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Math Help - Parabola exercise and checking a log-equation

  1. #1
    Junior Member Fnus's Avatar
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    Parabola exercise and checking a log-equation

    The parabola exercise is this one:

    'A parabola is given by the equation y = ax^2 + ax + 1, where a is a positive real number different from zero.
    (a): Determine a so that the parabola has exactly one intersection point with the x-axis.
    (b): At the point P(x0,y0) the parabola has a tangent which is parallel to the line with the equation y = 8x - 800. Find the value of X0'

    And then I had this log equation, and if someone would check if it's right, that'd make me very happy. (:

    <br />
\log_5 (2x+7) - 3 = \log_5 x<br />

    <br />
\log_5 (2x+7) - \log_5 125 = \log_5 x<br />

    <br />
\log_5 ((2x+7)/125) = \log_5 x<br />

    <br />
(2x+7)/125 = x<br />

    <br />
2x+7 = 125x<br />

    <br />
x = 118<br />
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  2. #2
    Moo
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    Hi !

    Quote Originally Posted by Fnus View Post
    The parabola exercise is this one:

    'A parabola is given by the equation y = ax^2 + ax + 1, where a is a positive real number different from zero.
    (a): Determine a so that the parabola has exactly one intersection point with the x-axis.
    (b): At the point P(x0,y0) the parabola has a tangent which is parallel to the line with the equation y = 8x - 800. Find the value of X0'
    Question a).

    ax^2+ax+1=0

    A quadratic equation has one and only one solution if it is proportional to a perfect square.

    This means that you want ax^2+ax+1 to be a perfect square.

    The discriminant \Delta should be equal to 0.

    \Delta=a^2-4a

    \implies a(a-4)=0 \implies a=... \ or \ a=...
    ----------------------------

    Question b)
    two parallel lines have the same slope. Here, the slope is 8.

    A tangent line to the curve representing a function f(x) has this equation :
    y=f'(m)(x-m)+f(m)

    So actually, you have to solve for f'(m)=8.

    Is it clear enough ? Because I didn't understand what X0' was...

    And then I had this log equation, and if someone would check if it's right, that'd make me very happy. (:

    <br />
\log_5 (2x+7) - 3 = \log_5 x<br />

    <br />
\log_5 (2x+7) - \log_5 125 = \log_5 x<br />

    <br />
\log_5 ((2x+7)/125) = \log_5 x<br />

    <br />
(2x+7)/125 = x<br />

    <br />
2x+7 = 125x<br />

    <br />
x = 118<br />


    Just check, in this kind of exercises, that the answer fits in the domain of the equation.

    that is to say 2x+7>0 and x>0
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