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Help needed in finding Tangent to a graph

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    If f(x)= 3x^(2) - x + 4, find the values of m for which line y=mx+1 is a tangent to the graph of f?


    2. Relevant equations
    None


    3. The attempt at a solution
    Well first you have to get the derivative of the function which is 6x-1....Then what? There are no points given.....I tried solving in terms of x and plugging it back in the equation but it didn't work:

    x=(m+1)/6 and y=6x^(2)-x+1....I tried substituting these but nothing happened......


    Any ideas?
     
    Last edited: Aug 31, 2009
  2. jcsd
  3. Aug 31, 2009 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. Try drawing a graph of the original function -- that should help you visualize the problem and how to get to the solution...
     
  4. Aug 31, 2009 #3

    okay, I solved the question through graphing! I got 5 and -7, however, I still can't figure out the algebraic solution!
     
  5. Aug 31, 2009 #4

    rock.freak667

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    Homework Helper

    Say you solve simultaneously and get following after simplifying ax2+bx+c=0.

    We can find the roots by

    [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    If b2-4ac>0, we get two real and distinct solutions.

    eg. x= (-1±1)/2 → (-1+1)/2 or (-1-1)/2

    if b2-4ac=0 we get one solution.

    if b2-4ac<0, we get two complex solutions.

    A tangent to a graph touches the graph at how many points? Thus how many solutions do you expect?
     
  6. Aug 31, 2009 #5

    berkeman

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    Staff: Mentor

    Well said, rock.
     
  7. Aug 31, 2009 #6
    A way to do this (among others) is to let (a, f(a)) be a generic point on the curve. Find the derivative of f and evaluate this when x = a. This is the tangential slope to the curve at (a, f(a)). The equation y = mx + 1 does tell us the tangent passes through (0, 1) (the y-intercept) so we can use the Point-Slope Form:

    [tex](y-f(a))=f'(a)(x-a)[/tex]

    and the fact that (0, 1) must be a solution of this equation to back-solve for a.

    This would provide the equation.

    --Elucidus
     
  8. Sep 1, 2009 #7
    Thank you all for your help. It turns out that I kept using a different number while solving and therefore my substitution was wrong which kept giving me wrong answers!!!! :P

    I've solved the question algebraically. You just need to substitute the derivative in for m and equate both equations which will give you: 3x^(2)-3
    Then solving for X, you get x=+-1. Plugging these 2 values into the derivative will give you the 2 values of m which are 5 and -7.
     
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