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Math Help - Find the equation of the line tangent to graph f

  1. #1
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    Find the equation of the line tangent to graph f

    I need to find an equation of the line that is tangent to the graph of f and parallel to the given line.

    Function:
    f(x)=(x^3)+2

    Line:
    3x-y-4=0


    So far I rearranged the equation of the line to get... y=3x-4

    So I plugged lines with the slope of 3 into my calculator. The graph y=3x seemed to look tangent to f(x) and parallel to the line. However, I need to solve this problem without a calculator, and without guessing by the looks of a graph.

    Thanks to anyone who tries to help me.

    --Cori
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    Quote Originally Posted by liz155 View Post
    I need to find an equation of the line that is tangent to the graph of f and parallel to the given line.

    Function:
    f(x)=(x^3)+2

    Line:
    3x-y-4=0


    So far I rearranged the equation of the line to get... y=3x-4

    So I plugged lines with the slope of 3 into my calculator. The graph y=3x seemed to look tangent to f(x) and parallel to the line. However, I need to solve this problem without a calculator, and without guessing by the looks of a graph.

    Thanks to anyone who tries to help me.

    --Cori
    recall that the derivative gives the slope of the tangent line at any arbitrary point.

    find f'(x) and set it equal to three and solve for x. that would give you the required x-coordinate. find the y-coordinate and then derive the tangent line from that

    try it. if you have any questions, ask
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    Ok, So I set f'(x)=3x^2=3
    Then I solved for x and got 1

    But how do I solve for y? Do I plug the 1 back into the derivative equation?

    Like this?
    f'(1)=3

    So the pt = (1,3) ?

    Then I'm stuck on how to find the equation for the line...

    Please let me know if I'm on the right track.

    Thanks so much!!!

    --Cori
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    Ok, So I set f'(x)=3x^2=3
    Then I solved for x and got 1

    But how do I solve for y? Do I plug the 1 back into the derivative equation?

    Like this?
    f'(1)=3

    So the pt = (1,3) ?

    Then I'm stuck on how to find the equation for the line...

    Please let me know if I'm on the right track.

    Thanks so much!!!

    --Cori
    first of, x = \pm 1

    you find the y-coordinate by plug the x-value you got into the original equation. you want a point on the graph, correct? why would you plug it into the derivative equation?

    when you find the desired coordinates, let's call them (x_1,y_1), then we can use them along with the slope m in the point-slope form to find the equation of the line

    Point-Slope Form: y - y_1 = m(x - x_1)

    and then rewrite it in the form y = mx + b when done
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    Ok, I see what I did. I don't know what I was thinking.

    So I took the x= +/- 1 and plugged it into the original...

    f(1)=(1)^3 +2 = 3
    f(-1)=(-1)^3 +2 = 1

    (1,3)
    1=x_1
    3=y_1

    I used both the coordinates to get slope = 1

    Then pt. slope form

    y-3=(1)(x-1)
    y=x+2 <-- final answer

    Is that right??

    Thanks SOOOOO much!!!

    Edit: Wait nevermind, that isn't parallel to the line it needs to be parallel to. Ahh. I'm so confused now.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    Ok, I see what I did. I don't know what I was thinking.

    So I took the x= +/- 1 and plugged it into the original...

    f(1)=(1)^3 +2 = 3
    f(-1)=(-1)^3 +2 = 1

    (1,3)
    1=x_1
    3=y_1

    I used both the coordinates to get slope = 1

    Then pt. slope form

    y-3=(1)(x-1)
    y=x+2 <-- final answer

    Is that right??

    Thanks SOOOOO much!!!

    Edit: Wait nevermind, that isn't parallel to the line it needs to be parallel to. Ahh. I'm so confused now.
    what are you talking about? the slope is 3, we said that from the beginning. our objective was to find the correct x and y coordinates so that we can plug them into the point-slope form. use +1 or -1 for x. these correspond to two different lines, either will give you a solution
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    Problem: I plug in the 1 and -1

    f(1)=3(1)-4 = -1
    f(-1)=3(-1)-4 = -7

    Coordinates: (1, -1) and (-1,-7)

    Slope=3

    But whenever I use pt slope I keep getting 3x-4, the original line equation. I don't know what to do from here.

    Thanks for all your patience. I made some very silly mistakes.

    --Cori
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    Problem: I plug in the 1 and -1

    f(1)=3(1)-4 = -1
    f(-1)=3(-1)-4 = -7

    Coordinates: (1, -1) and (-1,-7)

    Slope=3

    But whenever I use pt slope I keep getting 3x-4, the original line equation. I don't know what to do from here.

    Thanks for all your patience. I made some very silly mistakes.

    --Cori
    the original curve is y = x^3 + 2

    if we use x = 1:

    y = (1)^3 + 2 = 3

    so we have (1,3)

    using (x_1,y_1) = (1,3) and m = 3 we have by the point-slope form:

    y - 3 = 3(x - 1)

    \Rightarrow y = 3x ------------> slope of one of the tangent lines with the required conditions


    Now, this may be a bit trivial, but try it with x = -1



    Moral: don't lose track of what equations you must use. you are mixing up the tangent line with the original equation with the conditional line that we are given
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  9. #9
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    The other line is y=3x+4.

    I really think I get it now. Thanks SOOO much for all your help!! I couldn't have figured it out without you!

    --Cori
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    The other line is y=3x+4.

    I really think I get it now. Thanks SOOO much for all your help!! I couldn't have figured it out without you!

    --Cori
    yes you could.

    you're welcome anyway
    Last edited by Jhevon; September 2nd 2007 at 11:51 PM.
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