In Mathematica, how do I find a point where a curve crosses itself?
111079
a curve continuing indefinitly without change will intersect its beginning and will not cross ,,,,just continue over and over ,,FOREVER
1160851
http://www.citlink.net/~math/geometry/ch1/1-4_notes.pdf#search='segment%20addition%20postulate'
735957
Well. You can keep drawing tangents until one which you find is the steepest. From your info, i have the idea that the curve is in the shape of a bell. So perhaps you can try drawing tangents at the mid-range of the curve.\n\nHope this helps=)
772313
You draw it under the curve\n\nThe tangent line will only touch the curve at one point for this type of smooth upward bending curve.(At least for the part of the graph shown).\n\nYour graph plots total displacement verses time. If the graph was a straight line then this would indicate a constant "velocity" but since the curve bends upwards you are dealing with a positive acceleration.\n\nPractically, from your graph, suppose you wanted to know the approximate slope of the tangent line at t=5. Use the two closest data points: t=4, d=8.4 and t=6, d=16.5. Then the slope is about (16.5-8.4)/(6-4) = 4.05 at t=5. it will be slightly greater than this. You can use a straight edge to make a visual adjustment and then you can find the slope by dividing the displacement value at t=5 by (5 minus the value of t where the straight edge crosses the time axis).\n\nThe other way is to use the data to deduce, a closely as possible, the formula of the function being graphed. For example if the displacement was directly proportional to the square of the time, d = k t^2 for a determined value of k, then the slope of the tangent line at any time, t, would simply be 2 k t which is the time derivatiive of the function.
697951
find out what happens to Y (what it approaches) as:\nx goes to infinity\nx goes to negative infinity\nthen find out what happens to X as:\ny goes to infinity\ny goes to negative infinity\n\nalso find the intercepts (if they exist)
774094
- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
389156
actually, you slolve each of the three parenthesis and that tells you when the equation equals zero...\n\nX=-2, x=1 and x=5.\n\nNow look at each interval to determine whether the curve is greater than or less than zero in that interval.  For example, \nless than -2: x=-3 is less than -2.  So substitute x=-3 and solve... (-1)(-4)(-8)=-24 which is less than 0.\nbetween -2 and 1: x=0 is between these numbers.  So substitute x=0 and solve...(2)(-1)(-5)=+10 which is greater than 0.\nbetween 1 and 5: x=2 is between these numbers.  So substitute x=2 and solve....(4)(1)(-3)=-12 which is less than 0. \ngreater than 5: use x=6: (8)(5)(1)=+40\n\nSo the number line should show an open circle at -2 and an arrow pointing left, and open circles at 1 and 5 with a line connecting the two circles at 1 and 5.
602663
No, a line cannot intersect a circle in at most one point, unless you're talking about tangent. A line can be tangent to a circle at only one point.
623834
I think you mean that: Given a figure (function in the form of a curve), and the value of ordinate (y), does a value of abscissa (x) exist corresponding to that y?  \n\nIf the figure is an open curve, like y=x or y=ax^3, the answer is yes.\n\nBut there may be many other cases. For the curve y=x^2, which is a parabola, x may assume any real value; but y is always non-negative.\n\nAgain, x^2+y^2=a^2, is a circle and a closed figure; x can take up values between -a and +a. y also takes values in the same range.\n\nFor this curve, given a value of y, within the admissible range of -a to +a, there are two values of x. \nFor y=+a or y=-a, there will be only one value of x; and for y>a or y<-a, there will be no value of x.\n\nFinally, for periodic curves like y=sin x, for admissible values of y (between -1 and +1) there exist infinitely many values of x.\n\nSo, you see there is not one answer but many depending on the nature of the figure!
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
947081
It will look like a < with lines of slope 1 and -1.  The leftmost point will be at (0,-1).
759663
a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
458048
Or you can do this:\n\nhttp://www.xahlee.org/SpecialPlaneCurves_dir/Parabola_dir/parabolaTangents.png\n\nto form a parabola out of many straight lines.
963751
find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
826422
Find the slope of the line that connects those points.
910766
The only interesting point in this curve will be the inflexion point which you find by doing the derivative twice and equating to zero.\n\nf'(x) = 3x^2 + 1\n\nf"(x) = 6x = 0 <===> x = 0\n\nf(0) = -1\n\nThe curve will have an inflexion point at (0,-1).\n\nf(0) < 0, f(1) > 0 then:\n\nThe curve will also have a single real root in [0,1] (x = 0.6823)