Results 1 to 2 of 2

Thread: Finding the "normal line"

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Finding the "normal line"

    I understand in theory how to do this...it's just the particular problem that's throwing me.


    Question: Find the equation of the normal line to the graph of $\displaystyle f(x) = e^x - 1$ at the point where it crosses the x-axis.



    My Attempt:

    So $\displaystyle f\prime(x) = e^x$

    f(x) crosses the x-axis at pretty much x = 0 I think.

    so do I plug in 0 for x and solve in the original equation like this?
    $\displaystyle
    f(x) = e^0 - 1
    = 0$

    Giving the point (0,0)

    then plug that into y = mx + b

    I get

    $\displaystyle 0 = e^x(0) + b$
    so $\displaystyle b = 0$


    Does that mean the tangent line is just $\displaystyle y= e^x(x)$?

    Making the normal line $\displaystyle y = -\frac{1}{e^x}(x)$ ?





    Is that right? Or am I doing this completely wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by lysserloo View Post
    I understand in theory how to do this...it's just the particular problem that's throwing me.


    Question: Find the equation of the normal line to the graph of $\displaystyle f(x) = e^x - 1$ at the point where it crosses the x-axis.



    My Attempt:

    So $\displaystyle f\prime(x) = e^x$

    f(x) crosses the x-axis at pretty much x = 0 I think.

    so do I plug in 0 for x and solve in the original equation like this?
    $\displaystyle
    f(x) = e^0 - 1
    = 0$

    Giving the point (0,0)

    then plug that into y = mx + b

    I get

    $\displaystyle 0 = e^x(0) + b$
    so $\displaystyle b = 0$


    Does that mean the tangent line is just $\displaystyle y= e^x(x)$?

    Making the normal line $\displaystyle y = -\frac{1}{e^x}(x)$ ?

    Is that right? Or am I doing this completely wrong?
    Note that $\displaystyle f^{\prime}(0)=e^0=1$.

    So $\displaystyle y=x$ is the tangent line at that point, which then implies $\displaystyle y=-x$ is the normal line at that point.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Apr 24th 2011, 07:01 AM
  2. equation of "normal line"
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 29th 2011, 10:24 AM
  3. Replies: 2
    Last Post: Mar 2nd 2011, 09:01 PM
  4. Replies: 1
    Last Post: Oct 25th 2010, 04:45 AM
  5. Finding the "normal line" of a graph.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 4th 2009, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum