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Math Help - Equation of the Tangent

  1. #1
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    Smile Equation of the Tangent

    Hello! I have this question about tangents, which I have solved partly. But I have some difficulties in solving the ii - part. I would appreciate your help

    Here is the question:
    Consider the function h(x) = x^(1/5)
    i) Find the equation of the tangent to the graph at the point where x = a, (a doesn't equal 0). Write the equation in the form y = mx + c.

    I have solved this part. My approach was first to find the derivative of the function and then replace the x by a. And then convert it into the form
    y = mx + c, which gave me this result:
    y = 1/5 a^(-4/5) a + 4/5 a^(1/5)

    The part I couldn't solve:
    ii) Show that this tangent intersects the x-axis at the point (-4a, 0).
    I would really appreciate your help.
    Thanks,
    Amine
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  2. #2
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    Quote Originally Posted by Aminekhadir View Post
    Hello! I have this question about tangents, which I have solved partly. But I have some difficulties in solving the ii - part. I would appreciate your help

    Here is the question:
    Consider the function h(x) = x^(1/5)
    i) Find the equation of the tangent to the graph at the point where x = a, (a doesn't equal 0). Write the equation in the form y = mx + c.

    I have solved this part. My approach was first to find the derivative of the function and then replace the x by a. And then convert it into the form
    y = mx + c, which gave me this result:
    y = 1/5 a^(-4/5) a + 4/5 a^(1/5)

    The part I couldn't solve:
    ii) Show that this tangent intersects the x-axis at the point (-4a, 0).
    I would really appreciate your help.
    Thanks,
    Amine

    you have a little confusion: the tangency point is (a, a^(1/5)) and since the slope of the tangent line at this point is (1/5)a^(-4/5), then the tangent line's equation is:

    y - a^(1/5) = (1/5)a^(-4/5)[x - a] ==> y = (1/5)a^(-4/5)*x +(4/5)a^(1/5)

    It's almost the same you got, but for some reaosn you put a where x should be!

    Now just put y = 0 in the equation aboe and find out what the value of x is.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    you have a little confusion: the tangency point is (a, a^(1/5)) and since the slope of the tangent line at this point is (1/5)a^(-4/5), then the tangent line's equation is:

    y - a^(1/5) = (1/5)a^(-4/5)[x - a] ==> y = (1/5)a^(-4/5)*x +(4/5)a^(1/5)

    It's almost the same you got, but for some reaosn you put a where x should be!

    Now just put y = 0 in the equation aboe and find out what the value of x is.

    Tonio
    Hey,
    so should I say 0 = (1/5)a^(-4/5)*x +(4/5)a^(1/5)
    and then solve for x?
    Thanks a lot for your help
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  4. #4
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    Quote Originally Posted by Aminekhadir View Post
    Hey,
    so should I say 0 = (1/5)a^(-4/5)*x +(4/5)a^(1/5)
    and then solve for x?
    Thanks a lot for your help

    Exactly

    Tonio
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