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Find f`(x) for the tangent line of the graph

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose line tangent to graph of y=f(x) at x =3 passes through (-3, 7) & (2,-1).
    Find f'(3), what is the equation of the tangent line to f at 3?

    2. Relevant equations

    I found the slope of which equals -8/5

    Im not sure how to find the equation... do I do y-3=-8/5(x-3)? or is that wrong?
  2. jcsd
  3. Feb 26, 2012 #2
    The general form of a straight line is
    y=m x + c
    Try fitting that to your points
  4. Feb 26, 2012 #3
    The tangent line passes through those two points.
    Since it is a LINE, you may use the equation to solve for the slope:

    m= (y2-y1) / (x2-x1)

    Once you have found the slope, use any of of the two coordinates to solve for 'c', the constant/y-intercept.

    Once you have the constant, simply write it in the form y=(#)x + (#).
    That is the equation of the tangent line at that point x=3.
  5. Feb 26, 2012 #4
    y=-8/5x+8.2 is this answer correct?
  6. Feb 27, 2012 #5


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    Science Advisor

    Why do you keep asking? It is simple arithmetic to check whether or not (-3, 7) and (2, -1) satisfy the equations you give.

    Also, use parentheses to make your meaning clear. Many people would interpret "-8/5x" as "-8/(5x)". And I am not clear whether "8.2" means multiplication or "8 and 2 tenths". If you mean y= (-8/5)x+ (8)(2)= (-8/5)x+ 16 then if x= -3, y= (-8/5)(-3)+ 8.2= 24/5+ 8.2= 4. which is not 7. And if you mean y= (-8/5)x+ 8.2= (-8/5)x+ 82/10= (-8/5)x+ 41/5, then which x= -3, y= (-8/5)(-3)+ 41/5= 24/5+ 41/5= 65/5= 13, again, not 7.
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