How would i graph the function y = -x^2 + 3 ?
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...
904114
First, put your original equation as Y1\n\nFor Y2, type  nDeriv(Y1, x, x)\n\npress the buttons   (math)(8)   to get   nDeriv\nthe buttons (VARS)(1)(1)  to get Y1\n\n\nFor Y3, type  nDeriv(Y2, x, x)\n\nthen just display graph Y3\n\nIt will take quite a long while to graph, better if you manually differentiate the original equation once, then use the nDeriv to do second derivative.
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
902917
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
736099
plot the coordinates then do the vertical line test.  if the vertical line intersects more than two points on the graph then it is not a function.  each set of coordinates you have listed are functions (i free hand it with no graphing paper or calc) so forgive me if im wrong.  also refer to www.coolmath.com. it is a good site for breaking down math
971807
(0,0)\n(3,0)\ndraw the graphs and then find them out!\nif u can't, then mail me!
119382
y=log_4 X\nand\ny= -log_3 (X+s) + 2\n\nRemember that logarithmic function y = log_b x is the inverse of the exponential function y = b^x, so you just graph its inverse (the exp function) and the graph of your log function is its mirror image with respect to the line \ny = x (The line that passes through (0,0) and has a slope of 1)\n\nIn your first given,\ny = log_4 x\nIts inverse function is\ny = 4^x\n\nNow graph y = 4^x\n\nNext, graph y = x and use this Line as your guide mirror to draw the log graph. This means that the shape of the log graph is the exact mirror image of the exp graph.\n\nIn your second given\ny= -log_3 (x + s) + 2\nJust rearrange the terms\ny - 2 = -log_3 (x + s)\n-(y - 2) = log_3 (x + s)\n\nThe inverse is\n-(y - 2) = 3^(x + s)\nGraph this one, then use y = x again as your "mirror."  Simple.\n\nMay this work of mine help you in your studies ;-)
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
821928
I only see equalities... and what are you supposed to do with them?\n\nAssuming you need to graph them, first put them in terms of y:\n1) y = -x + 2\nThis has a slope of -1 (diagonal downward) and a y-intercept of 2.\n\n2) y = -2\nThis has no slope (horizontal line) and goes through the y-intercept of -2.\n\n3) y = x + 2\nThis has a slope of 1 (diagonal upward) and a y-intercept of 2.\n\nThe lines are shown in the attached diagram... Now what do you want to do with them?
1119524
well using the y = function, you may need to create a positive and negative graph\n\nfor a circle with a center of 0, 0:\n\nsquare root ( x^2 - r^2)  (replace r with your desired radius)\nand\n- square root( x^2 - r^2)
958831
Draw the graph of y = x^3\nThe (x-1) part shifts the graph to the right by 1.\nThe +4 part shifts the graph up by 4.\n(0,0) lies on y = x^3\n(1,4) lies on y = (x-1)^3 + 4\nNote that the x-coordinate increased by 1 and the y-coordinate increased by 4.
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
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!  :)
830542
y = sqrt[x/(x-5)]\n\nStart by getting a few points:\nf(0) = 0  (it passes through the origin)\nf(5) is undetermined (divide by zero)\n\nSee what happens when 0 < x < 5.  For x = 3, x/(x-5) is negative, therefore the function is undefined over that range.\n\nFor x > 5, x/(x-5) is positive, so the function exists.  When x gets very large, x >> 5 (e.g., x = 1 million), x/(x-5) approaches one, so f(x) approaches sqrt(1) = 1 from above.  Therefore, y = 1 is an asymptote, but f(x) is always larger than one for very large x.\n\nNow look at x < 0.  x/(x-5) is always positive, and it's always less than one.  A good point to use is:\n\nf(-4) = sqrt[-4/(-4-5)] = sqrt(4/9) = 2/3\n\nand another good point is\n\nf(9) = sqrt[9/(9-5)] = sqrt(9/4) = 3/2\n\nWe know enough to plot the graph:\n\n1.  Draw the axes; draw a dotted-line asymptote at y=1; and draw another dotted-line asymptote at x=5.\n\n2.  Let's do negative x first.  Start at the origin, draw the curve through (-4, 2/3), and continue the line out to the left, using y=1 as an asymptote.\n\n3.  The function is undefined for 0 < x <= 5.\n\n4.  For x>5, start the curve way up on the x=5 asymptote, bring it down through (9, 3/2), and continue it out to the right, using y=1 as an asymptote.
351583
how about instead of (a+b)^4 you plot (x+y)^4?\n\nit's quite a funny 3D plot.\n\nLooks like a conehead balancing on his head.\n\nSeriously... draw a cone rounded off on the top and then flip the paper upside down and you'll get an approximation of (a+b)^4.
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
727314
From y = x2 to y = x2 - 6 is: shift the graph 6 units 'down'.\nFrom y = x2 to (x + 5)^2 is: shift the graph 5 units 'to the left'.\n\nTh