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Parabola Question

  1. Sep 18, 2007 #1
    Im In a 1st Year University math class and its been a while since i did this stuff and cant remember if im doing this question right so any help would be much appreciated.

    The question is to find all the info from the equation y=2(x-4)^2-3

    I used the b+/- b^2 -4ac... equation and ended up getting 16+/- square root of 24/4.

    This gives an odd number (5.22474, 2.77525) so i dont think it is right. When i apart the original equation to make into ax^2+bx+c=0 form i did 2(x^2-8x+16)-3=0 and i am not sure if that is correct either or if it should be multiplies without brackets.

    Thanks for any help
     
  2. jcsd
  3. Sep 18, 2007 #2

    Dick

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    If your problem is to find the x-intercepts (roots of the quadratic), then you did it correctly
     
  4. Sep 18, 2007 #3
    great thanks for the help!
     
  5. Sep 18, 2007 #4
    was trying to graph this and isnt the vertex (-4,-3), so then how could those x intercepts be right since they are both on positive side of the x axis, and the vertex is in the negative side?
     
  6. Sep 18, 2007 #5

    Dick

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    Because you didn't get the vertex right. It should be at the minimum value of y, right? x=-4 is not the minimum, try x=+4.
     
  7. Sep 18, 2007 #6
    that makes sense i think theres a typo on the assignment then cause it says (p,q) are the coordinates of the vertex in the form of y=a(x-p)^2+q

    and then it gives y=2(x-4)^2-3
     
  8. Sep 18, 2007 #7

    HallsofIvy

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    Why would you think there is a typo? (p,q) is the vertex of y= a(x-p)2+ q because when x= p, y= a(p-p)2+ q= a(0)+ q= q while for any other value of x, x-p is non-zero, (x-p)2 is positive so y= a(x-p)2+q is q plus some positive number: (p, q) is the lowest point on the graph.

    Comparing the general for m y= a(x-p)2+ q with y= 2(x-4)2-3 isn't it clear that p= 4 and q= -3?
    (Notice that it is (x-p)2 and (x-4)2, not (x-(-4))2!)
     
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