Results 1 to 3 of 3

Math Help - point on tangent

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    point on tangent

    Show that the tangent to the curve y=(x^2+x-2)^3+3 at the point (1,3) is also the tangent to the curve at another point.

    This is what I did so far...
    I found the derivative of the curve and I got:
    y'=6x^5+15x^4-16x^3-36x^2+16x+12
    then I found
    f'(1)=-3
    and then I put
    -3=6x^5+15x^4-16x^3-36x^2+16x+12
    and now I am trying to solve for x, BUT I cannot find any factors of x... so? am I doing it right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member eXist's Avatar
    Joined
    Aug 2009
    Posts
    157
    Well when I take the derivative (Using the chain rule) I get:

    y' = 3(x^2 + x - 2)^2(2x + 1)

    and when I find the slope at 1:

    y'(1) = 3(1^2 + 1 - 2)^2(2(1) + 1) = 3(0)^2(3) = 0

    So, I'm finding 0 as the slope of the tangent line at the point (1, 3).

    Then you have to ask yourself what values of x make this true:

    0 = 3(x^2 + x -2)^2(2x + 1)

    We found one solution, that's when x = 1, so find the other, it's already factored.

    Good Luck - Chad
    Last edited by eXist; August 6th 2009 at 05:28 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    2
    Quote Originally Posted by skeske1234 View Post
    Show that the tangent to the curve y=(x^2+x-2)^3+3 at the point (1,3) is also the tangent to the curve at another point.

    This is what I did so far...
    I found the derivative of the curve and I got:
    y'=6x^5+15x^4-16x^3-36x^2+16x+12
    then I found
    f'(1)=-3
    and then I put
    -3=6x^5+15x^4-16x^3-36x^2+16x+12
    and now I am trying to solve for x, BUT I cannot find any factors of x... so? am I doing it right?
    Wow, if you did this by hand this must have took a century to work it all out. Check your derivative I think you may have an error somewhere with all those positive and negative and combining like terms may have messed something up. Let's take an easier approach at the derivative of this thing.

    y = (x^2+x-2)^3 + 3

    If you recall the chain rule. The rule can be found in the link below as well as some examples.
    (The Chain Rule)

    3(x^2+x-2)^2 * (2x+1) + 0


    y'= 3(x^2+x-2)(2x+1)

    Please post if you have more questions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 16th 2011, 04:57 AM
  2. Circle, tangent, and end point
    Posted in the Geometry Forum
    Replies: 4
    Last Post: June 6th 2010, 09:31 AM
  3. Tangent passes through a particular point.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 17th 2009, 04:06 AM
  4. Replies: 3
    Last Post: February 20th 2009, 05:28 AM
  5. Tangent at point...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 31st 2008, 02:46 PM

Search Tags


/mathhelpforum @mathhelpforum