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Thread: parabolas and line intersection

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    parabolas and line intersection

    Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x)?
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    Quote Originally Posted by mathhelp92 View Post
    Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x)?
    let the lines be $\displaystyle y = 2x + k$, where $\displaystyle k$ is the y-intercept

    $\displaystyle 2x+k = x(6-x)
    $

    $\displaystyle 2x+k = 6x - x^2$

    $\displaystyle x^2 - 4x + k = 0$

    to have real solutions, $\displaystyle b^2 - 4ac \geq 0$

    $\displaystyle 16 - 4k \geq 0$

    $\displaystyle k \leq 4$
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    Quote Originally Posted by skeeter View Post
    let the lines be $\displaystyle y = 2x + k$, where $\displaystyle k$ is the y-intercept

    $\displaystyle 2x+k = x(6-x)
    $

    $\displaystyle 2x+k = 6x - x^2$

    $\displaystyle x^2 - 4x + k = 0$

    to have real solutions, $\displaystyle b^2 - 4ac \geq 0$

    $\displaystyle 16 - 4k \geq 0$

    $\displaystyle k \leq 4$



    THANK YOU...but this way I will get the roots of it
    i need the equations for the line that have a slope of 2 and they pass through the parabola once, never n twice
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    Quote Originally Posted by mathhelp92 View Post
    THANK YOU...but this way I will get the roots of it
    i need the equations for the line that have a slope of 2 and they pass through the parabola once, never n twice
    To have 0 real solutions, the discriminant is negative. Solve for k.

    To have 1 real solution, the discriminant is 0. Solve for k.

    To have 2 real solutions, the discriminant is positive. Solve for k.
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    Quote Originally Posted by Prove It View Post
    To have 0 real solutions, the discriminant is negative. Solve for k.

    To have 1 real solution, the discriminant is 0. Solve for k.

    To have 2 real solutions, the discriminant is positive. Solve for k.


    So if k > 4, so then in the equation y=mx+b it will be y=2x+4..I THINK
    so thats 2 roots

    so how do we solve for only one root or no roots

    because if we make both equations equal to each other, we will get 4 every time :s this is where i am confused

    WHAT WE ARE TRYING TO SOLVE FOR IS THE VALUE OF B IN y=mx+b for 3 equations soo 3 different values of b
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    Quote Originally Posted by mathhelp92 View Post
    So if k > 4, so then in the equation y=mx+b it will be y=2x+4..I THINK
    so thats 2 roots

    so how do we solve for only one root or no roots

    because if we make both equations equal to each other, we will get 4 every time :s this is where i am confused

    WHAT WE ARE TRYING TO SOLVE FOR IS THE VALUE OF B IN y=mx+b for 3 equations soo 3 different values of b
    I've already told you the answer.

    If there aren't any intersection points then the value of $\displaystyle b$ is ANYTHING LESS THAN 4. So use a letter like $\displaystyle k$ and say that $\displaystyle k <4$.

    If there is only one intersection point then the value of $\displaystyle b$ IS 4.

    If there are two intersection points then the value of $\displaystyle b$ is ANYTHING GREATER THAN 4. So use a letter like $\displaystyle k$ and say that $\displaystyle k > 4$.
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