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Curve function help

  1. Jun 10, 2010 #1
    I am trying to figure out this curve function for the given data.

    y intercept = 1
    i have x intercept values of 4 and -4,

    it is a parablola.

    I can remember how to integrate, but when drew out the curve I could not figure out the function of it for the life of me.

    Let me know if you need more info. thanks.
  2. jcsd
  3. Jun 10, 2010 #2

    Gib Z

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    You know the equation is a quadratic, and you know its two roots. Not too hard if you think about it.
  4. Jun 10, 2010 #3
    so x=-4 and x=4.

    then i get (x+4)(X-4) = x^2 -16

    then what with the y intercept.

    it's been a long time since i have done any math.
  5. Jun 10, 2010 #4


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    It's actually y=a(x-4)(x+4) where a is some constant multiplier. Notice that this still makes it a quadratic with the specifications that its roots be 4 and -4.

    Now what value of a will make the quadratic pass through y-axis at 1? Notice that this point is (0,1).
  6. Jun 10, 2010 #5
    -1/16th correct? - 1/16 * (x^2 -16) going to give you 1 at the y intercept.

    would this be the actual curve function? -1/16X^2 + 1?

    When I integrate that for the area, why is the value negative? this was what i recieved after integration when b = 8 and a = 0. answer -2.667
    Last edited: Jun 10, 2010
  7. Jun 10, 2010 #6


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    Yes that's correct. But you could've also found the value of a by plugging in y=1 and x=0 into y=a(x-4)(x+4) and solving for a.

    By b and a do you mean the limits of integration [tex]\int_a^b[/tex] ? If so, you're getting a negative answer because there is more area under the x-axis between [tex]4\leq x \leq 8[/tex] than there is above the x-axis at [tex]0\leq x\leq 4[/tex].
  8. Jun 10, 2010 #7
    yes those are my limits. with my sketch, the curve is above the x axis from 0 to 8. How is there more area under the x-axis when I thought the area I was trying to get is above the x axis. Does that make any sense?
  9. Jun 10, 2010 #8


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    Weren't we just previously finding a quadratic that has roots of -4 and 4, and cuts the y-axis at 1? What does this tell you about where the quadratic is above the x-axis, and it's not between 0 and 8.
  10. Jun 10, 2010 #9
    i'm sorry, I just was taking the difference between the two to have a total of 8. Yes that is correct.

    back to my previous question regarding why the area is negative when I thought all the integral did was deteremine the area above the x axis. and between the curve.
  11. Jun 10, 2010 #10


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    Well just because the difference between -4 and 4 is 8, doesn't mean you integrate from 0 to 8. Integrating from 0 to 8 is asking for the area between the x-axis and the curve in between x=0 and x=8, but wherever the curve is under the x-axis, this is taken as negative area.

    The integral does exactly that, but the curve isn't above the x-axis for all of the values of x between 0 and 8, as per my previous post.
  12. Jun 10, 2010 #11


    Staff: Mentor

    You keep omitting one side of the equation that defines the relationship between x and y. It should be [color = red]y =[/color] (-1/16)x2 + 1
    What area are you trying to find? You haven't given a description of it. If you are trying to find the area between the parabola and the x-axis, you should not get a negative value for the area. In fact, you should never get a negative value for an area. If you do, that means you set up the integral incorrectly.
  13. Jun 11, 2010 #12

    I hope this helps, I integrated your function.

    Integrate (-1/16x^2 +1), from -4,4. = -1/48x^3+x, from -4, 4.

    So. [(-1/48)(4^3)+4]-[(-1/48)(-4^3) - 4]

    = [-4/3 +4 ] - [4/3 -4]
    = 8-(8/3)
    = 16/3

    This should give you the area under the graph from -4 --> 4
    hope this works. Let me know if it still gives you a problem.
  14. Jun 11, 2010 #13
    thank you. I see where I went wrong. I was still using limits of 0 and 8. Thanks for the help.

    Ok. now another question relating to the same topic. What if the y-intercept is negative. Then would you get a negative area?

