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Math Help - Question - tangents and their equation

  1. #1
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    Question - tangents and their equation

    Hello everyone

    I am studying tangents and come across this kind of problem where I am going to find the equation for one or more tangent with a specific directioncoefficient for an function.

    Example:

    Find the tangents for the funtion y = 3x^2 - 2x + 1 that has an directioncoefficient of 4.

    How do I solve this kind of problem? I am halfway through, I think.

    First we get the derivate for the function, which is:

    y = 6x - 2

    The we check and see for what X value Y equals 4, correct?

    4 = 6x - 2

    1 = x

    And after that? Assuming that the above is correct.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Hello everyone

    I am studying tangents and come across this kind of problem where I am going to find the equation for one or more tangent with a specific directioncoefficient for an function.

    Example:

    Find the tangents for the funtion y = 3x^2 - 2x + 1 that has an directioncoefficient of 4.

    How do I solve this kind of problem? I am halfway through, I think.

    First we get the derivate for the function, which is:

    y = 6x - 2

    The we check and see for what X value Y equals 4, correct?

    4 = 6x - 2

    1 = x

    And after that? Assuming that the above is correct.
    i have no idea what the direction coefficient is. but assuming what you did so far is correct, this is probably the way to go.

    we know we want the tangent at x = 1. plugging x = 1 into the original equation we get y = 2. so we want the tangent at (1,2) with slope 4. use the point-slope form, and solve for y.

    we have: y - y_1 = m(x - x_1)

    where (x_1,y_1) = (1,2) and m = \mbox{ slope } = 4
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  3. #3
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    i have no idea what the direction coefficient is.
    Heh, maybe it isn't called "direction coefficient" in english.

    Ex. y = 7x has an (something something) of seven. \frac {\delta y} {\delta x} = 7
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  4. #4
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    Quote Originally Posted by λιεҗąиđŗ View Post
    Heh, maybe it isn't called "direction coefficient" in english.
    Hej,

    this is for your private math dictionary:

    a) If you are refering to a straight line then the "direction coefficient" is called slope

    b) if you are refering to the graph of a function then the "direction coefficient" is called gradient
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  5. #5
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    Hej,

    this is for your private math dictionary:

    a) If you are refering to a straight line then the "direction coefficient" is called slope

    b) if you are refering to the graph of a function then the "direction coefficient" is called gradient
    Guten Tag

    Thanks, ok so it's the slope we're after.

    Direction coefficient sounds more scientific than slope though.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by λιεҗąиđŗ View Post
    Direction coefficient sounds more scientific than slope though.
    indeed it does! i'm going to start using that to confuse the people i tutor

    Jhevon: Silly girl, just find the direction coefficient!

    Silly Girl:
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  7. #7
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    Hehe...

    Right-o, back to the problem.

    I have talked to an fellow and he expained the general formula for finding the tangents equation.

    t(x) = f(a) + f(a)(a + x)

    T(x) = (3a^2 - 2a + 1) + (6a - 2)(a + x)

    T(x) is the tangent of course.

    4 = 6x - 2

     1 = x

    A is 1 in this specific problem.

    Let's insert 1 and see what happens...

     T = (3(1)^2 - 2(1) + 1) + (6(1) - 2)((1) + x)

    T = ( 3 - 2 + 1) + (6 + 6x - 2 - 2x)

    T = 4x + 6

    Please correct me if you find any errors, this isn't what the book want for an answer.
    Last edited by λιεҗąиđŗ; September 30th 2007 at 10:53 AM.
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