Results 1 to 10 of 10

Math Help - Find equation of line f(x) parallel to given line

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    29

    Find equation of line f(x) parallel to given line

    This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

    Function
    <br />
 f(x) = x^3<br />
    f'(x) = 3x^2

    Line
    <br />
3x-y+1 = 0<br />
    y = 3x+1

    From here we know that m=3 since y=mx+b

    Then I made f'(x)=3
    3x^2=3
    x^2=1

    To get the value of x.
    x=1

    Now I go back to the given line and plug in:
    y=3(1)+1
    y=4

    My point is (1,4)

    I did most of it as I was writing this, I think my light bulb turned on.

    y-y1=m(x-x1)
    y-4=3(x-1)
    y=3x+1

    Did I do it right or is there another way of doing this?
    Last edited by drkidd22; July 30th 2009 at 04:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by drkidd22 View Post
    This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

    Function
    <br />
 f(x) = x^3<br />
    f'(x) = 3x^2

    Line
    <br />
3x-y+1  <br />
is this 3x-y+1 = 0 ?
    y = 3x+1

    From here we know that m=3 since y=mx+b

    Then I made f'(x)=3
    3x^2=3
    x^2=1

    To get the value of x.
    x=1

    x = -1 also

    Now I go back to the given line and plug in:
    y=3(1)+1 3(1) + 1 = 4, not 5
    y=5

    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    Quote Originally Posted by drkidd22 View Post
    This is another one of those problems that I kinda get the idea of how to do it but then just can't think of anything else. So, to the problem.

    Function
    <br />
 f(x) = x^3<br />
    f'(x) = 3x^2

    Line
    <br />
3x-y+1<br />
    y = 3x+1

    From here we know that m=3 since y=mx+b

    Then I made f'(x)=3
    3x^2=3
    x^2=1

    To get the value of x.
    x=1

    Now I go back to the given line and plug in:
    y=3(1)+1
    y=5

    My point is (1,5)

    I did most of it as I was writing this, I think my light bulb turned on.

    y-y1=m(x-x1)
    y-5=3(x-1)
    y=3x+2

    Did I do it right or is there another way of doing this?
    red part is wrong
    if I understood the question you should enter the x=1 in the <br />
 f(x) = x^3<br />
             function, instead of the line equation. the line equation will give you the line equation, if you correct the red part
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    29
    is this 3x-y+1 = 0 ? yes

    Then I made



    To get the value of x.


    x = -1 also why is this?

    3(1) + 1 = 4, not 5 My bad, but then if you say x = -1 it would be 3(-1)+1= -2?


    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2009
    Posts
    29
    this is the derrivative of f(x)=x^3 which you need to find the slope, since the slope is 3 that why I made =3 to get x = 1
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by drkidd22 View Post
    is this 3x-y+1 = 0 ? yes

    Then I made



    To get the value of x.


    x = -1 also why is this?


    if
    x^2 = 1 , then x^2 - 1 = 0 ... (x+1)(x-1) = 0

    this says that there will be two distinct lines tangent to the curve y = x^3 that are parallel to
    y = 3x+1

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2009
    Posts
    29
    I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by drkidd22 View Post
    I'm not complettly clear on this problem yet. So what is the equation of the line parallel to the given line? and how is it done?
    you've established that the slope of a line parallel to y = 3x+1 is 3.

    f'(x) = 3

    3x^2 = 3

    x = \pm 1

    now ... get the two points on the original curve where f'(x) = 3

    y = x^3

    y = (1)^3 ... (1,1)

    y - 1 = 3(x - 1)

    y = 3x - 2 is the first line


    y = (-1)^3 ... (-1,-1)

    y + 1 = 3(x + 1)

    y = 3x + 2 is the second line

    now look at the attached graph ... (I made the colors match)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2009
    Posts
    29
    Thanks alot, that clears it up. May I ask what program/software you used to make the graph?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by drkidd22 View Post
    Thanks alot, that clears it up. May I ask what program/software you used to make the graph?
    Graph
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 18th 2011, 10:36 AM
  2. Replies: 5
    Last Post: November 4th 2009, 06:49 PM
  3. Find the equation of a parallel line
    Posted in the Algebra Forum
    Replies: 6
    Last Post: July 1st 2009, 02:11 PM
  4. Find the equation of a parallel line
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 31st 2009, 11:54 AM
  5. Replies: 5
    Last Post: February 12th 2008, 08:21 PM

Search Tags


/mathhelpforum @mathhelpforum