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I tried to find the zero of an equation and I put (x-8)^2 into my y= and hit graph.i went to second trace set my bounds, but i got the err no sign change. btw, my window reads;
Xmin= -10
Xmax= 20
Xscl= 1
Ymin= 0
Ymax= 100
Yscl= 1
Xres= 1
When you get the sign change error it is either that you have mixed up the left bound and the right bound or that you are marking the left and the right bound both above the x-axis or both below the x-axis. That's why you get the sign change error. Your range values must have different signs.
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Error 07 Syntax
Means you made a typing error while entering the equation. Using the minus sign instead of the negation sign (change sign) (-), can produce a syntax error. Maybe you need to explicitly type the multiplication sign
y= (-) 1.5*x +3
Try entering it as 3 - 1.5*x
For future reference, some variable referenced in your equation has been archived. You can archive/unarchive variables and other objects in the memory menu available by pressing 2nd [MEM].
INVALID DIM error message may occur if you are trying to graph a function that does not involve the stat plot features. The error can be corrected by turning off the stat plots. To turn the stat plots off, press 2nd then stat plot, and then select 4:PlotsOff. By the way it is easy to graphing function y=x^3-x^2-10x-8. See captured images: full graph and detail of the left side max
As you can see this function have zero values in the points -2, -1 and 4
Press the Y= editor and type in the function after Y1= or Y2=.
Once finished, press the GRAPH button.
If you do not see the portions of interest (where y=0) use the ZOOM or WINDOWS features to Zoom in on the area where the function has zero value.
After you re-frame the graph, [2nd][TRACE] to open the (CALC) menu options.
Select the one that says Zeros.
You will be asked to enter the left bound of the interval where the calculator will search for a root.
Enter the left bound and press ENTER.
You will be asked for the right bound.
Move cursor to the right to be on the other side of the point where the function vanishes and press the ENTER button.
You will be asked for an initial guess of the zero (root). Enter a value between the left bound and the right bound, and press ENTER.
Wait for the calculator to return with one root.
If the calculator does not find a zero, your interval is incorrect ( no zero in it) or the initial guess is too far from the root, or the interval is too large. Narrow it or change the initial guess.
When you find the first root, move cursor along the graph to the neighborhood of the other root (if it exists) and restart the process.
Try rewriting your equations in the equivalent forms that follow. Y1=3-X Y2=(1-X)/3
If you do not get a syntax error it means that you were not using the correct minus sign. If the - is the first symbol after the equal sign you must use negation (-) or change sign key. Similarly, after the division / key you must use the negation or change sign (-)
First lets make sure you have a manual. go to:
http://ec1.images-amazon.com/media/i3d/01/A/man-migrate/MANUAL000010670.pdf
Go to chapter 3 and it explains the details in graphing
If you were trying to find the roots (zeros) of an algebraic equation
with the solve( command or the interactive solver you might have
supplied an interval where the function is always positive or always
negative. If the expression is always of the same sign on an interval,
the interval does not contain a root.
If you extend the interval, the procedure may not
converge fast enough and the calculator may not find the solution.
Abetter way to do it is to sketch the
graph of the function and use the graph to choose a reasonable interval
that extends on both sides of the root, and a better initial guess.
If
you were doing some financial
calculations,
You attempted to calculate the I% variable when FV,
N*PMT, and PV are all positive, or all negative.
You tried to calculate Irr( when neither CFlist,
nor CFO is positive
You tried to calculate Irr( when neither CFlist,
nor CFO is negative
To view both graphs, the = signs on both y1 and y2 must be contained in black boxes. If you put your cursor over the = sign on the second equation and hit enter, it turns that graph off and you won't be able to see anything but the equation in y1 line. If there is not black box around the = sign of any equation in any of the lines you have typed an equation into, you will not see the graph.
The calculator only displays from -10 to 10 on the x and y axes, and both of your lines are outside of that range. However, it is easy to change that.
1. Hit the gray "WINDOW" button that is directly under the screen 2. Change "Xmax" to 150 3. Change "Ymax" to 200
Remember to change both of those values back to 10 when you are done with this problem!
Here's how to graph it: 1. Hit the "Y=" button 2. Enter your first equation after Y1, and your second equation after Y2 3. Look over both the equations to make sure you typed them correctly 4. Hit the "GRAPH" button 5. If you don't see the intersection you will need to adjust Xmax and Ymax again 6. Hit Hit "2ND" and the "TRACE" to bring up the "CALC" menu 7. Scroll down to "intersect" and hit enter 8. Hit enter 3 times (You use the first and second curve options when you have more than 2 lines) 9. It should display the coordinates at the bottom 10. I got (76, 142.1) for the intersection
Thanks for asking this question! I used to do this a much harder way, but while I was calculating the answer to your question I hit a wrong button and accidentally found this method!
try a simple equation like x^2-1 which should come out with x = 1,-1
no sign change would indicate that the graph of the curve never crosses the x axis, if you graphed the above equation and calculated the zeros, it would come out with -1, 1 which it should, but if you put in x^2+1, the graph would never cross and therefore have no sign change from pos to neg or vice versa, the left-rt are the left and right bounds or the lower and upper bounds which was corrected in later calculators, Please let me know the outcome of graphing both of the above equations and then putting them in the solver. you could also use the quadratic equation to prove the answers if you liked, hope this helps
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