Sketch the curve by finding the domain, x- and y-intercepts, symmetry, intervals of increase and decrease,?
918456
domain\nall real x\nx intercept\nputting y=0\n0=x+rt{x]\n-x=rtx\nx^2=x\nx(x-1)=0\nx=0 or 1\nx intercept=(0,0) and (1,0)\ny intercept\nsetting x=0\ny=0\ny intercept=(0,0)\nno symmetry exhibited
818186
.  The y intercept is simply where the curve intersects the y axis, which means x=0.  Just plug in zero for x and find what the function is equal to.\n.  The local maxima and minima occur where the first derivative (the slope of the curve) is equal to zero.\n.  Points of inflection are where the rate of change in slope (the second derivative) goes from positive to negative or vice versa, so take the second derivative.\n.  The value of the function is increasing with x where the slope, (the first derivative) is positive.  The function is decreasing where the slope and first derivative are negative.\n.  The curve is concave up when the slope is increasing with x, and concave down where the slope is decreasing; so define the intervals over which the second derivative is positive or negative, and you will have define the concave up and concave down regions.
180320
Draw any line passing through origin. Its equation is y = mx and the slope ‘m’ is y/x.\n\nNow move the line up (or down) through a distance‘d’ with out changing the slope. The line cuts the y-axis at a distance d from the origin. This is y-intercept.\n\nThe x-coordinates of all the points on the line remain the same whereas all y-coordinates increase (or decrease) by‘d’. But the slope remains the same and x coordinates also remain the same. Therefore we give a correction to y coordinate to keep m constant.\n \nDenoting +d or -d by ‘b’, the slope of the new line’ ‘m’ is (y- b) /x. \n\nThus the equation of any line in slope intercept form is \n\nm =(y - b)/ x. or y = mx + b.
25053
i don't understand your question too well because it isn't a question but i answer with\nslope intercept form(replace y with F(x)\nmakes it in function notation\nY=mx+b\nm is the slope rise over run\nx is the variable that you can plug in\nb is where the line crosses the y axis\n\nstart with a point on the y axis that is the the y intercept(b)\n\nfrom there do a slope\nexample: y=1/4x+2\n\n\n+++++++++!++++++++++ \n+++++++++!++++++++++ \n+++++++++!+++x++++++ \n+++++++++x++++++++++ start with a point at 2y\n+++++x+++!++++++++++ \n-----------------+ ------------- then move up one and right 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and a dot down one and left 4\n+++++++++!++++++++++ \n+++++++++!++++++++++ and connect the points, then you would have graphed a segment of the line and just extend that and you solve the problem for your graph
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
697951
find out what happens to Y (what it approaches) as:\nx goes to infinity\nx goes to negative infinity\nthen find out what happens to X as:\ny goes to infinity\ny goes to negative infinity\n\nalso find the intercepts (if they exist)
468700
I'll calculate these, but I'll apologize now to those others who might already be doing this if I post my replies after yours.....\n\n1. x-intercept:2,y-intercept 1\nx-int = 2 at point (2, 0) = (x1, y1)\ny-int = 1 at point (0, 1) = (x2, y2)\n\nslope (m) = (y2-y1)/(x2-x1)\nm = (1-0)/(0-2)\nm = -1/2\n\ny = mx + b (now, b is the y-intercept)\ny = (-1/2)x + 1\n\ny = (-1/2)x + 1 (substitute other point for verification)\n0 = (-1/2)*(2) + 1\n0 = 0\n\ny = (-1/2)x + 1\n\n------------------------------------------------------------\n2. x-intercept:5, y intercept 5\nx-int = 5 at point (5, 0) = (x1, y1)\ny-int = 5 at point (0, 5) = (x2, y2)\n\nslope (m) = (y2-y1)/(x2-x1)\nm = (5-0)/(0-5)\nm = -1\n\ny = mx + b (now, b is the y-intercept)\ny = -x + 5\n\ny = -x + 5 (substitute other point for verification)\n0 = (-1)*(5) + 5\n0 = 0\n\ny = -x + 5\n\n------------------------------------------------------------\n3. x-intercept:-4,y-intercept -1 \nx-int = -4 at point (-4, 0) = (x1, y1)\ny-int = -1 at point (0, -1) = (x2, y2)\n\nslope (m) = (y2-y1)/(x2-x1)\nm = (-1-0)/(0-(-4))\nm = -1/4\n\ny = mx + b (now, b is the y-intercept)\ny = (-1/4)x - 1\n\ny = (-1/4)x - 1 (substitute other point for verification)\n0 = (-1/4)*(-4) - 1\n0 = 0\n\ny = (-1/4)x - 1
333191
Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
101438
Y=MX+B\nY=-X ------> Y=-1X+0 \nSLOPE IS -1 AND THE Y-INTERCEPT IS 0
531033
Draw it your self on x-y plane, then determine the coordinates
637800
Try this website www.algebrahelp.com/calculators/equation/3dgraphing/pg2.htm
608711
i think it's a parabola.. im not sure. u should try plotting it on a graph.. substitute  the x and y values and see what form comes out..
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
1095247
For a fixed a and n, and allowing b and c to vary, the curve would be a hyperbolar centered at the origin. You will only be taking the rational points on that curve, though.
793059
give values of -3,-2,-1,0,1,2,3 to x and find the corresponding values of y\nyou get the ordered pairs\nnow plot
746165
Remember the fomula...\n\ny = mx + b\n\nm = Slope\n\nb = Y-intercept\n\nx and y = points\n\nSo plug everything in...\n\n1 = 4(2) + b\nMulitply\n1 = 8 + b\nSubtract 8 on each side\n-7 = b\n\ny = 4x - 7