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Math Help - Find the equation of the tangent line(s)

  1. #1
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    Find the equation of the tangent line(s)

    Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)
    Hint:Find the slope first then use y-y=m(x-x)

    The equation is  f(x)= 2x^3-3

    i dont really get this equation anyone know were to start
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  2. #2
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    Find the slope of x+6y+5=0
    Find the derivative of f(x)

    what is the relationship of the slope between 2 lines that are perpendicular?

    Set that slope equal to f'(x)

    Hope this helps. Ask if you are still confused.
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  3. #3
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    Find the slope of x+6y+5=0

     m=-1/6

    Find the derivative of f(x)

     6x^2

    what is the relationship of the slope between 2 lines that are perpendicular?

    I believe the slopes the same

    Set that slope equal to f'(x)

     -1/6=6x^2

    Still confused
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  4. #4
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    Quote Originally Posted by john doe View Post
    Find the slope of x+6y+5=0

     m=-1/6

    Find the derivative of f(x)

     6x^2

    what is the relationship of the slope between 2 lines that are perpendicular?

    I believe the slopes the same

    Set that slope equal to f'(x)

     -1/6=6x^2
    Slopes are the same when they are parallel.
    When 2 lines are perpendicular, their slopes are opposite reciprocal of each other.

    So you would set  6x^2=6

    Find x and put it back into the original equation to find y and that will be your point.

    Now just use the point-slope equation since you know the point and the slope.
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  5. #5
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    hows this look

     y+1=6(x-1)
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  6. #6
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    Quote Originally Posted by john doe View Post
    hows this look

     y+1=6(x-1)
    There should be 2 lines.
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  7. #7
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    i dont understand
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  8. #8
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    Quote Originally Posted by john doe View Post
    i dont understand
    haha sorry it's late here, I forgot to add in the explanation.

     6x^2=6

    x=1, -1
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  9. #9
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    i must be losing it because i dont know what that means



    my graph kinda turned out with a parabola and a tangent line
    going though at 1,-1





    its kinda 3 aclock
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  10. #10
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    Let's back up a bit.

    Here is your question: Find the equation of the tangent line(s),which is perpendicular to the line x+6y+5=0, to the graph of F(x) at the tangent point(s) (x,y)


    so we need to find at where does F(x) have slopes of 6.

    So we did F'(x)=6x^2=6
    the solution would be x=1 and -1 because x is squared

    That means the function F(x) has 2 points where the slope is 6 that is at (1, F(1)) and (-1, (F(-1)).

    This means there are 2 lines which satisfies the question.

    Hope this clears things up!
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  11. #11
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    I think i see it at
    so that would look like a x^3 graph shifted to the bottom
    theres a slope 6 that hits the graph 2 times 1 and -1
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  12. #12
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    Quote Originally Posted by john doe View Post
    I think i see it at
    so that would look like a x^3 graph shifted to the bottom
    theres a slope 6 that hits the graph 2 times 1 and -1
    Hmm, I'm not quite sure what you mean.

    The slope of x+6y+5=0 would be -1/6

    Perpendicular slope=6

     f'(x)=6x^2=6
     x= 1,-1

     f(1)=-1
     f(-1)=-5

    so the 2 lines would be

    y+5=6(x+1)
    y+1=6(x-1)

    All I was trying to tell you after your first solution is that there is more than one answer.
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  13. #13
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    alright thax for the help

    i dont fully understand it but

    i appreciate all the time your talking

    this late at night to explain to me

    (ill take a look at it tomorrow)
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  14. #14
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    In general, let f:[t_{0},\infty)\to\mathbb{R} be a differentiable function.
    Then at the point s\in(t_{0},\infty) tangent line of f is g(t):=f^{\prime}(s)(t-s)+f(s) for all t\in[t_{0},\infty).
    And let h(t):=at+b for some a,b\in\mathbb{R} with a\neq0 be a line, which is supposed to be perpendicular to the line g.
    Then, we have to determinate s in such a way that g is perpendicular to h; that is, a*f^{\prime}(s)=-1.
    If f^{\prime} as inverse function (f^{\prime})^{-1} (as in this case), then
    s=(f^{\prime})^{-1}(-1/a)
    and so
    g(t)=-\frac{1}{a}\big(t-(f^{\prime})^{-1}(-1/a)\big)+f\big((f^{\prime})^{-1}(-1/a)\big).
    The rest is just simple calculation.
    I hope you will find this clearifying.
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