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Quick tangent line question (calc final tomorrow)

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known
    i have to find all tangent lines in the equation below that pass though the point (0,0)
    [tex]y=x^{3}+ 6x^{2}+8[/tex]

    i took the derivative and got this.

    then i substituted the points x=0 and y=0 into the derivative equation and the got 0=0.

    i am kind of stuck here cause i havent done a problem like this in a while. what should i do next?
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2


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    The derivative equation is y'=3x^2+12x, that's the slope of the tangent line. Another way to express the slope of the tangent line is
    (delta y)/(delta x)=(y-0)/(x-0)=(x^3+6x^2+8)/x. Equate the two.
    Last edited: Dec 11, 2007
  4. Dec 11, 2007 #3
    i know that the derivative is the slope/ but how can i find the equations of all of the tangent lines that go thru the origin?
  5. Dec 11, 2007 #4


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    There are two different way to express the slope of the tangent line. i) use the derivative, ii) use the difference (delta y)/(delta x) for two points on the line, like (x,x^3+6x^2+8) and (0,0). Isn't that what I just said? Equate them.
  6. Dec 12, 2007 #5
    Well since you found the slope and you have the point (0,0) why not just do the point slope forumla? That will give you the equation of the tangent line. Right?
  7. Dec 12, 2007 #6
    First of all ask yourself how many tangent lines can you find in one point.
    Second, you have found the formula for the slope of the curve in ANY point, but you need to find the one for X=0, and then construct a line that has that slope and passes through the point that has been given to you.
  8. Jan 14, 2009 #7
    Equation of a line: [tex] y_1-y_0 = m(x_1 - x_0) [/tex]
    Two points on the line: [tex] (0, 0) (x,y) [/tex]
    slope: [tex] m = y' [/tex]
    original equation: [tex] y = x^3+6x^2 +8 [/tex]

    Start filling in the blanks.
  9. Jan 14, 2009 #8


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    No, that's y', not y.

    The problem does not say the curve passes through (0,0) (nor is it true). There is no point in putting x= 0, y= 0 in any equation.

    Any line that goes through (0,0) is of the form y= mx and has slope m. The tangent line must go through a point on the graph and must have m equal to the derivative. At that point [itex]y= mx= x^{3}+ 6x^{2}+8[/itex] and [itex]y'= m= 3x^2+ 12x[/itex]. Solve those two equation for m and x (it's only m you need but different values of x may give different slopes and different tangent lines). Try multiplying both sides of the second equation by x and setting the two values for mx equal.
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