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Math Help - quadratic function

  1. #1
    Senior Member euclid2's Avatar
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    quadratic function

    A quadratic function has roots  3 and  -5 which passes through the point  (1,5) . Under what conditions will it intersect with a line of slope  -4 Once? twice? Never?
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    Since it is a quadratic equation, it has the form of a parabola. You know where it intersects the x-axis, and you also know 1 other point, which tells you which way the parabola opens up. From there you can draw a sketch of the graph and then see where it would intersect a line of sloope -4.
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    Senior Member vincisonfire's Avatar
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    Because it has root 3 and -5,  y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)
     y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}
     y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}
    Set  -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b and solve for b. All value smaller than this b will intersect since the parabola is tipped down.
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  4. #4
    Senior Member euclid2's Avatar
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    Quote Originally Posted by vincisonfire View Post
    Because it has root 3 and -5,  y = \alpha (x-3)(x+5) = \alpha (x^2+2x-15)
     y(1) =-12\alpha=5 \implies \alpha = \frac{-5}{12}
     y = \frac{-5}{12}(x^2+2x-15) = -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4}
    Set  -\frac{5}{12}x^2-\frac{5}{6}x+\frac{25}{4} = -4x+b and solve for b. All value smaller than this b will intersect since the parabola is tipped down.
    How did you come up with this and how do i solve for b?
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