How do you graph a line given its equation?  What does the line represent?  What is different about graphing a
117178
Well, your third question is incomplete..\n\nBut I can answer the first two:\n\nIf you have an equation, assuming we're on cartesian coordinates, the simplest thing to do is to solve the Y variable to a couple or more X variables.\nLike:\nY=3x + 1\n\nx=0, Y= 3(0) + 1 = 0 + 1 = 1    (0,1)\nx=5, Y= 3(5) + 1 = 15 + 1 = 16  (5,16)\nand so on...\n\nThen you plot the coordinates on the graph and unite the resulting dots with a curve. In my example it should yeild a straight line.\n\nGraphs help describe trends on phenomenons that are tied to the equations.
807533
Draw the line y = 1; this is a horizontal line through y = 1.\nThink about all the points below this line (since you have y < 1).  \n\nDraw the line y = 2x - 3; this has a y-intercept of -3 and a slope of 2, so it goes through (0, -3) and goes upward to (1, -1), (2, 1), etc.)\nThink about all the points above this line (since you have y > 2x - 3).\n\nBecause the question says *and* you should only consider the points that are in both regions (below y = 1 and above y = 2x - 3).  Shade in the intersection of these two regions.\n\nI've attached a graph of what it should look like...
771447
It is an equation, the result of the equation is the function. The result may be plotted and it would be a linear function. The plot is a function of the form (equation). It is linear only if it works out to a stright line, which this does.\n\n"A linear function of x is a function of the form y = ax + b, where a and b are constants.  It's called a "linear" function because a plot of the x and y coordinates for which this equation is true takes the shape of a straight line"
935919
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
869140
Just graph it like an equality, and then your answer is everything either above (>) or below (<) the line.  For example, in the last inequality, the answer is everything above the line y = -2, not including the line itself.
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.
203791
The x-intercept is when the line crosses the x axis. When your equation's graph crosses the x axis, it means that y is zero at that point. right?\n\nSo to find the x-intercept you assume y=0 and solve for x.\n\nNow to find the y-intercept, it is when x=0. If you imagine it graphically, when it touches the Y-axis, X's value is zero.\n\nSo to find the y-intercept you assume x=0 and solve for y.\n\nExample: y= 3x-6\nx-interp: assume y=0\n      0= 3x -6      \n 3x = 6     \n x = 2\n\ny-interp: assume x=0 \n     y= 3*0 -6        \n          y= -6
620528
There are two equations there, not one.\n\nFor the first equation, plug the numbers on your X-axis into "x", calculate "y", then draw the points on your graph.  I'll get you started:\n\nx=0, y = -3\nx=1, y = -1\nx=2, y = 1\netc.\n\nThe second equation simply means that x = 1.5 and y = 3.  This is just a single dot on a graph.
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The equation of a line is y=mx+c\nHere when x = 6, m = 4/3, y = -2\nTherefore c = y-mx = -2 - (4/3)*6 = -10\n\nHence y = (4/3)x-10\n\nHope this helps!
921939
Good Evening\n\nI think we've been avoiding this question because we can't graph here... however to get you started, create an x|y chart for each line, plug in a couple of values, and connect the dots  (for a line you only need two points, but some books recommend finding three, so you can verify you didn't make a mistake somewhere)\n\nx+y = 8\nx|y\n----\n4|4\n3|5\n2|6\n\nx - y = -4\nx|y\n0|4\n1|5\n2|6\n\nGraph the two lines, and they should intersect at (2,6) since it appears in both of the charts...\n\n~i~
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No, you must use algebra to plot lines. How else would you explain the variables x and y?\n\nSlope-intercept form looks like this:\n\ny = mx+b\n\nOr, in the case of vertical lines:\n\nx = my+b\n\nWhere m is the slope and b is the x- or y- intercept, depending on which one you are using.
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The slope of a line is defined as:\n\nm = (change in y) / (change in x)\n\nor:\n\nm = (y2 - y1) / (x2 - x1)\n\nGiven the points (-3, 3) and (0,-3) means:\n\nx1 = -3\nx2 = 0\ny1 = 3\ny2 = -3\n\nPlugging these numbers into the equation for m yields:\n\nm = (-3 - 3) / (0 - -3)\n\nm = (-3 - 3) / (0 + 3)\n\nm = (-6) / (3)\n\nm = -2\n\nNote that a slope of 0 means that there is no change in y, or that y1 = y2
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The form y = mx + b is called the "slope-intercept" form of an equation describing a straight line.\n\nIn this form, m is the slope (how much the y-value changes for each unit change of the x value) and b is the point along the y-axis where the line crosses it (where x-value is zero).\n\nSo, as m increases, the line bocomes closer to vertical and when m decreases, the line becomes closer to horizontal.\n\nSimilarly, when b increases the line's slope remains the same but the line moves up on the graph, while if b decreases then the line moves down.\n\nHere is a great website for playing with this form of a linear equation in slope-intercept form:
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Before graphing, treat the inequality as an equal sign, and put the equation in a form you know how to graph.  The easiest is probably the slope intercept form, y = mx + b, where m is the slope and b is the y intercept.\nGraph the line that you see, but because it is an inequality, graph it as a dashed line. If the equation is greater than or equal to, or less than or equal to then graph it as a solid line.\nPick a point that is not on the line and plug it into your equation. If the line doesn't cross the origin, that would probably be a good point to choose.  Plug in the point you have chosen, and if the inequality holds true for that point, shade in the direction of the point from the line.
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From a table -- see which y-value corresponds to x=0\nFrom a graph -- see where the graph crosses the y-axis\nFrom an equation -- get the equation in y=mx+b form; the y-interecept is b
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If you have an equation that is Y=__X-__ or Y=__X+__, then the second space is the Y-intercept and the first space is the slope.\n\nSo, for your equation, Y=3X-5, 5 is the Y-intercept, and 3 is the slope.  The Y-intercept is where the line crosses the Y axis, which is 5.  So, on the Y axis (which is X=0), count up 5 spots and that is your intercept --> (0,5).\n\nThen you have to use the slope, which remember is the number in the space before the X --> 3.  3 is the same as the fraction 3/1, and in the fraction of a slope, the top (numerator) is how many spaces up or down you move from the original point, and the bottom number is how many spaces you move left or right from the original point.  So, your first point being (0,5), move up 3 and to the right 1 (up and right because both 3 and 1 are positive).  You should be at (3,6).  \n\nMaking an input/output chart makes plotting your graph a lot easier.  Make a chart with two rows, the top row being X and the bottom row being Y.  To the left of your chart, label X and label Y.  (So that you remember your slope, I would write X --> 3, Y --> 1).  Put your intercept point (0,5) in the chart: 0 to the right of X and 5 to the right of Y.  Now you can see that you just add the slope!  0+3=3 and 5+1=6.  So, your next point is (3,6)!  From there, keep going to the right, adding as you go.  Add another 3 to 3 and another 1 to 6.\n\nGood luck!
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.