How to change y=cx^d into a form for straight line graph?
960996
log y = d log (cx)\n\nlogy = d log c + d logx\n\nlet logx = X\nlog y = Y\n\nand d logc = C\n\nthe equation becomes\n\nY=dX+C\n\nthis is of the form y=mx+c which is a straight line
976124
dy/dx = (dy/dt)/(dx/dt)\ndy/dt = 1/(2√t)\ndx/dt = 1\n\ndy/dx = 1/(2√t)\nx=t\ny=√t\n\nSubstituting t=1/4:\n\ndy/dx = 1\nx = 1/4\ny = 1/2\n\nUsing point-slope form:\n\ny-1/2 = 1(x-1/4)\n\nSimplifying:\n\ny=x+1/4
796114
Graph equations are of the form y=mx+c\n\nm is the slope or gradient\nc is the y intercept.\n\nSo, for y=-x+1, \nm= -1\nc = 1\n\nAnd for 2y=3x, you have to rearrange it to make it look like y=mx+c.\n\nSo divide by 2.\n\ny=3x/2\n\nm= 3/2 \nc =0\n\nHope you understood my explanation :)
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
445014
(a) just substitute:  C = 10X + 40, where X = number of units produced.\n\n(b) It's a straight line with horizontal axis X and vertical axis C.  So, plot (0,40) and say, (5,90) and draw a straight line.\n\n(c) It's a straight line also thru the origin with a slope of 50.  Here, the vertical axis is R but it's the same scale as the C.  The line will go thru the points (0,0) and (1,50).\n\n(d) On the graph, it will be where the two lines intersect (1,50) or set the two equations equal to each other and solve:\n  \n     50R = 10X = 40  and 40R = 40  so X = 1
660602
d/dx[y/x^2] gives y'/x^2 - 2y/x^3 on expansion
891649
(y-3x) cannot be graphed because there is no equality...\nyou could graph y=3x... that would be a line that passes through (0,0) with a slope of 3
1112986
linear: graph is a straight line that is not vertical.  table shows values steadily increasing or decreasing.\n\nquadratic: graph looks like a "U" - either up (v) or down (^) and either wide opening or small opening.  table shows values decreasing/increasing rapidly, then slowly, then the opposite direction, such as:\nx:  1  2 3 4 5 6 7 8  9\ny: 16 8 4 2 1 2 4 8 16\n\nexponential: similar to the quadratic, but it increases more rapidly, has y=1 when x=0, and does not "go the opposite direction" like the quadratic does.  it starts out very little, like a fraction, then shoots up and doesn't come back.  the table is like that: starts out little, increases a little, increases more and more rapidly, then its too big.\n\nabsolute value: looks like a "V", not like a quadratic that looks like a "U".  absolute values have an indifferentiable point, or in other words the bottom tip of the "V".  it's like a linear equation that is positive on one side where it should be negative.  the table is like a linear, but always positive where it should be negative.\n\nsquare root: looks like half a "U" turned 90 degrees clockwise on its side and only positive.  starts along the x-axis with y=0 at some x, then increases rapidly for a short time, then increases very slowly afterwards.\nx: 1  2  3   4    5    6      7     8\ny: 0 .5 .9 1.2 1.4 1.5  1.55 1.57\n\nnote that the sample tables are not representing real functions, they just graph a general idea of what it should look like.
1175462
Graph the equations. They should look like a big "U". The axis of symmetry is an imaginary line going right through the vertex and splits the graph in half. The point at where the axis crosses the x axis is what the axis of symmetry is. \n\nPlugging in is easy. make a chart for x. Now choose like 4 different easy values to put into the equation as x. 0,1,2,3. Use your calculator to find the answers, and make a separate side of the chart as y. the answers you get are your y's. Then you graph it. \n\nI don't have a calculator right now, so I'm leaving that to you.
442406
Make a table of values.  Then plot the points.  Your graph should be a "V" shape (not a parabola which is U shaped)---whether upright, sideways, or downwards.
653148
All equations for a line will have the form:\n\ny = mx + b,\n\nwhere x & y are coordinates, m is your slope, and b is your y-intercept.\n\nTherefore, substituting in your slope and y intercept, you get:\n\ny = -2/3x + 4\n\nThis equation can then be used to graph the line, since given any x, the corresponding y can be found. :)
707539
x = (y^2 + 2y -1) / 2\ndx=ydy+dy\n1=(y + 1 )dy/dx\ndy/dx=1/(y+1)\nd2y/dx2 = -1 / (y+1)^2
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
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
737893
Take your quadratic equation.\n\nThe graph cuts the x axis when y = 0\n\nSo simply put your equation = 0, and then solve it either by using factorising or the quadratic formula:\n\n(-b plus or minus sqrt(b^2 +4ac)/2a)\n\nif I remember right, (could be minus 4ac)
432398
y = -1/2 sin·­­­¹ pi x\nTo graph this, it is easier to rearrange to make x the subject, then turn the graph paper on its side and plot x as a function of y, just remember that y is now positive going left and x is positive going up.\n-2y = sin·­­­¹ pi x\nsin (-2y) = pi x\n1/pi sin (-2y) = x\n\nnow sin (-2y) = sin 2y   (check your trig properties)\n\nso we want to graph\nx = 1/pi sin 2y\n\nnow for the graph  y = a sin bx\nthe amplitude is a and the period is 2pi/b\nso the amplitude is 1/pi\nthe period is 2\n\nPlot your sin function along the y axis. Then turn the paper the right way up. Remember that to be a function, it can only have one value of y for each value of x, you only want a single half S shape, going from y = -.5 to .5, this way the graphs remains a 1 to 1 function.
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)