how do you graph linear functions?
542210
Solve for y: y=3x\n\nThe slope is 3, and the y-intercept is (0,0). (y=mx+b: m is slope, b is y-intercept)\n\nThe slope is the change of y over the change of x.\n\nStart from (0,0) and go up 3 units, right 1 unit.  Connect the points and continue the line.
1153957
The website below seems to be what you're looking for.  The spreadsheet you can download uses the NORMDIST function to do the graph.  Check out the help feature for the function to see its equation.\n\nI assume you know how to calculate the mean and standard deviation for your data.  The AVERAGE and STDEV Excel functions are good for this.
985180
Pick 2 numbers for X, then plug them into your equation to get F(X), which is the Y value. Then you have the equations and put them onto your graph.\n\nTo plot the function, put your Y-intercept in (-3) then from there use the slope (-1/2) to plot other points. On the function, if you did it right you should come across your points that you made.\n\nHope this helps. Good Luck!!
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
989452
The parent function is y = 6^x. That function contains the point (0,1). Since f(x) = 6^(x+1) is a translation of 6^x one unit to the left, the point (0,1) maps to (-1,1).  \n\nThe graph of f(x) = 6^(x+1) looks like the graph of 6^x translated one unit to the left.
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
333191
Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
599765
When graphed, a linear equation will form a straight line while a non linear equation will not be a straight line
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?
148719
You just draw a graph yourself.\n\n! ain't gonna graph for you. Rather ! love calculus.
25053
i don't understand your question too well because it isn't a question but i answer with\nslope intercept form(replace y with F(x)\nmakes it in function notation\nY=mx+b\nm is the slope rise over run\nx is the variable that you can plug in\nb is where the line crosses the y axis\n\nstart with a point on the y axis that is the the y intercept(b)\n\nfrom there do a slope\nexample: y=1/4x+2\n\n\n+++++++++!++++++++++ \n+++++++++!++++++++++ \n+++++++++!+++x++++++ \n+++++++++x++++++++++ start with a point at 2y\n+++++x+++!++++++++++ \n-----------------+ ------------- then move up one and right 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and a dot down one and left 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and connect the points, then you would have graphed a segment of the line and just extend that and you solve the problem for your graph
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.
982059
In column A, put values of x.  I usually put something like -10, then make each cell below it be -9, -8, etc. up to perhaps 10.  The values you pick will probably depend on your function.\n\nIn column B, put your y values.  You can make this a formula based on column A.  So if the equation was y = x^2, you would write =A1^2.  Copy this formula down the rest of the column.\n\nNow select those two columns and click the graph wizard.  Pick an XY (Scatter) plot, and pick one with connecting lines.  It should give you what the graph will look like.
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.
763572
Hi!\n\nSo, what you are looking at is a linear equation.  You can tell this because there is only one instance of the variable x, and it is not raised to any power.  So, the graph will take the form of a line.\n\nAll linear equations take the form f(x) = mx + b\n\nWhere m is typically called the "slope" and b is called the "y-intercept".   The y-intercept is the point on the graph where the function intercepts the y-axis (the one that goes from top to bottom on the graph).  So, right away you can tell that the y-intercept in your equation is -5 (negative 5).  What this means is that the graph intercepts the y-axis at the -5, or in other words, there is a point on the graph at the location (0,-5).  You can put a dot there.\n\nThe next thing you need to see is the slope.  In you equation, the slope is -2 (negative 2).   Slope really means "rise over the run", or in other words, how much the graph rises for every point it goes to the right.  A slope of negative 2 means that for every increase in x of one unit, the y value decreases by negative 2.   \n\nSo, starting at the point (0,-5) which we determined above, go down 2 units and to the right one unit (putting you at the point (1,-7).  That's another point.  You can do it again and find the next point which is (2,-9) and put another point.  You now can put your ruler on the graph and draw the rest of it.\n\nIt's good to understand how this type of equation works.  You could have also made a table of x values and y values, and simply computed a series of y values for a series of x values (that you picked).  Then you could have drawn the graph also.  But understanding this type of equation will help you in the future.
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).
825543
ok , \nyou need to plot a graph of the equation 2x^2-4x-3 .\nto b more clear: 2*x^2 - 4*x - 3\n\nFor that u need to take the horizontal axis as x = 0,1,2,3,...\nand vertical axis (y axis ) as 2x^2-4x-3\n\nFor every value of x obtain a value for y,\neg.\nfor x = 0 ; 2x^2-4x-3 = 2 * 0^2 - 4*0 - 3 = -3\nfor x = 1 ; 2x^2-4x-3 = 2 * 1^2 - 4*1 - 3 = 2 - 4 - 3 = -5\nfor x = 2 ; 2x^2-4x-3 = 2 * 2^2 - 4*2 - 3 = 8 - 8 - 3 = -3\n.\n.\n.\n.\nand so on for as many points you need to plot\n\nnow with values of y as -3,-5,-3 corresponding to x = 0,1,2 plot appropriate points and join them to obtain the graph.\n\nto plot a point , take the first point (0,-3) thats for x = 0 , y = -3,\nand so on (1,-2) & (2,-3).\n\nI hope this helps :)