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Calculus limits

  1. Apr 14, 2008 #1
    Determine the equations of both lines that are tangent to the graph of f(x)=x^2 and pass through the point (1, -3)

    I can find the limit of the graph, which is 2x but I cannot understand how you get two equations
  2. jcsd
  3. Apr 14, 2008 #2
    Well, you're right the derivative of x^2 is 2x. What does this tell you?

    It tells you that if you consider the point (x,x^2) (which lies on the graph) and if you further consider the tangent to the graph at this point, the slope of the tangent will be 2x.

    For example, the point (3.9) lies on the graph because 3^2 = 9. If you draw a tangent to the curve at the point (3,9) ths tangent has slope 6 because 2*3 = 6.

    Well now you're asked to find equations for some line. What do you know about the lines in questions? Well, they should pass through the point (1,-3). What else do you know? Well, they should be tangent to the curve. Quite a lot of information you can use to determine the equations of the lines.

    So I suggest the following.

    Write down the general equation for a line


    What does it mean that this line passes through (1,-3)?

    Now you have the line and the parabola. These two cuves might intersect in two points, in one point or they might not intersect at all. Which of these three possibilities corresponds to the line being a tangent?

    If the line y=mx+b is a tangent to your parabola, it has to be a tangent at some point. It will be important to calculate that point. You know that the slope of the line must be equal to the derivative of the parabola at that point.


    There is an even easier method. Can you write down the general equation for a line passing through (1,-3)? This will be a one-parameter family of lines, meaning there is only one piece of information necessary to specify the line. You can think about that as the direction of the line, you know one point, namely (1,-3) then it seems reasonable that the line is determined by one number, for example its slope :smile: Then you can take the general expression of this line and calculate where it interesects with the parabola, which will be some quadratic equation in x. Then you should think about how many solutions you want .. zero, one or two. This gives you an equation for the parameter of the line with two solutions and you're done.
    Last edited: Apr 14, 2008
  4. Apr 14, 2008 #3
    thank you so much for your help :)
  5. Apr 15, 2008 #4
    i am a bit stuck because i do not understand how to plug in the numbers so ive got this far:

    the limit is 2a so now i can say... 2(1)
    m=2 at that point

    now in y=mx + b form say
    y=2x + b
    -3= 2(1) + b

    therefore one of the lines is 2x-5. How would i get the other equation (also, the answers i believe are y=-2x-1 and y=6x-9) thanks
  6. Apr 15, 2008 #5
    The derivative is 2a (or 2x). this is right. Your conclusion that the slope of the tangent must be 2 is wrong. Remember the slope of the tangent is the derivative of the curve at the point where the tangent "touches" the curve. This is NOT (1,-3) - in fact (1,-3) does not even lie on the parabola.

    Start with y = mx + b und plug in the point (1,-3) - because you want the line to pass through this point ... this gives -3 = m+b or b = -3 -m, so the equation of the line becomes

    y = mx - 3 -m = m(x-1) - 3

    Can you calculate the intersection points of these lines with the parabola? Remembering that the lines are tangents, how many intersection points do you want to have?

    This is a method which does not use derivatives at all - the best would be to understand both ways. :smile:
    Last edited: Apr 15, 2008
  7. Apr 16, 2008 #6


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    To be tangent to y= x2, your line must touch it at some point, say (x0, x02), and have slope 2x0.

    Now, what is the equation of the line passing through (1, -3) with slope 2x0?
    If you know that line also passes through (x0, x02) what equation do you get by substituting those values of x and y? Solve that equation for x0.

    You will get two lines because that equation for x0 is quadratic and you will get 2 values for x0.
  8. Apr 17, 2008 #7
    A formulation of the parabola problem

    I`ve had some difficulty coming up with a clear formulation of the parabola problem. I`ve come up with the following which seems to be reasonably clear.

    Let functions y(x) and m(x) be functions for the parabola and the slope of the parabola


    The origin of line segments that terminate on the parabola (not necessarly at the points of tangency) is


    The function n(x) is the slope of these line segments


    The values of x at the points where the lines are tangent to the parabola are the
    values where the slope of the lines and the slope of the parabola are the same.
    These values are x=-1 and x=3, which are solutions of the equation


    The equations of the corresponding lines, using the point/slope formula are:

    y-y0==m(-1) (x-x0)
    y-y0==m(3) (x-x0)
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