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Math Help - Algebra, points of intersection

  1. #1
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    Algebra, points of intersection

    Why does 'a' have 2 values if y= x is tangent to parabola y = x^2 + ax + 1. I thought there would be only 1 solution.

    Similarly, why are there 2 valus of 'm' when y = mx - 0.5 is tangent to y = x^2 + 5x?

    There is only 1 solution for b when y = -x is tangent to y = x^2 + x + b ??
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  2. #2
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    Re: Algebra, points of intersection

    Please explain how did you proceed?
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  3. #3
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    Re: Algebra, points of intersection

    I used the discriminant to find the unknown. I equated the discriminant to 0 as the line is a tangent to the parabola.
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  4. #4
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    Re: Algebra, points of intersection

    In order that a line, y= mx+ n, be tangent to a parabola, y= ax^2+ bx+ c, the line must first intersect the parabola: the point, (x, y), must satisfy mx+n= ax^2+ bx+ c. But in order that it be tangent, that must be a double root which means that the discriminant of the equation ax^2+ (b- m)x+ c- n= 0 must be 0. The discriminant, (b- m)^2- 4a(c- n) is quadratic in both b and m and so that equation may have two roots, while c, in y= x^2+ x+ c is not quadratic.

    Geometrically, increasing a, in the parabola, y= x^2+ ax+ 1, moves the vertex to the left and downward. There will be two different values of a, one positive, the other negative, that will be tangent to y= x.
    I have attached a graph.
    Algebra, points of intersection-parabolas.jpg

    For the second problem, y= mx+ 0.5 can be any line passing through (0, 0.5). That point is NOT on the given parabola so there exist two different lines through it that are tangent to the parabola, one above y= 0 and one below y= 0.
    Algebra, points of intersection-newparabolas.jpg

    Here is a graph of the last one:
    Algebra, points of intersection-parabola.jpg
    Last edited by HallsofIvy; February 17th 2013 at 06:41 AM.
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