Results 1 to 5 of 5

Math Help - Tangent Graph Calc BC Problem

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    79

    Tangent Graph Calc BC Problem

    Let f(x)= 6-x. For 0 < w < √6 , let A(w) represent the area of the triangle created by the coordinate axes and the line tangent to the graph of f at the point (w, 6-w). Find A(1).

    Here is the Diagram of what I'm attempting to work with

    http://img472.imageshack.us/img472/50/a5nt.jpg

    A start to this problemw ould greatly be appreciated.

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by r2d2 View Post
    Let f(x)= 6-x. For 0 < w < √6 , let A(w) represent the area of the triangle created by the coordinate axes and the line tangent to the graph of f at the point (w, 6-w). Find A(1).

    Here is the Diagram of what I'm attempting to work with

    http://img472.imageshack.us/img472/50/a5nt.jpg

    A start to this problemw ould greatly be appreciated.

    Thank you in advance!
    The equation of a line tangent to f at a point b is given by y = f'(b)(x - b) + f(b). So if f = 6 - x^2, then f(1) = 5, f'(1) = -2 and the line is given by y = -2(x - 1) + 5 = -2x + 7. The y-intercept of this line is 7, and the x-intercept is at 7/2. So the area of the triangle is 1/2 * 7 * 7/2 = 49/4 = 12.25.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    79
    Thank you for your response.

    I understand all except that in the tangent line equation, why would the "x" in y=mx+b be (x-b)? That part seems somewhat confusing.

    Thanks

    And From there, would you plug 1 into the tangent line equation to find A(1), which would be 5?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2009
    Posts
    80
    Quote Originally Posted by r2d2 View Post
    Thank you for your response.

    I understand all except that in the tangent line equation, why would the "x" in y=mx+b be (x-b)? That part seems somewhat confusing.

    Thanks

    And From there, would you plug 1 into the tangent line equation to find A(1), which would be 5?
    It's the point-slope form of a line: if you want a line through the point (a, b) with slope m, the equation of the line is y = m(x - a) + b. Try graphing it.

    A(1) is 12.25; see the last part of my first response.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    79
    Ah, I understand now. Thanks a lot for your help. I really appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multivariable Calc Image-Graph Problem.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 14th 2010, 11:15 AM
  2. Replies: 0
    Last Post: November 2nd 2010, 10:38 PM
  3. Replies: 8
    Last Post: October 13th 2009, 06:04 PM
  4. Calc Equation or Tangent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 29th 2009, 03:55 PM
  5. calc I: graph dec and cc-up
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 25th 2008, 08:22 PM

Search Tags


/mathhelpforum @mathhelpforum