Using Graphing stategy find: Y=X/X^2-9?
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
857762
Not sure if it's the shortest or nicest way, but I was able to get to a solution using substitution u=x^2, and v=y^2, then adding 1/4 to the v side to make it a quadratic equation, solving for v, then re-substituting, and solving for y.\n\nI got y = sqrt ( sqrt (x^2-x^4+1/4) + 1/2 )\n\nYou should be able to plot that ...
733627
Rewrite the equations as:\ny = -x + 2 and\ny = (x+10)/2 = 1/2*x +5\n\nNow, if you insert x=0 in each equation, you will get a y-value.\nIf you insert y=0 in each equation, you will get an x-value.\nFor each line, you can use those two intercepts to graph the line.
795670
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.
239990
There are no values of x for which the function is less than 0.  The function value is between 0 and 1 when x<=0.\n\nPlot a few values to sketch the graph: x=-2,y=1/9; x=-1,y=1/3; x=0,y=1; x=1,y=3; x=2,y=9
514202
All you have to do is combine like terms and then solve for x or y.  Then for the solving and graphing, refer to your textbook.  I don't feel like telling you HOW to graph it.
958895
If you mean y = 2 - x then theabove answers are right.\n\nI'm wondering, though, if you might mean you have a graph of some function f(x) and want to draw a graph of the combined function g(x) = f(2 - x). If so you simply flip the original graph around the y axis (this gives you -x) and then shift the whole thing 2 to the right (this is the 2). You can check individual points if you want to make sure: g(2) = f(0), g(3) = f(-1), ...
897443
2x + y = - 9\n\n2x + y - y = - 9 - y\n\n2x = - 9 - y\n\n2x/2 = - 9/2 - y/2\n\nx = - 9/2 - y/2\n\n- - - - - - - -\n\ny values\n\n2x + y = - 9\n\n2x + y - 2x = - 9 - 2x\n\ny = - 9 - 2x\n\n- - - - - - - - - - - - -\n\n2).\n\nx values\n\n-x - y = - 8\n\n- x - y + y = - 8 + y\n\n- x = - 8 - y\n\n- 1(- x) = - 1(- 8) + (- 1)(y)\n\nx = 8 - y\n\n- - - - - - - - - - - -\n\ny values\n\n- x - y = - 8\n\n- x - y + x = - 8 + x\n\n- y = - 8 + y\n\n- 1(- y) = -1(- 8) + (- 1)(x)\n\ny = 8 - x\n\n- - - - - - - - - - - -\n\n3).\n\nx values\n\n- 3x + 4y = 2\n\n- 3x + 4y - 4y = 2 - 4y\n\n- 3x = 2 - 4y\n\n- 3x/ - 3 = 2/- 3 - 4/- 3\n\nx = - 2/3 + 4/3\n\n- - - - - - - -\n\ny values\n\n- 3x + 4y = 2\n\n- 3x + 4y + 3x = 2 + 3x\n\n4y = 2 + 3x\n\n4y/4 = 2/4 + 3x/4\n\ny = 1/2 + 3x/4\n\n- - - - - - - - - -\n\n4).\n\nx values\n\n- 2x - y = - 3\n\n- 2x - y + y = - 3 - y\n\n- 2x = - 3 - y\n\n- 2x/ -2 = - 3/- 2 - y/-2\n\nx = 3/2 + y/2\n\n- - - - - - - - - \n\ny values\n\n- 2x - y = - 3\n\n- 2x - y + 2x = - 3 + 2x\n\n- y = - 3 + 2x\n\n- 1(- y) = -1 (- 3) + (- 1) (2x)\n\ny = 3 - 2x\n\n- - - - - - - -s-
769049
Option a: use one of the fancy calculators around nowadays.\nOption b:\nOk, we need at least a reasonable guess to get going. Lets draw a graph. But how do you draw a graph of 2cos(x) + x? We don't. We draw a graph of 2cos(x), and another graph of -x, and see where they are equal - because thats the point we're looking for. Don't forget to use radians, since you can't add a number to a degree.\nThe solution looks like it is around -1 or so. Now we can do a bit more guessing.\nPutting -1 into that equation gives 0.0806. Lets try -0.9. Whoops - that gives 0.34, which is getting even worse. How about -1.1? Ahha! That gives -0.19, so we know the answer is between -1.1 and -1.0.\nWe can keep narrowing it down in this way. Try -1.05: thats negative, so we're between -1.0 and -1.05. Try -1.02: thats positive, so we're between -1.02 and -1.05. Try -1.03: thats negative, but nearly 0. So we're between -1.02 and -1.03.\nWe're nearly there - try -1.025. thats positive, so we're done! We know the answer is between -1.025 and -1.03, so it must be -1.03 to the nearest 100th.\nThats quite a bit of work, but its not too bad. There are faster ways to do it, but they're a bit more complicated.
484451
iso late the y by subtracting 6x from each side, so you are left with 2y = 9-6x, then divide each side by the coefficient of y (2) making your equation: y=9/2-3x...the y intercept of the line is 9/2 and the slope is -3 (so start at the point on the graph 9/2, then move 3 squares down, and over to the right one)
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
702364
LOGa(y/x) = LOGay-LOGax= 4.79-3.58 = 1.21
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
833077
In each, get y by itself.  I'll do #1 for you\n\nadd the absolute value part to both sides to get y = |9x-36| - 10\n\nAs long as x >= 4, the absolute value does not affect the outcome so the equation is simply y = 9x - 36 - 10 or y = 9x - 46.\n\nSo draw that graph starting with x = 4, then to the right.\n\nIf x<4 the absloute value of 9x-36 becomes its opposite which is -9x + 36.  So you have y = -9x + 36 - 10 or y = -9x + 26\n\nSo draw that graph starting at 4 and going left.  You should end up with a V-shaped graph with the point of the V at (4,-10)
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).
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.
827182
for a graph, go to http://www.calculator.com/calcs/GCalc.html\n\ntype in \n\ne^x\nclick enter\ne^(2x)\nclick enter\n\nand then seperately\n\nlog(x)\nclick enter\nlog(2x)\nclick enter\n\nby the way\n\ne^(2x) = (e^x)^2\n\nso in comparison to e^x, the problem was squared\n\nas for\n\nlog(2x) = log(2) + log(x)\n\nso in comparison to log(x)\n\nlog(2x) was moved to the the left log(2) units