Sketch the graph of {x:x^2<4} on a number line??
367523
x^2 < 4, swap the < for a = for a sec.  That gives you x^2 = 4, and x = +/- 2.  So on a number line, draw an open circle at -2 and 2 (open circle because the original problem was < and not <=).  Now comes the logical part - if you plug in something to the far left of -2, does it satisfy your original equation?  Nope.  How about far right of 2?  Nope again.  What about between -2 and 2?  Yep.  So fill in or darken the line between the open circles.
325867
First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
1183576
just pick two points from the graph\nif you wanna check plug them in the equation\n\n\n\nand when the line crosses the Y axis Y does NOT =0\nX=0 when line crosses Y axis\n\nchoose any set of #s that will not give you the square root of any negative. because that would result in an imaginary #, not a real
365741
I think all you have to do is plug in a few values for C to get a graph.  For example, try 0, 5, 10, 15, 20.  It's continuous so you can connect the dots.\n\n(0,0), (5,87.5), (10,350), (15,787.5), (20,1400)\n\nTotal labor for 20 is 1400.
1066492
x = -4\n10points thank u!
158569
Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
617377
first put it into the form (x-h)^2=4p(y-k)\nCenter: (h,k)\nFocus: (h,k+p)\nFocal Width: 4p\n\nPlot those points and you should be good!\n(if you're having trouble getting it into that form try using the complete the square method)
775797
http://www.coolmath.com/graphit/
406468
D = {x | x < -3 U x > 4 }
950608
get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.
769460
I think that you want x element of [-6,4)\n\nYou wanted closed on the left and open on the right, is that true?\n\nJust remember that for intervals:\nopen   ( )\nclosed [ ]\n\nAnd you can match them up: [-6,4)
317895
Generally, you break an inequality such as this into two parts and graph each separately.  So you'd be graphing x=-1 and x=3.\n\nThe type of line depends on the sign: if it's a <= or >=, then use solid lines to indicate anything on the line satisfies the original inequality.  If it's just < or >, then use a dashed line.  (This is similar to open and closed circles on a number line.)  Then, you pick points in the different regions that you created with the lines and determine if they satisfy the inequality.  If so, the region gets shaded.\n\nThere is no y dependency, so you basically have two vertical lines for your graph for this problem.  Checking the regions, only the area between the lines satisfies the inequality, so that's the only region that gets shaded.
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.
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!  :)
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
506906
In here if you just type x^2 everyone understands that to be x squared.\n\nIn MS Word, you can check superscript under edit font
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...