    These are my new variables. x=-4.5 and x = 4.5
    y-intercept = - .25

    y = 1/81x^2 - .25

    Integrate (1/81x^2 +1), from -4.5,4.5 = 1/243x^3+x, from -4.5, 4.5

    So. [(1/243)(4.5^3)-4.5]-[(1/243)(-4.5^3) + 4.5]

    = [.375 -4.5 ] - [-.375 + 4.5]
    = -8.25

    when I do this by hand I get -8.25, when I check with my calculator I achieve -1.5 Did I do something wrong?
    Last edited: Jun 11, 2010
  15. Jun 11, 2010 #14


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    Why did you integrate 1/81*x2+1 when you need to integrate 1/81*x2-1/4
  16. Jun 11, 2010 #15


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    You should never get a negative area. If you get a negative number for the area, you have set up the integral incorrectly.

    For example, if the problem is to find the area between the graph of y = x2 - 2x and the x-axis, the integral for the area is
    [tex]\int_0^2 [0 - (x^2 - 2x)]~dx = \int_0^2 - x^2 + 2x~dx = \left.\frac{-x^3}{3} + x^2\right|_0^2[/tex]
    = -8/3 + 4 = 4/3

    If you naively set up this integral as
    [tex]\int_0^2 x^2 - 2x~dx [/tex]
    you will get -4/3, which is not the area.

    The graph of y = x2 - 2x has intercepts at x = 0 and x = 2, and lies below the x-axis between these intercepts. The typical area element has an area of (0 - ( x2 - 2x) [itex]\Delta x[/itex], or ( -x2 + 2x) [itex]\Delta x[/itex].
  17. Jun 11, 2010 #16
    I did. that was a typo.

    For Mark:

    So even though the line is y = x^2 - 2x

    you changed the way it was in the integral so that x^2 in negative, and 2x is positive? I'm confused, why would you do that.
    Like you demonstrated you will get a negative area, when integrating your origional formula.

    So all I need to do is integrate my curve function after I multiply it by a negative.
    Last edited: Jun 11, 2010
  18. Jun 13, 2010 #17


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    If you have some function y, when you graph -y everything is flipped in the x-axis. This means when you take -(x2-2x), the graph is flipped such that the roots are the same, but the curve between the roots is now above the x-axis and thus the integral gives you a positive value for the area.

    You could also take the absolute value of the integral, so you don't have to worry about whether the graph lies below or above the x-axis. (But be warned, if the graph lies above and below in some region, the area above the x-axis will cancel out the "negative area" below the axis).
  19. Jun 13, 2010 #18


    Staff: Mentor

    No, I didn't multiply these by -1. I made sure that the dimensions of the rectangle I was using had positive lengths. To do this, I subtract the smaller number from the larger number. On the interval [0, 2] the graph of y = x^2 - 2x lies below the x-axis. If I want the distance from a point (x, y) to the point directly above it on the x-axis, y - 0 will give me a negative value. 0 - y = 0 - (x^2 - 2x) will give me a positive value.

    For example, (1, -1) is a point on this curve. This point is 0 - (-1) = 1 unit away from the x-axis. That's all there is to it.
  20. Jun 15, 2010 #19
    I have been working on this equation off and on but nothing seems to be correct.

    I have x intercepts of -4.5 and 4.5
    The y-intercept is .020833


    x^2 - 20.25
    thus giving (-1/81x^2 + .020833)
    the equation I have developed is (-1/81x^2 + .020833)

    i still recieved a negative area of -0.562503.

    Is this due to their being more area on the opposite side of the x-axis?
  21. Jun 15, 2010 #20


    Staff: Mentor

    Again, you are leaving off one side of the equation. By not working with equations, you are making things much harder for yourself than they need to be. If the equation is y = (x-4.5)(x+4.5), you don't get a y-intercept of .020833.

    If you are given that the x-intercepts of a quadratic function are 4.5 and -4.5, the equation will be y = a(x-4.5)(x+4.5). Since the y-intercept is .020833, the point (0, .020833) must satisfy the equation y = a(x-4.5)(x+4.5). Use this point and solve for a.

    This is NOT an equation. An equation has an = symbol in it.
    You should not get a negative area. Please reread what I said in post #15.
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