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Calculus Problem (involving tangent line)

  1. Feb 15, 2004 #1


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    Here is exactly how the problem reads:

    53. Find an equation of the tangent line to the graph of F at the indicated point.

    y = x^4 - 3x^2 + 2 (1,0)

    I'll be honest. I do not understand how to do these type problems. I've been reading this section of the book over and over to no avail. For some reason, I still cannot grasp how to do this sort of problem. So I'm posting this question here hoping that someone can ignite a lightbulb in my head so I can figure this kind of problem out.

    Any help on getting started is greatly appreciated.
  2. jcsd
  3. Feb 16, 2004 #2


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    For these problems, I suggest that you first find the value of the derivative at the indicated point. This is the slope of the tangent line at this point. Since you know that the indicated point is a point on this line, you can write (y-y1) = m(x-x1), where (x1,y1) is the ordered pair given as the indicated point and m is the slope of the tangent line at that point. If you wish, you can solve that equation for y to get it into the standard y = mx + b form. If you need any more help, please show what you're trying on this problem and we'll try to pick it apart some more.
  4. Feb 16, 2004 #3


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    What, precisely, is it that you don't understand?

    Do you have trouble finding the derivative?

    Do you know the fundamental definition of derivative: that the derivative of a function at a given value of x is the slope of the tangent line?

    Do you know how to find the equation of a line given the slope and one point it passes through?

    In this case you are given that y = x4 - 3x2 + 2. You surely should know the "power rule": the derivative of xn is nxn-1 so that the derivative of x4 is 4 x3 and the derivative of x2 is 2x1= 2x. You can find the derivative of 2 either by writing it as "2x0" and using that formula: the derivative is 2(0)x-1= 0 or, more simply, by recalling that the graph of y= 2 is a horizontal straight line and so has slope 0.
    Now put them together: the derivative of y is 4x3- 3(2x)= 4x3- 6x. In particular, when x= 1 the derivative is
    4(1)3- 6(1)= 4- 6= -2. The tangent line passes through (1,0) and has slope -2.

    Any (non-vertical) straight line can be written in the form y= mx+ b where m is the slope. The line you are looking for can be written in the form y= -2x+ b. You know that when x= 1, y= 0 (it passes through (0,1)) so 0= -2(1)+ b. Okay, what is b? What is the equation of the tangent line?
  5. Feb 16, 2004 #4


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    I got it figured out late last night. In fact, as I was turning out the lights to go to bed, it hit me. So I was up another 5 minutes doing the problem, which is correct.

    Thanks for the advice/help y'all.

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