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Graph of y=x^2

  1. Feb 14, 2006 #1
    Interesting feature of this graph. Consider 2 points on the parabola, I'll take (-2,4) and (4,16). By multipling the positive x values (2*4=8), you can get the y-intercept of the line from (-2,4) to (4,16). Proof: The line including (-2,4) and (4, 16) is written as y=2x+8. Thus, the y-intercept is 8. My question is why does this work? I've been trying to figure it out for a while, and I am completly stumped on this one... Any help would be greatly appreciated! :smile:
  2. jcsd
  3. Feb 14, 2006 #2


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    First find the general equation for the slope b in terms of x1, y1, x2, and y2. Then use the fact that y1 = x1^2 and y2 = x2^2.

    Actually it's not the product of the absolute value of the x values, it's the opposite of the product of the x values.
  4. Feb 14, 2006 #3


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    Start from the 2 point formula for a line.

    [tex] \frac {y - y_1} {x - x_1} = \frac {y_2 - y_1} {x_2 - x_1} [/tex]

    The formula for your parabola is

    [tex] y = x^2 [/tex]

    So we can write

    [tex] y_1 = x_1^2 [/tex]
    [tex] y_2 = x_2^2 [/tex]

    Use this information in the 2 point formula to get

    [tex] \frac {y - y_1} {x - x_1} = \frac {x_2^2 - x_1^2} {x_2 - x_1} [/tex]

    Note that the numerator on the Right Hand Side is the differenc of squares and can be factored to get

    [tex] \frac {y - y_1} {x - x_1} = \frac {(x_2 - x_1) (x_2 + x_1)} {x_2 - x_1} [/tex]

    Cancel like factors in the RHS
    [tex] \frac {y - y_1} {x - x_1} = (x_2 + x_1) [/tex]

    Now rearrange this to get

    [tex] y - y_1 = (x - x_1) (x_2 + x_1)[/tex]
    Simplify to get:
    [tex] y = x (x_2 + x_1) - x_1 x_2[/tex]

    Clearly you are correct for the simple parabola, in addition it can be seen that the slope of the line is the sum of the x coordinates.
    Last edited: Feb 14, 2006
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