best fit line on a graph?
364185
The most important reason for the best-fit line being straight is that you can find the slope of the line very easily.  Depending on what you are graphing, the slope can be very significant.  For example, if you plot distance vs. time, the slope is the velocity.  If you plot velocity vs. time, the slope is the acceleration.  In general, the slope will tell you the rate at which the responding variable is changing, which is almost always important.  In fact, this is very often the reason for making the graph in the first place, and at the very least provides an important tool to analyse experimental data.
973670
http://www.amazon.com/s/ref=nb_ss_gw/103-4798383-9379069?url=search-alias%3Daps&field-keywords=Precalculus+with+Limits%3A+A+Graphing+Approach
72398
I found a nice and simple online grapher at http://www.walterzorn.com/grapher/grapher_e.htm
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.
164074
alright, finally a question i can answer, ok first of all u need some bivariate data (meaning data in the first two lists). exponential regression is going to find the equation of the line that best fits the data. on my calc there is linear regression (straight line), quadratic regression (parabola looking thing), cubic regression (cubic looking thing), quartic regression (weird looking parabola), and ooo i see what ur talking about. hmm i think its like finding an equation of the line then looks like an exponential equation. and example of this is graphing y=2^x. try that and u'll see what i mean
949382
whichever is closest, no particular pattern.\n\n/doc
150119
Have you tried the ezplot routine?  You can get help from the context help window.  Mathlab is the bomb.
248595
I was going to say graph paper to. Then you can keep a straight line. Very simple solution. If your talking about bigger graphs then it would be a good idea to sketch the drawings at the job and bring them back to a drawing table if thats what your looking to do.
466006
Pick a few easy points like pi/2, pi, 3pi/2, 2pi.  You know the period.  Then just graph it.
731311
I use GrafEq for my graphs, you could give it a try at http://www.peda.com/grafeq/
249763
by plotting the points, use a graphing calculator for best answers
709666
Use a graph paper. Vector normally given in the form of (i,j). Right now, assume your graph x-axis is your value of 'i' and y-axis is your value of 'j' plot them in and you will get it.\n\nRemember to size your scale as you do it in a normal mathematical way so that your whole vector line will be on 1 piece of graph paper.
979952
X=[1,2,3,4,5];\nY=[4,10.72,17.42,37,55.7];\n\np=polyfit(X,Y,2);\n\nxPlot=0:.1:6;\nyPlot=polyval(p,xPlot);\n\nplot(X,Y,'o',xPlot,yPlot)\naxis([ 0 6 -inf inf])\nlegend('Data','Second order fit')
911974
Go with a line graph, that way you can also find out (or estimate) temperatures at times when you didnt measure it.
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
1144569
for graphing, dee-dee-dee
354280
Take a look at this webpage:\n\nhttp://www.walterzorn.com/grapher/grapher_e.htm\n\nType: \nx min = 0\nx max = 20\ny min = 0\ny max = 10000\n\nFunction f(x):\n1000*exp(0.1x)\n\nThen click plot graph.\n\nVoila!