The calculator is only graphing a half circle when it should be a full circle, specifically not in the negative y-values of the plane.

How many points do you want to use?

No. The graph of a circle is a graph, one which can have two different values for certain arguments. The graph of the top or bottom half of a circle can be considered a graph of a function.

The solution set of a linear inequality in two variables (x,y) is one half of the Cartesian plane. The half-plane may or may not contain the boundary line defined by the = sign. Thus it is expected that the calculator shades the relevant half-plane. If you are dealing with nonlinear inequalities the calculator shades inside or outside the curve.
When drawing multiple inequalities the Casio offers you the possibility to shade the regions where ALL conditions (ALL inequalities) are simultaneously satisfied. The other option is OR.
If you are not interested in the shading do not draw inequalities, draw equations (with =).

Check the Theta variable. Give theta a domain wide enough to cover the upper half-plane as well as the lower half-plane. I cannot tell you what limits to use because I do not know the function you are trying to plot.

The maximum value of the sine function is positive one, the minimum value is negative one. Within any limited range the maximum and minimum will fall between the two extremes.
If you have a graphing calculator, the easiest way to to find the maximum and minimum values of a sine curve is to plot it. Some calculators will even automatically find the max and min values. Others you'll have to eyeball it.
If you don't have a graphing calculator, you can calculate the value at a series of points and hone in the max and min values.
If you need further assistance, please reply to this post and specify the make and model of your calculator.

Pi is a mathematical constant whose value is the ratio of any Euclidean plane circle's circumference to its diameter; this is the same value as the ratio of a circle's area to the square of its radius. It is approximately equal to 3.14159265 in the usual decimal notation.

Use the DRAW program.
Press [2nd][PRGM] (DRAW)
Select [9:Circle (]
Command echoes on main screen
Complete the command with coordinates of the center and the radiius
Close parenthesis and press [ENTER]
The circle is displayed.

Ex: Circle(0,3,2) draw a circle with cntere at (0,3) and radius equal to 2.

Alternatively: If the conics application is on the calculator run it.
Press [APPS]
Select [Conics]
Select [1:CIRCLE]
Choose the canonical form (X-H)^2 + (Y-K)^2=R^2 where H, and K are the coordinates of the center and R is the radius.
You will be prompted to enter the values for H,K, and R

You can also use the general equation for a circle aX^2 +aY^2+BX+CY +D=0
Enter the values of a, B, C and D.Change the values to get what you want.

That is normal behavior. The log functions are not defined if the argument (x value ) is negative.

Hello,
I take your word for it. If graph is out of range you should zoom out.
But first try [ZOOM][6:Ztandard]. If you still see nothing, press [WNIDOW] and set XMIN=-100, XMAX=100, YMIN=-100, YMAX=100.
Otherwise you must know a few things about the behavior of the function you try to hraph
1. If X is always positif, set XMIN=0
2.If X is always negative, set XMAX=0
3. If Y has only positive values, set YMIN=0
4. If Y has only negative values, set YMAX=0

You can also calculate Y for X=0. Let Y(0) be that value. Then you try to have the Y-Values centerd around it: YMIN=Y(0)-10 and YMAX=Y(0)+10.

Hope it helps.

Hello,
Here how you do it taking as exemples y1=x^3 -5 and y2=abs(x^3-5)
Turn calculator ON by pressing [AC/On] . Use arrows to select [Graph] icon and press [EXE]
On line Y1= enter X[^]3 -5 [EXE]
On line Y2= press [OPTN] [F5:NUM][F1:abs] [(] X [^]3 -5 [)] [EXE]
The graphs will be drawn one after the other. You might have to adjust the window limits.
Here is an ouput drawn on another calculator.

For negative values of the x-variable, Y1 has negative values, while in the same range, Y2 has positive values, equal and opposite to those of Y1. In that range, Y1 and Y2 are symetric of each other with respect to the X axis.
For positive values of the x-variable, Y1 and Y2 are identical.

Hope it helps