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Math Help - help with a slope of the tangent to a curve

  1. #1
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    help with a slope of the tangent to a curve

    Here's the problem -

    400-6x^2

    Find the slope of the tangent to this curve at this point (1, 394)

    HELP!!
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  2. #2
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    how are you supposed to find the slope? ... straight power rule derivative, or derivative from the limit of a difference quotient?
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  3. #3
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    The question is how to find the equation of the line tangent to a specified curve, f(x), at a specified point, (a, f(a)). The answer is quick and easy:

    <br /> <br />
y \ = \ f(a) \ + \ f'(a)(x-a),<br /> <br />

    where f'(a) is the derivative of f(x) evaluated at a.
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  4. #4
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    so the derivative of 400-6x^2 is -12x

    then where do I go?
    Last edited by lava1980; October 30th 2008 at 01:38 PM. Reason: wrong problem - i'm an idiot
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  5. #5
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    Quote Originally Posted by lava1980 View Post
    so the derivative of 400-6x^2 is -12x

    then where do I go?
    Just follow the equation I posted, where  (a,f(a)) \ = \ (1, 394).
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  6. #6
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    I'm thinking way too hard...

    y = f(1) + f'(-12)...

    but what's x?

    Quote Originally Posted by lava1980 View Post
    so the derivative of 400-6x^2 is -12x

    then where do I go?
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  7. #7
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    Quote Originally Posted by lava1980 View Post
    I'm thinking way too hard...

    y = f(1) + f'(-12)...

    but what's x?
    Here, x is a variable. It stays as x. Remember, you are trying to find the equation for a line.
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  8. #8
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    Okay, that makes sense.

    y =f(1) + f'(-12)(x-1)

    What's after that?

    Quote Originally Posted by Varitron View Post
    Here, x is a variable. It stays as x. Remember, you are trying to find the equation for a line.
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  9. #9
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    Quote Originally Posted by lava1980 View Post
    Okay, that makes sense.

    y =f(1) + f'(-12)(x-1)

    What's after that?
    Actually evaluate f(1) and f'(-12), distribute and group like terms, and you are done.
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  10. #10
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    But I have to find the slope of the tangent to the curve.

    Does y = 394 + what?

    Quote Originally Posted by Varitron View Post
    Actually evaluate f(1) and f'(-12), distribute and group like terms, and you are done.
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  11. #11
    Junior Member Khaali91's Avatar
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    i think what he is trying to say is...

    remember the derivative is going to be the slope...
    if the slope is -12 then m=-12 and u have (x,y)-> (1, 394)

    so solve for b in y=mx+b

    394=-12(1)+b

    then u have the equation for the line tangent to the point

    correct me if im wrong, i just learned this last week =/
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  12. #12
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    Quote Originally Posted by lava1980 View Post
    But I have to find the slope of the tangent to the curve.

    Does y = 394 + what?
    Firstly, I just realized one of your terms is wrong. It's not supposed to be f'(-12), it's f'(a) = f'(1) = -12. So, since f(1) = 394:

    y = 394 - 12(x-1) = 394 - 12x + 12 = -12x + 406

    The slope is obvious since it is in the form y = mx + b.
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  13. #13
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    This seems correct, but I can't get other points on the graph to fit the equation...

    f(x) = -12x
    if x = 1, then the slope = -12
    if the slope is -12 then m=-12 and (x,y) > (1, 394)
    so solve for b in y=mx+b
    394=-12(1)+b
    406 = b

    y = -12x + 406

    Any ideas?

    Quote Originally Posted by Khaali91 View Post
    i think what he is trying to say is...

    remember the derivative is going to be the slope...
    if the slope is -12 then m=-12 and u have (x,y)-> (1, 394)

    so solve for b in y=mx+b

    394=-12(1)+b

    then u have the equation for the line tangent to the point

    correct me if im wrong, i just learned this last week =/
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  14. #14
    Junior Member Khaali91's Avatar
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    Quote Originally Posted by lava1980 View Post
    Here's the problem -

    400-6x^2

    Find the slope of the tangent to this curve at this point (1, 394)

    HELP!!

    only the point (1,394) will be on the graph of 400-6x^2 AND y=-12x+406

    the tangent means it touches the graph 400-6x^2 once so no other point will be on it.
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