Let P=(1,1) be a point on the degenerate cubic  curve y^2=x^3.  Compute nP for n=1,2,...,5.  What do you think
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
1134068
No, \nfor f(x)=(x+5)/(x-9), you substitute 1 for x, then solve for f(x) to obtain the y-value at x=1. So:\n\n                                 f(1)=((1)+5)/((1)-9)= -6/8\n\n                               which shows that at x=1, y= -6/8 = f(1)\n\nThe point (1, -5/9) is not on the graph.
994152
First, I will say that all the gory details for the following are listed at wikipedia.org.  I'm not going to repeat them, but try to explain some of the missing steps.\n\nGiven a general cubic equation x^3+ax^2+bx+c=0, we can reduce the equation to the form \n\ny^3+py+q=0, \n\nwhere y=x+(a/3) and where p and q are somewhat complicated expressions in a,b,c (not horrible, but not worth writing out here.) This is so because, with x=y-(a/3),\n\n(y-(a/3))^3 + a*(y-(a/3))^2 + b(y-(a/3) + c\n=y^3 - a*y2 + a* y - (a/3)^3 + a* y^2 + a bunch of other stuff\n\nand so the terms in y^2 will cancel.  Work by some Italian mathematicians in the 16th century led them to believe that this was a good form into which to put the equation, because they could then solve it.\n\nNow, try substituting the expression v-u into the reduced equation.  Why?  Possibly some trial and error led the original discoverers to this notion.  What does happen is that when you make this substitution, you get the polynomial\n\nv^3 - 3v^2u + 3vu^2 - u^3 + pv - pu + q = 0\n\nRegroup a bit to get \n\nv^3 - u^3 + (v-u)*(p-3uv) + q=0.\n\nThen we could select v^3 - u^3 + q=0 and p-3uv=0.  This is a set of simultaneous equations which we can, in principle, solve for a set of values (u,v).  One way to do so is by substitution; write u=p/(3v) and substitute into the other equation.  This yields a quadratic equation in v^3 which is easy to solve, and which in turn can be solved for v.  This gives you the value of v-u, which gives you a solution to the reduced equation.  Using the relation y=x-(a/3) gives you a solution to the original equation.  I'm not writing out all the gory details because you can get all of those from a site like wikipedia or mathworld.wolfram.com.\n\nHaving said all that, it's unlikely you'd ever need to use this for anything.  One useful fact is that ANY cubic equation with real coefficients has at least one real solution (thanks to the intermediate value theorem) so knowing this root (or a good approximation) will allow you to reduce your problem to a quadratic equation, which is easy to solve.\n\nAnd yes, I'll include a link to a site which does list the gory details.
934522
y' = [(x-1) - x] / (x-1)^2 = -1/(x-1)^2, so no critical points.  y' < 0 for all x where y is defined, so it's always decreasing.\n\ny'' = 2/(x-1)^3, which is positive when x>1, and negative for x<1.  Therefore the curve is concave up for x>1, and concave down when x<1.\n\nThere is a vertical asymptote at x=1, and lim x->inf x/(x-1) = \nlim x->inf 1/(1-1/x) = 1 (same for x -> -inf) so there is a horizontal asymptote at y=1.
724027
sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
694939
Your parabola will be of the form f(x) = k*(x-int1)(x-int2) where int1 and int 2 are the intercepts given and then you adjust k so that f(x) passes through that minimum.  So f(x) = k*(x+2)*(x-3).  \n\nThe minimum is going to happen half way between the two intercepts, or at x=.5, and f(.5) = k*2.5*(-2.5) = -6.25k which is supposed to be -5 from the problem.  So then k must be .8 (since .8 * -6.25 = -5) and your polynomial is:\n\nf(x) = .8(x+2)(x-3) = .8x^2 - .8x - 4.8
197811
get the points where curves insersect theirselves\ny^2 = -3y^2 + 36\n4y^2 = 36\ny^2 = 9\ny = +3, -3\nso the x value to those y values is 9\n---------------------------------------------\narea = integrate (-3y^2 + 36 - y^2, y) from -3 to 3\narea = -4/3 y^3 + 36y, evaluate from -3 to 3\narea = ((-4/3)(3)^3 + 36(3)) - ((-4/3)(-3)^3 + 36(-3))\narea = (-36 + 108) - (36 - 108)\narea = 144
325102
sphere points: A, B, C as vectors from sphere center O.\n\ncompute:\n\nX = A x B\nY = A x C\n\n\nU = X x (A + B)\nV = Y x (A + C)\n\ncenter points:\n\nP = ± U x V\n\n(if necessary normalize P to sphere radius)\n\n--------------\n\ncross product: A x B\n\nsketch:\nhttp://static.flickr.com/48/129027255_d097e01a2d_o.jpg\n\nfurther reading: spherical Voronoi Diagrams, projective geometry\n\nthanks Neodiogenes. my solution was incomplete. i corrected it.
