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Math Help - Help with curve and line

  1. #1
    Member helloying's Avatar
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    Help with curve and line

    Find the range of values of c for which the line y=c(x-3) does not intersect the curve y+2x=x^2+6

    i equate the two eqn together. x + 6 -2x= cx -3c

    and i put b-4ac<0

    so i get (-2-c) - (24-12c)<0

    then i am stuck i dont know how to find the range of values for c
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  2. #2
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    Quote Originally Posted by helloying View Post
    Find the range of values of c for which the line y=c(x-3) does not intersect the curve y+2x=x^2+6
    The function y = c(x - 3) is a straight line, and the function y = x^2 - 2x + 6 is a parabola. You're needing to find values of c so that the straight line passes the parabola off to one side.

    To find where the lines do not intersect, we find first where they do intersect; the solution will be the other values of c.

    c(x\, -\, 3)\, =\, x^2\, -\, 2x\, +\, 6

    cx\, -\, 3c\, =\, x^2\, -\, 2x\, +\, 6

    0\, =\, x^2\, -\, (c\, +\, 2)x\, +\, 6\, +\, 3c

    Applying the Quadratic Formula, we get:

    x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 4(1)(6\, +\, 3c)}}{2(1)}

    x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 4(1)(3)(2\, +\, c)}}{2}

    x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 12(c\, +\, 2)}}{2}

    x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)((c\, +\, 2)\, -\, 12)}}{2}

    x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)(c\, -\, 10)}}{2}

    So there will be valid solutions (and thus intersections) for (c + 2)(c - 10) > 0. To find the solution, you need to solve the quadratic inequality.

    The zeroes of (c + 2)(c - 10) are obviously at c = -2 and c = 10. The parabola corresponding to y = (c + 2)(c - 10) will open upward, so the quadratic will be positive (that is, the parabola will be above the x-axis) on the ends, before the first zero and after the second zero.

    So which intervals give intersections? Then which interval does not?

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