On A TI-86 Calculator, how do i set a range for the X(or Y) scale on a graph?
89290
TI-86? im guessing that its quite different that an 84 or 83plus.
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.
271763
on a graph...when it's 1/2 or 0.5 then graph the point half way between the two lines (between 0 and 1)...if it's 1/3 then split the box into three sections...etc etc.
1143408
After turning ON the calculator press on the GRAPH button.  Then press F1 on top left corner.  Enter you equation.  Press GRAPH button again and then press F5 located on top right corner and your calculator should be graphing it.  If you like to change to graphing mode from radians to degrees, press 2ND button (yellow button on the left side) and then MORE.  Move down using cursors until it says RADIAN/DEGREES.  You can choose either option and then press ENTER.  If you have any other question about the calculator, let me know.
372526
some calculators can do this\n\nmathematica software will do it
399118
y-int= (0,-5)\nx-int= (-1.2,0)
790619
you can use the normal calculator of windows for these conversions.\n\nOptions are only available in a scientific view of the calculator.
637834
this is paralell to the x axis. It`s 7
39072
you can get x lower, because it can be all real numbers.  However, your answer is correct, because if you put a negative number in for x, you get a positive answer.  So the range is y is greater than or equal to -3.
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
906710
It allways depends on the graph you are drawing, theres no setup, that would match all graphs, that u can make the Ti draw. I mostly use the zoom functions of this Ti, its way faster then to set up the view everytime.
816764
This is equivalent to y < -3.  Since y = -3 would be a HORIZONTAL (not vertical) line, you'd graph a dotted line going straight across at -3 on the y-axis.  Since you want the y-values that are LESS than that, you'd shade below that line.
1119524
well using the y = function, you may need to create a positive and negative graph\n\nfor a circle with a center of 0, 0:\n\nsquare root ( x^2 - r^2)  (replace r with your desired radius)\nand\n- square root( x^2 - r^2)
788027
The graph is a V.  From the 0,0 draw a line with slope = 1 to the right of the y axis and draw a line with slope = -1 to the right of the y axis.  For y<|x|.  All you have to do is shade in the area below the V.  If y>|X|, all you have to do is shade in the area above the V.  Less than, shade below; greater than, shade above.  That should answer all the questions you've submitted.\n\nAnd no, absolute values do not form parabolas.  Some do, not all.  Like y = sqrt[|x-1|].  This forms 2 graphs each going away from each other.\nNice going NerdyKid.  You're not a nerd, you're a dork.
997106
It is possible to numerically calculate integrals on an 83, but it takes the calculater some time because it literally adds up the area under the curve. You have to plot the function in the y =. Then press CALC (2nd TRACE), choose option 7, enter the lower limit, enter the upper limit. The TI-83 only does definate integrals.
809000
There are many ways to approach this question the easiest being if you have a graphing calculator just plug in a few points ie. (23,9200) (24, 9216)<look below for explanation> into stats and then use the stats operation quadreg and plug that formula into your y values and look at the table for the biggest y or you can look at the graph and use the maxium feature.\n\nif you don't have a graph calc then you need to grab a few points ... I'll use (23,400) (24, 384) (25, 368) (26,352) and (27, 336)... now multiply the x and y for each and then put that in the y value for each x.... ie. (23,9200) (24, 9216) (25, 9200) (26, 9152) (27, 9072)\n\nmake make a small t-chart (with 2 extra columns the first d and then d') with your first two as your new x then y points.... with the first extra column(d) take the difference of each y(for the first entry you will not be able to have an answer) ie. 16; -16; -48; -80; now in the second extra column(d') take the difference of the d column and multiply it by -1 ... -32, -32, -32\n\nnow you ready to start making your quadratic... take d'(32) and take it to the -2 power<I may be wrong here and it might be the -1 power, but you can just test and come back if wrong> then place that number in front of x^2 (I will refer to this number as a)\n\nok now take this formula ax^2 +bx +K(this is the constant but in this problem you have none) now plug in a value you have for x from your second points and solve for b now this is your quadratic, you should test it on a point or two, but use this to create a huge t-chart of different points and look for the when the value gets smaller, and the point before it gets smaller is your point\n\nwe actually ran into the answer while doing the problem which is common in math books but the point (24, 9216) or when you have 24 trees is the maximum yield
827182
for a graph, go to http://www.calculator.com/calcs/GCalc.html\n\ntype in \n\ne^x\nclick enter\ne^(2x)\nclick enter\n\nand then seperately\n\nlog(x)\nclick enter\nlog(2x)\nclick enter\n\nby the way\n\ne^(2x) = (e^x)^2\n\nso in comparison to e^x, the problem was squared\n\nas for\n\nlog(2x) = log(2) + log(x)\n\nso in comparison to log(x)\n\nlog(2x) was moved to the the left log(2) units