790996
if cuts the x axis when p(x) = 0\n\nthat is 2x-3 = or x= 3/2\n\n2nd point x-5 =0 or x = 5\n\n3rd point x+1 = 0 or x = -1
716583
If you know the vertex then you know\n\na (x-xv)^2=y-yv  , use P to find a, the shape \n\nat P:\na(xp-3)^2 =yp-1\na(-1-3)^2 = 3-1\n16a = 2\na=1/8\n\nSo, the parabola is \n\ny = 1/8*(x-3)^2 + 1\n\nCheck: \nat V  y= 1 ok\nat P y = 1/8(-4)^2 + 1 = 16/8+1 = 3 ok
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
1166634
Say the point (1,2) is on f(x) and the point (2,5) is on g(x).\nNow since (1,2) is on f that means f(1) = 2.\nSimilarly, g(2) = 5 since (2,5) is on g.\nYou can now find (gof))(1)\n(gof)(1) = g[f(1)]  = g{2) = 5\n(1,5) is a point on (gof)(x)\nThe y-value of one function has to match the x-value of the other one.\nLet (4,7) be on f and (3,4) be on g.  You can find (fog)(3)\n(fog)(3) = f(g(3)) = f(4) = 7\n(3,7) is a point on (fog)(x)
844444
Remember that the vertex of the parabola for y = x² is at (0,0).  You can confirm this by putting x = 0 in and seeing that you get y = 0.\n\nIf you instead replaced x with (x - 3):\ny = (x - 3)²\n\nNotice how it would take a value of x = 3, to get y = 0.  In other words, the whole graph, including the vertex would move 3 units to the right.  The new vertex would be at (0, 3).\n\nSimilarly, if you replaced y with (y - 2):\ny - 2 = x²\n\nThis is the same as:\ny = x² + 2\n\nSo every value would move up 2.  The graph would move two units up and the vertex would similarly move to (0, 2).\n\nCase 1:\nParabola moves 3 units to the right -- vertex now at (3,0)\n\nCase 2:\nParabola moves 2 units up -- vertex now at (0, 2)\n\nExtra credit question for you:\nWhat would happen if you did both?\n(y - 2) = (x - 3)²
653036
when you have 2 equations and you have to solve for a point(s) those points are the intersections of thir plots.\n\nSince their are intersections, y from first eq has to be equal to y from second equation for one or more common x values.\n\nKnowing this you can do y1 - y2 =0\n\nIn this porblem you have:\ny1=e^(-x)+1\ny2=ln(x)+3\n\nso \n\n( e^(-x)+1 ) - ( ln(x)+3 ) = 0 \n\nequals to:\n\ne^(-x)-ln(x)-2=0\n\nNow any x that satisfy that eq is a intersection point, therefore an answer of your problem.\n\nFor this eq (and many others) Newton method for finding zeros (or roots) is the best option. (check link on source)\n\nLooking at plots, this curves has only one interception point, close to x=0.3\n\nBy aplying this algorithm, starting with a value of x0=0.5 you will find:\nx1=0.231...\nx2=0.281...\nx3=0.286...\nx4=0.286713167526\nx5=0.286713169911\n\nyou can aproach as much as you need.\n\nFinally, to find y1 value (or y2 because are the same) \npick one of the ecuations and solve for the x value you\nfind.\n\ny2=ln(x)+3\n\ny2=ln(0.286713169911) +3\ny2=1.75072702906073\n\nthen (x,y)=(0.286713169911 , 1.75072702906073)
51024
Sorry it's not easy to draw polynomial division here.\nAnyway: the first remainder (if I calculated without errors!) is -a^2-a^3\nThe remainder of the second division is a^3-2a-3.\nYou then have the equation:\n-a^2-a^3=a^3-2a-3\nthat is\n2a^3+a^2-2a-3=0\n\nYou can also write this as\n2a^3+a^2=2a+3\n\nNow you can find a graphical solution to the equation by searching the intersection between the cubic y=2t^3+t^2 and the linear y=2t+3, and considering the value of the ascissa of such intersection.\n\nOtherwise, any other method for solving cubic equations can work.
384702
The first 3 are all straight lines, so you just need to get 2 points in each.\n1. y = |x| + 3 for x = 1 then y = 4, for x = 2 y = 5.\n\n2. y = 3/2x - 2 for x = 2 then y = 1, for x = 4 y = 4\n\n3. y = x - 5 for x = 5 y = 0 for x = 6 y = 1\n\nFor #4. This is a curve shaped like and upside down U that passes through 0,0, for all values of x, y will always be negaive.\n\nThe best way is to substitute values of x and get values of y and sketch.\n\nHope I was some help
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.