y+3<0, sketch the graph of the solution?
816764
This is equivalent to y < -3.  Since y = -3 would be a HORIZONTAL (not vertical) line, you'd graph a dotted line going straight across at -3 on the y-axis.  Since you want the y-values that are LESS than that, you'd shade below that line.
799272
is it asking you to graph that. otherwise it is not solvable. if it's asking you to graph it then you have to plug into your graphing calculator y=x-9
844444
Remember that the vertex of the parabola for y = x² is at (0,0).  You can confirm this by putting x = 0 in and seeing that you get y = 0.\n\nIf you instead replaced x with (x - 3):\ny = (x - 3)²\n\nNotice how it would take a value of x = 3, to get y = 0.  In other words, the whole graph, including the vertex would move 3 units to the right.  The new vertex would be at (0, 3).\n\nSimilarly, if you replaced y with (y - 2):\ny - 2 = x²\n\nThis is the same as:\ny = x² + 2\n\nSo every value would move up 2.  The graph would move two units up and the vertex would similarly move to (0, 2).\n\nCase 1:\nParabola moves 3 units to the right -- vertex now at (3,0)\n\nCase 2:\nParabola moves 2 units up -- vertex now at (0, 2)\n\nExtra credit question for you:\nWhat would happen if you did both?\n(y - 2) = (x - 3)²
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
582742
Graph the line 3x + 4y = 12. \nIt divides the plane into one section where 3x + 4y < 12 and one where 3x + 4y > 12\n\nCheck to see see which side of the line (0,0) is on. It satisfies the inequality,  so shade the half plane that contains (0, 0) it's in the solution set.\n\nGraph the line x + 3y = 6. Again, check (0, 0) it satisfies that inequality, so shade everything on that side of the second line.\nNow draw the lines x = 0 and y = 0 (y axis and x axis)\nThose are the boundrys of your solution.
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola
590565
y = mx + b\n\n0 = x + 3\n\ny can be anything, since it is not in the equation, and x will always be -3.
490656
Make Y=0\nyou have to have m and x though to solve b\nExample:\nY=mx+b\n0=3(4)+b\n0=12+b\n -12 on both sides of equal sign\n-12=b\nhope this helped!
986609
y=(x+3)^2=x^2+6x+9\ny'=2x+6=0\n2x=-6\nx=-3\ny(-3+3)^2=0\nso the vertex is (-3,0)
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.
809234
you would plot it on a graph at (0,5) \nor "y-int=5"
101438
Y=MX+B\nY=-X ------> Y=-1X+0 \nSLOPE IS -1 AND THE Y-INTERCEPT IS 0
770341
You need to get y by itself.\n\nadd -3x to both sides so you get 2y> 3x+4\nnow divide 2 --> y>3/2x+2\n\ngraph that as you would if it were y=3/2x+2\n(starting at 2 on the y axis. then move up 3 points, and 2 to the right.)\n\nMake sure your line is DOTTED. It is only solid if the inequalilty is less/greater than or equal to.\n\nto check which side is to be shaded, plot points on both sides. try (0,0) if the line doesn't cross through that.\n\n:)\n\nIf the answer is correct, shade the region where (0,0) or the coordinates that work are.
374950
Plug in x, solve for y as follows:\nIf x=0 then solve for y and get y= 3/2\nif x=10 y= -10/2+ 3/2 = -7/2\nCreate a graph using the horizontal as the x scale and the vertical as y.  Plot the above and you will see that it's a straight line.
241669
Y=x^3(x+2)(3x-1)\n we have three possibility for x\nx^3=0 than X=0\nx+2=0 than X=-2\n3x-1=0 than x=1/3\n\nso the answer is a
223761
if your equation is y=x^2, then the graph that you draw is going to be a parabola. You can see this by having a table first. The table should consist of 2 values...x and y. So you say when \nx=-3, y=9,\nx=-2, y=4,\nx=-1, y=1\nx=0,  y=0\nx=1,  y=1\nx=2,  y=4\nx=3,  y=9\nso now if you have a square line paper, just plot it and you will see that it will give you a parabola. Another thing is that when you draw it, first just plot the points and then when you connect the point connect them using curves and not a line,...that's why u get a u shaped graph and not just a combo of lines.
338193
The equation represents a straight line graph whose gradient is 3 , and intercepts the the ordinate axis(i.e y-axis) at the -2 position.\n\nthe easiest way to arrive at this is to draw a table of values. Such that you find the values of y, using values of x from -6 to +6.\n\nYour graph will confirm my story, \nBut from mere looking at the equation I can make a good prognosis, when i compare it with Y=mx + c .where Y represents the value of the equation, X represents the value of x which made y what it is.\nM stands for the gradient of the graph, while c stands for the intercept on the Y-axis.