Results 1 to 8 of 8

Math Help - Derivatives and Tangents

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    11

    Derivatives and Tangents

    Can someone describe how to solve this problem to me?


    Determine a, b, and c so that the curve y=ax^2+bx+c shall be tangent to the line 3x-y=5 at (2,1) and shall also pass through (3,-1).

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

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    You need three conditions to uniquely determine a, b and c. We have:

    y(x) = ax˛+bx+c, tangent line y = 3x-5 at (2,1) and point (3,-1).

    So, the conditions are:

    1) y(2) = 1
    2) y'(2) = 3
    3) y(3) = -1

    Do you understand why? If so: solve the system.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, BubblegumPops!

    Determine a,\,b,\text{ and }c so that the curve y\:=\:ax^2+bx+c shall be tangent
    to the line 3x-y\:=\:5 at (2,1) and shall also pass through (3,-1).

    Since (2,1) is on the curve: . a\!\cdot2^2 + b\!\cdot\!2 + c \:=\:1\quad\Rightarrow\quad{\color{blue}4a +2b + c \:=\:-1}

    Since (3,-1) is on the curve: . a\!\cdot\!3^2 + b\!\cdot\!3 + c\:=\:-1\quad\Rightarrow\quad{\color{blue}9a + 3b + c \:=\:-1}


    The line y \:=\:3x-5 has slope 3.
    The slope of the tangent to the curve: . y' \:=\:2ax + b
    At x=3\!:\;y'\;=\;2a\!\cdot\!3 + b \:=\:3\quad\Rightarrow\quad{\color{blue}6a + b \:=\:3}


    Solve the system of equations for a,\,b,\text{ and }c.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2007
    Posts
    11
    hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Quote Originally Posted by BubblegumPops View Post
    hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.
    4a+2b+c=-1 ...[1]

    9a+3b+c=-1 ...[2]

    6a+b=3 ...[3]

    [2] - [1]:

    5a+b=0 ...[4]

    [3] - [4]:

    a=3

    \implies b=-15

    \implies c = 17
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2007
    Posts
    11
    Thank you! I think i have it, but I don't understand how you got ...[1]. I thought that should equal 1, not -1.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Sorry, you're right. It should be 4a+2b+c=1. Solving them similarly will yield (a,b,c)=(5,-27,35)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2007
    Posts
    11
    Thank you so much for your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  2. Tangents
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 5th 2009, 04:11 PM
  3. Derivatives and Tangents
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 18th 2009, 04:34 PM
  4. Derivatives and tangents
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 6th 2009, 10:21 PM
  5. Tangents
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 27th 2008, 06:36 PM

Search Tags


/mathhelpforum @mathhelpforum