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Math Help - Parabola and tagent of the x-axis

  1. #1
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    Parabola and tagent of the x-axis

    i have this question where i need to find the value of b where the formula of a parabola is tagent to the x-axis but i have no idea how to proceed.

    y=4x^2+bx+25

    thanks in advance.
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  2. #2
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    The wording of your question was confusing. I think you are trying to find the value of b that makes the stated parabola tangential to the x axis.

    You need the curve to intersect the x axis and have a gradient of 0 at the point of intersection. Since you posted in the pre calculus part of the forum I assume you aren't able to use calculus to get a pair of simultaneous equations to solve.

    Instead you'll have to use the properties of the graph. A parabola has only 1 turning point. If it is tangential to the x-axis then it cant cross it. taken together this means that the equation f(x) = 0 has only 1 solution, ie the graph has a double root.

    If a parabola has a double root it must be possible to factorise it as follows:
    y=(ax + c)^2

    expanding the brackets:
    y=a^2x^2 + 2ac +c^2
    you know that the equation is
    y=4x^2 + 2b +25

    comparing coefficients:

    coefficients of x^2
    a^2 = 4  \rightarrow a=2

    coefficients of x^0(constant term)
    c^2=25 \rightarrow c = 5


    coefficients of x
    b=2ac
    b= 20

    in case you dont believe me, the graph is plotted here:
    http://www.wolframalpha.com/input/?i=plot+4x^2+%2B20x+%2B25

    You can see it is tangential to the x axis as required.

    NB: the above solution is not unique. b=-20 works too. You can get this by taking the negative square root instead of the positive one when comapring coefficients.
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  3. #3
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    thanks, i didnt see the problem like that. sorry for the wording, i had to translate it from french. Im not to familiar with the english terms for math for each one of them.
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  4. #4
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    Quote Originally Posted by DAngel View Post
    i have this question where i need to find the value of b where the formula of a parabola is tagent to the x-axis but i have no idea how to proceed.
    y=4x^2+bx+25
    Simply put: for tangency to exist, sqrt(b^2 - 4ac) = 0;
    so b^2 = 4ac
    b^2 = 4(4)(25) : b = 20

    C'est bien? J'espere que oui.
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  5. #5
    Zap
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    The minimum value of a quadratic function f(x)=ax^2 + bx + c occurs at x=-b/2a
    therefore:

    x= b/8

    setting y=0

    0=4(-b/8)^2 + b(-b/8) + 25

    therefore

    0= b^2/16 - 2b^2/16 +25

    b^2/16 = 25

    b=20

    Sorry... new here and have still to decipher how to use the equation editor
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