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Homework Help: Equation of tangent lines

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Give the equation of the two lines through the point (-1, 3) that are tangent to the parabola y= x^2


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 3, 2010 #2
    What have you done so far?

    Also, are you sure you copied the question correctly?
     
  4. Feb 3, 2010 #3
    i really do not no where to start?? I did copy the question correctly.
     
  5. Feb 3, 2010 #4
    Equation of a line

    1. The problem statement, all variables and given/known data

    Give the equation of the two lines through the point (-1, 3) that are tangent to the parabola y= x^2

    2. Relevant equations



    3. The attempt at a solution

    Not sure where to start?? Do I find the deriviative of x^2? I know that equals 2x...
     
  6. Feb 3, 2010 #5

    rock.freak667

    User Avatar
    Homework Helper

    Re: Equation of a line

    So you know that dy/dx gives the gradient of the tangent at a point (x,y)

    dy/dx=2x.

    So what is the gradient of the tangent at the point (-1,3)
     
  7. Feb 3, 2010 #6

    Mark44

    Staff: Mentor

    Re: Equation of a line

    Start by drawing a graph..
     
  8. Feb 3, 2010 #7
    Re: Equation of a line

    I did draw the graph...still not helping....So if dy/dx is equal to 2x, would you plug -1 in for x?
     
  9. Feb 3, 2010 #8

    Mark44

    Staff: Mentor

    Re: Equation of a line

    No, not at all.

    Looking at your graph, about where do the lines through (-1, 3) hit the parabola so that they are tangent to it?
     
  10. Feb 3, 2010 #9
    Re: Equation of a line

    well there is only one line that goes through it.....and I dont know where it is tangent? it is negative...
     
  11. Feb 3, 2010 #10

    Mark44

    Staff: Mentor

    Re: Equation of a line

    I see two lines. When you say "it is negative" please be more more specific. What is negative?
     
  12. Feb 3, 2010 #11
    Re: Equation of a line

    it is negative as x approaches zero?
     
  13. Feb 3, 2010 #12

    Mark44

    Staff: Mentor

    Re: Equation of a line

    What is "it"? Please be more specific. I have no idea what you are trying to say.
     
  14. Feb 3, 2010 #13
    Re: Equation of a line

    the tangent line on the parabola at point (-1,3) as it approaches zero
     
  15. Feb 3, 2010 #14

    Mark44

    Staff: Mentor

    Re: Equation of a line

    How can a line be negative? In "as it approaches zero" what is it? Please stop using pronouns.

    The point (-1, 3) is NOT on the parabola!
     
  16. Feb 3, 2010 #15
    Re: Equation of a line

    You obviously don't know what you're being asked to find. That's fine, those questions used to throw me off too...

    This video will show you an example of what you're actually being asked to do, it will also give you a better understanding of why you're being asked to find this line and also why the derivative is actually being asked of you...

    http://www.5min.com/Video/The-Equation-of-a-Tangent-Line-169041763

    I advise watching as many videos on this site, made by this guy as possible. For calculus, they will give you an understanding of what you are doing & that's a great thing if your pre-calculus is weak.

    Also, this video will only give you one of the lines you're looking for, once you have fully watched this video (and the related ones if you can!!!) then I want you to find out what the "Normal" to a line. This is also called a Perpendicular line. This is what you're being asked to find but they don't explicitly state it as you're supposed to know this, but don't fret, just learn it now :)


    btw, the point (-1, 3) is not on the parabola, the point (-1,1) is, then (-2,4), maybe you wrote the wrong thing down here...
     
  17. Feb 3, 2010 #16

    Mark44

    Staff: Mentor

    Re: Equation of a line

    Or not. AFAIK the OP wrote the problem correctly. It doesn't matter that (-1, 3) isn't on the parabola. The goal is to find two lines through this point that are tangent to the parabola. I'm trying to get the OP to at least have a visual understanding of what's going on in the problem.

    I'm not being too successful, so far.:tongue:
     
  18. Feb 4, 2010 #17
    Re: Equation of a line

    I still do not understand why there is two equations?? I know the derivative is 2x... so that means the slope = 2.....the point (-1,1) is a pt. on the parabola. So, could I use this for point slope form?
     
  19. Feb 4, 2010 #18

    Mark44

    Staff: Mentor

    Re: Equation of a line

    Is the line from (-1, 3) to (-1, 1) tangent to the parabola? You can ignore what sponsoredwalk said about normal or perpendicular lines. This problem has nothing to do with them.

    Do you have a graph of the parabola and the point (-1, 3)? This point is inside the parabola. One of the tangent lines would have to be very steep, and with a negative slope. Can you show this on your graph?

    I think it will be extremely difficult or even impossible for you to solve this problem without having a clear image of what you are trying to do.
     
  20. Feb 4, 2010 #19
    Re: Equation of a line

    i see that (-1,3) is inside the parabola.....I just don't understand how the line passes through that point while being tangent to the parabola?? Wouldn't that line intersect the parabola then?
     
  21. Feb 4, 2010 #20
    The problem is that the function [tex]f(x) = x^2[/tex] is what's called a convex function, which means (among other things) that the graph of [tex]f[/tex] lies above all its tangent lines and below all its chords. You can think of a convex function as one that "bends upward" everywhere; you know [tex]f[/tex] is convex because the second derivative of [tex]f[/tex] is the constant function [tex]2[/tex], which is everywhere positive. But the point [tex](-1, 3)[/tex] lies above the graph of [tex]f[/tex], since [tex]f(-1) = 1 < 3[/tex]. This means that [tex](-1, 3)[/tex] does not lie on any tangent line to the graph of [tex]f[/tex].
     
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