why is it not nescesary to plot points of this graph:  y = a (x - h) 2 + k.?
600230
Do you mean (x-h)^2? If so, yes, this complies with the general equation of a parabola y = ax^2+c. \nThe value of y is proportional to the square on x, that has to be a parabola.
697027
I ususlly think that if an equation such as y = -1 doesn't have an x in it, then it won't hit the x-axis.  The only way this could happen is if the line is horizontal, through -1 on the y-axis.  Similarly, since x=4 has no y in it, it can't hit the y-axis and so must be vertical.
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), ...
331692
As long as it is a function of one variable and you plot enough points, yes.  If it is a funciton of TWO variables, you will need to draw a surface in 3D, which is definitely harder, especially since I can't draw.  :P\n\nAnyway, as long as it is only a function of one variable, all you have to do is make sure to plot enough points, so that you get a good idea of what the graph is doing, and then fill in the holes.  A much better way to graph complicated functions involves calculus; using derivatives to find the points where slope is zero and stuff like that.  However, as long as you plot enough points and are careful to notice any spots where your function is not defined (like if it would take the natural log of a negative number or divide by zero or something), you can get a fair approximation.\n\nJust know that if it is really complicated, you need to make sure to plot a bunch of points, especially around the origin and such, where it is not clear exactly which "sub-function" inside your function is overpowering the other ones.\n\nWhen you get to calc you'll learn (one would hope) a nice way to approximate a bunch of continuous functions without having to guess at whether you plotted enough points.\n\nBy the way, if you have a graphing calculator, that is all it really does; it plots points.  It plots a whole bunch of points, sure, but it's still just plotting points.\n\nHope this helped.
771247
for these you need to find the x and y intercepts(the point at which the line crosses the x axis is the x-intercept, and and the point at which the line crosses the y axis is the y-intercept), and then connect the two points to form a line. (you do this for every line.)\n\nfor example: x-y=8. <<is in the Ax+By=C form, where A, B, and C are coefficients\n             to find the x intercept, you put C over A or C/A. in this case it is 8/1=8. That is your x intercept. for the y intercept you put C over B or C/B--in this case  it is 8/-1 or -8. You graph that on a coordinate plane. You connect the two points and get the line x-y=8.\n\n\nYou do the same thing for line 2x+y=7. Your x-intercept is 7/2 and your y-intercept is 7/1 or 7. When you connect those two points and graph that line, there will be a point where those two lines intercect. Find the coordinates of that point and that is your solution.  \n\n\ntry the next one on your own, and you'll get it!!! once you find the x- and y-intercepts it's a piece of cake. good luck!!! need any more help, email me.
881866
Convert the equation given in the form of y=mx+c and  "m" indicates the slope .In this particular problem the given equation is y+x=4 (or is it y+x-4=0 ?),therefore re-write it as y=- x+4.The co-efficient of x is 1.Therefore,the slope is 1,\nAs for drawing the graph ,in the equation y=-x+4 ,you take different values of x and calculte the corresponding values of y.Let us take the values of x as 1,2 and 3 respectively.The corresponding values of y would be 3,2 and 1.So plot three points on your graph paper with co-ordinates of (1,3),(2,2)and (3,1).[each point indicating the value of x and corresponding value of y found from our precvious calculation].Now,draw a straight line joing these three points and extend them on either sides.This is the graph of the equation
134706
I'd plot the points and figure out the slope (vertical change/horizontal change).  In this case, it would be -1 [2 / -2].  Then trace out the line to see at what point it crosses the Y axis.  In this case, it would be -2.  Then plus those numbers in the the equation "Y = (slope)X + (intersection of Y axis)"  So, if I'm not mistaken, the solution should be "Y = -X - 2".
293210
the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
274876
by inspection I can understand another point will be...\n\n(0, 5) because if you add 1 to X you must add 2 to Y
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..
729410
You can  find two different values for x that are easy to find 1/7 of, such as 7 and 14; see what y would be at those points [ -4, -3]; then plot the points in either order and connect the dots.  It is smart to also plot a third point to reduce the chance of error, because if one of the three is wrong, they won't line up.
530897
Plot and join these points : (x,y) = (-1,-5);(0, -4);(1,-3);(2,-2);(3,-1);(4,0);(5,1).
497181
You only need two points to determine a line.\nYou should use three just to make sure you don't make a mistake.\nPick two values of x and find the corresponding values for y.\nThis one is pretty simple.\nLet x = 0, then y = 0 (since y = x)\nLet x = 1, then y = 1\nLet x = -3, then y = -3.\nPlot these points on the graph and connect them with a straight line.
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.
976808
c=9\ns=28/15\nr=10\ny=-7/4\nfor the rest, i'd assume you're supposed to graph them need two points to graph lines so\n(-1,13)and(0,61,3)\n(0,-14)(14,0)\ndon't know what you mean by 7+1/4m...doesn't tell enough\n(0,5)(2.5,0)\n(0,5)(5/2,0)\n(1,22/6)(22/6,1)\n(0,7)(-21,0)
873365
I don't see any x  in the equations but I think the problem is that xmax< = xmin or Y max<=Y min  as a result the calulator could not calulate it that way try making Xmin < Xmax and Ymin < Ymax  this should solve the problem.
955476
i am assuming you want an x and y table.  just pick any x values, and plug them in to get a y value.  (go easy on yourself, pick easy values like 0 and 1)  make sure you throw in some negative numbers as well,  just to make sure you have got the main idea of graph.  if you are graphing this, just start at your vertex and then plot your points you just found-you have the right idea, just remember the number in the parentheses is OPPOSITE, so it would be to the RIGHT 3 down 7.