answer again. Also just a graphing Question...maybe i am just over analyzing things but.. I have to graph/create a family of functions with the same X coordinate and with the initial function of y = X squared.
    so i could graph some thing like this. (0,1) (0,2) (0,3) ??? is it that easy? I have never heard of family of functions untill today so idk.
     
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