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Math Help - equation of the line tangent to a curve at its inflection point

  1. #1
    Junior Member LexiRae's Avatar
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    equation of the line tangent to a curve at its inflection point

    An equation of the line tangent to the graph of y=a^3+ax^2+2 at its point of inflection is:?

    i have no idea.
    i know you do something with the second derivative and find where it goes from concave upward to concave downward.
    but whats with the equation?
    thanks in advance!
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  2. #2
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    Quote Originally Posted by LexiRae View Post
    An equation of the line tangent to the graph of y=a^3+ax^2+2 at its point of inflection is:?
    I am sure that you mean a {\color{red}x^3} +ax^2 +2
    Slope: m=3ax^2+2ax.
    Second derivative: 6ax+2a
    Then inflection point: 6x + 2a = 0\; \Rightarrow \;x = \frac{{ - 1}}{3}.
    The point is \left( {\frac{{ - 1}}{3},\frac{{2a}}{{27}} + 2} \right) and slope is m=\frac{-a}{3}
    You can finish.
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