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Equation of the tangent line to the curve

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the tangent line to the curve: 7*x*e^(x)+8 at (0,8)

    2. Relevant equations

    Derivative I guess?

    3. The attempt at a solution

    I know you have to take the derivative of the equation given which I think is

    7*x*e^(x) + 7*e^(x)

    Then you plug it into slope intercept form: y-y=m(x-x)

    I did this and got y-8=(7*e^(x)*(x+1))*x, but apparently that's not right...

    What did I do wrong? Maybe i just typed it in wrong?
     
  2. jcsd
  3. Aug 30, 2007 #2

    learningphysics

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    Calculate the slope first by plugging in x=0 into the derivative equation... then when you have the number for the slope, plug it into the slope intercept form.

    The slope should just be a number.
     
  4. Aug 30, 2007 #3

    Dick

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    Yeah, it's pretty wrong. In the line equation m is a constant. Otherwise, it wouldn't be a line, now would it? Put (0,8) into the derivative before you plug it into the line equation.
     
  5. Aug 30, 2007 #4
    so essentially you have to use your x-intercept twice
     
  6. Aug 30, 2007 #5
    So the slope is just 7?
     
  7. Aug 30, 2007 #6
    Thus the final equation is y-8 = 7x ?
     
  8. Aug 30, 2007 #7
    it should probably be in slope-intercept form

    y=mx+b
     
  9. Aug 30, 2007 #8
    It says that point-slope form is fine.. but it would be y=7x+8
     
  10. Aug 30, 2007 #9

    learningphysics

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    Yup. That's right.
     
  11. Aug 30, 2007 #10
    Alright.. thanks a lot!
     
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