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Parabolas and Other confusing equations

  1. Mar 17, 2008 #1
    So here's the deal. I have three equations for three parabolas. I need to find the x and y inercepts, and a simplified radical to go with it.

    Here are the equations.
    Parabola one: y=(x-5)^2+6, Vertex: (5,6)
    Parabola two: y=2(x-1)^2-4, Vertex: (1, -4)
    Parabola three: y=-1(x+3)^2+8, Vertex: (-3, 8)

    Does anyone know how to set this problem up and what equations I should use?
  2. jcsd
  3. Mar 17, 2008 #2
    Are you trying to find the x and y intercepts of each of the three parabolas? The x-intercept is when y = 0 and the y-intercept is when x = 0, so just plug away.
  4. Mar 17, 2008 #3
    Yes I'm trying to find the x and y intercepts of each of the three parabolas. But I don't really understand your method. Are you saying just to get each x an y alone and solve?
  5. Mar 17, 2008 #4
  6. Mar 17, 2008 #5
    Hmm, okay thanks
  7. Mar 17, 2008 #6
    Being on the x & y intercepts refers to being on the x & y axis, respectively. If you want to be on the x-axis, what must the y-value be? Same goes for the y-int value.
  8. Mar 18, 2008 #7
    that makes no sense, and it doesn't answer my question. I know what x and y intercepts are. I know that if I want to be on the x axis the y value has to be 0 and so forth.

    I want to know an equation I can use to calculate them from those three equations.
    Do I use the quadratic formula? or another quation?
  9. Mar 18, 2008 #8
    Parabola one: y=(x-5)^2+6, Vertex: (5,6)

    Show me the steps you would take to find the x-int for this problem.
  10. Mar 18, 2008 #9
    There is no x intercept. It's above the x axis and it goes up.
    but if i didn't, I would probably do the quadratic formula
  11. Mar 18, 2008 #10
    In this case, in order to use the quadratic equation, you would need the values for a, b, and c. To get that, you would need to expand (x-5)2. That's absolutely possible. However, it is even easier in this case, because plugging in 0 for y would give you:

    0 = (x-5)2 + 6 => -6 = (x-5)2

    Do you see how you can solve for x very simply? In fact, if you had the equation in the form ax2 + bx + c and not in the vertex form in which they were given, one way of solving for the roots (x-intercepts), you may remember, is completing the square, which is what you did.

    Moreover, the quadratic formula itself was figured out by completing the square for the general form ax2 + bx + c, just using all letters instead of numbers.
  12. Mar 18, 2008 #11
    Ah I see.
    Thanks. That was much more helpful than any of the other answers I got.
  13. Mar 18, 2008 #12
    LOL ... wow, you need to read a little more into what others have suggested to you. Effort goes a long way, extends further than just solving a parabolic equation.
  14. Mar 19, 2008 #13
    rocomath, I think you were extremely helpful. Thank you for your input.
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