i need to graph this equation for school and also what the domain and range of it are thanks
137257
sec is 1/cos. So the domain is the set of all real numbers except when cos(2x - pi) is 0. To graph it, first plot the cosine graph corresponding to -2cos(2x - pi). Then, "flip" the wave along y=2 and y=-2 for when the graph is positive and negative. Basically, what you get are sets of parabolas going off to infinity alternating between being above and below the x axes. Then, erase your cosine graph and shift these "parabolas" up by 3, and you're good.\nSince -2cos can take any value between -2 and 2, -2sec can take any value except those between -2 and 2. Shift up by 3, and you get a range of all real numbers less than or equal to 1 and greater than or equal to 4.
500894
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
483599
go here- http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html\n\nand enter     2x^2   in the big blank field and a graph will pop up!
470696
lol you need a graph for the first part!!\nand for the rest use:\ndistance formula: d= [squared root of (x1-x2)squared + (y1-y2)squared]\nmidpoint formula:  [(x1+x2)/2 , (y1+y2)/2]\n\nAnswers:\nd= squared root of 68\n----------------------------\nm= [(-4+4)/2, (-1+ (-3)/2)]\nm= (0,-2)
478559
x=-t/2\n2x=-t\n-2x=t\ny=t^2\ny=(-2x)^2\ny=4x^2\nThe graph is a parabola with vertex at (0,0), and in the x-interval [0,5], the end falls at (5, 100).
464008
-3x^2+24x-49\ndy/dx=-6x +24\n-6x+24 =0, so x=4 \n\nSo the max occurs when x=4. ( We know it is a max because the coefficient of x^2 is negative.)\nso the max is -3(4^2) +2*4-49 = -48 +8 -49= -89\nThis means that the range (possible values of y) is all the real numbers =< -89 . Because -89 is the max value. In interval notation this is - infinity <y<= -89.\n\nThe graph is a parabola which is increasing upward until it hits the max when x = 4 and then decreases forever\nThus f(x) is increasing over the interval - infinity <x<4  and it is decreasing over the interval 4<x< + infinity. At x =4, f(x) is neither increasing or decreasing.\n\nThe vertex is located at (4, -89)\n\nSince f(x) is never positive, it never crosses the x-axis and thus has no real roots. The two roots are imaginary.\n\nI think that covers all your questions.
138225
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
444200
f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]\n= 3(x - 2)/(x - 3) (x ≠2, -2)\n= [3(x - 3) + 3]/(x - 3)\n= 3 + 3/(x - 3)\n\n1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→2 [3 + 3/(x - 3)]  \n= 0 \nand \nLim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→-2 [3 + 3/(x - 3)] \n= 13/5\nSo it has 'holes' at (2, 0) and (-2, 13/5)\n\n2. It is undefined when x = 3, 2 and -2\nSo domain is all x except x = 3, 2, -2\nRange all y except y = 0 and 13/5\n\n3. As x increases and decreases without bound \n3/(x-3) approaches 0\nso 3 + 3/(x-3) approaches 3\n\nie lim x → ±∞ [3 + 3/(x-3)] = 3\n\n4. When x < 3\nx - 3 < 0 \nSo 3 + 3/(x-3) < 3\n\nWhen x > 3\nx - 3 > 0 \nSo 3 + 3/(x-3) > 3\n\nSo for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3\n\n5. When y = f(x) = 0\n3(x - 2)/(x - 3) = 0\nie x = 2\nSo f(x) cuts x-axis at x = 2\n\n6. When x = 0\nf(x) = 3*(-2)/(-3) = 2\nSo y-intercept is y = 2\n\n7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.\n\nThat should get you a good start I think
551890
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.
502940
Domain is the X coordinates, and range is the y coordinates.\n\nWhen a problem says "what is the domain?"\nThey want to know if for any x value you put in, if there will be a real answer.\n\nRange, yeah, thats just the Y values. :D
467735
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.
453239
Range would be the set of values for f(x)\nDomain would be the set of values for x\n\nSo they want you to take your equation...\nf(x) = 4x - 9\n\n...and set the value of f(x) to -5:\n-5 = 4x - 9\n\nNow you need to solve for x as follows:\n\nAdd 9 to both sides:\n4x = -5 + 9\n\nSimplify:\n4x = 4\n\nDivide both sides by 4:\nx = 1\n\nx = 1, f(x) = -5\n\nSo your ordered pair is (1, -5)
519965
Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
532160
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.
436418
It is probably best to determine the range after you have the graph.  You will need to look at the local maximums, local minimums, and the horizontal asymptotes to determine the range.  (In other words, it is not a fast problem).
352452
[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).
650821
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.