how do you find the vertex and then graph it?
148902
its the middle of a angle like if da two lines combine the pointwere they combine is da vertex..........and you cant graph vertexs
1025113
An instructor always wants to know the following about a graph:\n\nThe x-intercepts, the y-intercepts, and for parabolas, where is the vertex.\n\nSet x=0 to find the y intercept.  Hmm (0,0)\n\nSet y=0 and solve to get the x-intercepts:\nIt factors into x(4-x) so, y=0 when x=0 and 4.\n\nYou can always get the vertex by using x= -b/(2a) or, if there are x-intercepts, it will always be halfway in between them.  In this case, the vertex is at x=2.  Plug that in to get y=4.\n\nSo, the important info is:  It opens downward.  The vertext is at (2,4).  The x-intercepts are at (0,0) and (4,0) and the y-intercept is at (0,0).
979117
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
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
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.
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...
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~
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).
861221
looks like the vertex is (-5, 3)
325867
First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
373083
Well, I'm not going to answer the question for you, but I'll tell you how to do it.\n\nWhen you're graphing the function, you set up a table with f(x) and x as the two columns, then solve for x for any set of numbers, like -3, -2, -1, 0, 1, 2, 3, and that's probably more than enough. Graph those on a piece of graph paper (f(x)=y, and then x). \n\nNow, to find the INVERSE, you simple switch the X and the Y. Instead of solving for x, you solve for f(x). So, your problem would be written x= -2/3y + 5. An easier way to do it, take the table you made before, and switch the columns, so that the solutions for x are underneath f(x) and the numbers you picked for f(x) are underneath the x column. Graph those coordinates. When you look at the graph, the inverse should be a mirror reflection of the first one, with  the graph of y=x as the line of symmetry. \n\nThat's how you find the inverse of any problem. Sorry if anything's unclear, if you need a better visual, I'm sure you could google up this exact method.
27352
the x coordinate of the axis of symmetry is given by -b/2a. here it is -(-12)/-6 = -2. plug this back into the equation and you get f(x) = 27 so your vertex is (2,27)
919538
it is narrow because a is more than 1.\nThe vertex=1,-1 you take the opposite of h which=x and y=-1\ndo a t table to find the max and min \nput the vertex in and put points for x and solve for y for each one using the equation
796882
well, you can pick an X and plot a Y (meaning, take values close to the Origin for X and put them into the equation (and inequality), and see what Y you get for each of them\n\nfor instance, if you pick 1 for X, and put it into the equation, you get y = 1. . so now you have one pair (one point you can already plot on your graph, the point (1, 1).\n\nNow pick 0, and you get a Y of 0, so you have the point ( 0, 0), so now you have two points.  \n\nnow you need to see what it does in negative territory.\n\nPick some negatives (like -1 ) and you get 1, so you know that the point ( -1, 1) is on the graph. . \n\nif you pick positive 2 for x, you get 2 for y. . so ( 2, 2) is on the graph. . \n\nif you pick -2 for X, you get positive 2 for Y, so ( -2, 2) is on the graph.\n\nnow you can plot the graph of that line. . it's two lines, perpendicular to each other that meet (and change direction) at the Origin (the point (0, 0).\n\nAs for the inequality, just graph it as though it is an equals sign instead of an inequality sign, and then then you know where the line is, and you find out which points are on which side of that line by picking X's and finding what Y value they give you (when you put them into the equation (or into the inequality).\n\nhope this helps, \n\ngood luck and God bless you.
531033
Draw it your self on x-y plane, then determine the coordinates
772016
x1 >= .2(x1 + x2)\nx1 >= .2x1 + .2x2\nSubtract .2x1 from each side\n.8x1 >= .2x2\nDivide both sides by .2\n4x1 >= x2\nx2 <= 4x1\nNow, assuming x1 is on the horizontal axis (x) and x2 is the vertical axis (y), then\n1) Place a dot at (0,0).  This is the y-intercept.\n2) The slope is 4.  Starting at the point (0,0), move up 4 and to the right by 1.  Place a dot there.\n3) Connect the two points with a straight line.\n4) The graph will be shaded below the line.\nI found this by picking any point that is not on the line.\nFor instance, the point (1,0) is not on the line.\n(1,0) means x1 = 1 and x2 = 0\nx2 <= 4x1\n0 <= 4  Since this is true, you shade the side of the line where this point lies.
350103
Here's what you do:\n\nYour equation is y= 3/2x^2 - 6x + 5\n\nNotice that:\n\nWhen x = 1, y = 1/2\nWhen x = 2, y = 6 - 12 + 5 = -1\nWhen x = 3, y = 27/2 - 13 = 1/2\n\nSo, you can figure out that the axis of symmetry is x = 2 and you can see that the vertex is a minimum, right?