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How do i input the equation f(x)= x^3 -3x+2 (-2,2). I need to graph and fine the local maximum and minimum

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  • Casio Master
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I am puzzled by the (-2,2) after the equation. I suspect it is the coordinates of some point. But I won't push the matter further, and I will just ignore that info. I am no mind reader.

  1. Turn calculator ON.
  2. If you do not see icons, press Menu.
  3. Use navigational arrows to highlight [GRAPH] and press [EXE]
  4. Type the function.
  5. Use the [X.T,theta] key below [ALPHA] to type the X variable or the [ALPHA][+]. In general you should use the [X,T,theta] because is context sensitive: If you draw parametric or polar graphs, pressing that key gives you the right variable (T, or theta)
  6. Use the caret [^] 3 to raise X to the third power and complete the function. and press [EXE]
  7. At the bottom of the screen you see some tabs.
  8. Press [F3:Type] and you will see all types of graphs you can draw. But since you want Y=, press [EXIT] to return to Y=editor
  9. Press the Function key to above which you see [DRAW]
  10. Adjust the Window dimensions if necessary
Finding the min and max
  • While the graph is still displayed press [SHIFT][F5:G-Solve].You see all the possible calculationsF1:root, F2:Max,F3:Min,F4:YiCPT (y_intercept),F5:Isct (intersection) F6:=> to show the others, among which is the F3:integral.
  • Press F3:min and hold your breath.
  • After a while you get your minimun. Be patient, even if the small square at the top right disappears.
  • Once you get the minimum, press [SHIFT][F5-G-Solve] and repeat for the maximum.

Posted on Feb 01, 2010

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1 Answer

How i can solve quadratic equations which have no real roots


If the quadratic equation has no roots, you cannot find the roots. The discriminate is negative, so if we attempt to use the quadratic equation, we get no roots.

For example, y=x^2 +0x + 3

a=1, b=0, c=3

(-b+/- sqrt(b^2 -4ac))/2a

Substituting in the numbers, we get

x= (-0 +/-sqrt(0^2 - 4(1)3))/2
x= (0 +/- sqrt (-12))/2

We cannot do the square root of -12. Therefore, there are no roots. This is the same as having no x-intercepts. The discriminant is b^2-4ac. In this case it is -12. Thus, there are no real roots.

However, you can still determine the maximum or minimum, the vertex, the axis of symmetry, the y-intercept and the stretch/compression. With this information you can graph the equation.

In this case, the y-intercept is 3, the vertex is at (0,3), the axis of symmetry is x=0, the minimum value of the function is 3.

Good luck.

Let me know if you have any questions.

Paul

Nov 04, 2015 | Casio FX991ES Scientific Calculator

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Determine the x and y intercept of the graph for each equation. 5x-y=-5 And -3x+2y=7


5x-y=-5 answer x=-1, y= 5 http://www.wolframalpha.com/input/?i=Determine+the+x+and+y+intercept+of+the+graph+for+equation%3A+5x-y%3D-5

-3x+2y=7 answer x = -7/3, y = 7/2 http://www.wolframalpha.com/input/?i=Determine+the+x+and+y+intercept+of+the+graph+for+equation%3A+-3x%2B2y%3D7

Feb 23, 2015 | Office Equipment & Supplies

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Minimum point of a plot graph


Two, since only one straight line exists that goes through the same two points.

Alternatively, you could define a straight line with one point and a slope.

Jan 10, 2014 | Office Equipment & Supplies

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Minimum or maximum values of a parabola


Follow these steps: graph the parabola and after that 2nd then CALC keypad. To find extreme value of the function select 3: min or 4: max options. Select the function and set left bound, right bound(using left and right arrows) by pressing ENTER. You can see images bellow for all these steps for the y=-2x^2+5x+3


1. step: graphing parabola
3_9_2012_7_34_06_pm.jpg


2. step: left or lower bound
3_9_2012_7_36_01_pm.jpg


3. step: right or upper bound
3_9_2012_7_37_33_pm.jpg


4.step: extreme value of the function
3_9_2012_7_39_04_pm.jpg

Mar 08, 2012 | Texas Instruments TI84PLUS Slvr Viewscreen...

1 Answer

I need to find decreasing and increasing intervals and I dont know how to do this on my TI 83


While your TI-83 plus cannot explicitly find increasing or decreasing intervals, it can find the minimum(s) and maximum(s) of a function.
Press the "Y=" button.
Enter the equation in "y=" format into the first open equation.
Press the "GRAPH" key.
If the maximum and minimum of the curve are not shown in the window it has, press ZOOM->0.
Press 2ND->CALC (under the "TRACE" key).
Press 3 if you are trying to find a min, 4 if you're trying to find a max.
It will ask for bounds. Enter a left and right bound by looking at the graph itself and choosing a left bound that is clearly to the left of the min/max, and a right bound clearly to the right of the min/max.
Then it will ask you to guess a value. You can skip this step by pressing ENTER instead of entering a value.
It will calculate the value. To find increasing/decreasing intervals, just know that every interval with a minimum as the left bound and a maximum as the right bound is increasing, and vice versa.

Feb 25, 2011 | Texas Instruments TI-83 Plus Calculator

1 Answer

My TI-83 Plus calculator won't graph anything whatsoever. Did I need new software. Please help me.


First - Are you graphing within the proper window? Hit the zoom button and change it to zoom Standard within the graphing window.

Second - Are you inputting the equation correctly? Make sure you are inputting your equations in a y= format. If possible, please reply with a screenshot of your graph screen.

Feb 07, 2011 | Texas Instruments TI-83 Plus Calculator

1 Answer

When I'm entering a quadratic function, I'm getting the graph of something that is not a parabola like it should be. However, when I graph x^2 by itself, it looks the way it should. What can I do to fix...


The actual graph depends on the constants (parameters) that are in the equation. Try playing with the dimensions of the window by zooming in or zooming out.
The most interesting part of the graph is the region near the maximum or minimum.
Here is a graph of the function y=X^2-5X 13 when drawn with standard window dimensions
3b585d8.jpg
Not very revealing. However if you press the Zoom key and select [2:ZoomIn] you see this (no changes) but the calculator is ready to accept your input concerning the new center of the graph.
bcd4c3e.jpg
At this stage, use the Arrow keys to move the center of the graph to another point as in the next screen capture.

0e09736.jpg
Notice the new center as a sign on the screen and its coordinates at the bottom. By pressing [ENTER] you get this
fc695b8.jpg
If you keep Zooming in you may end up having something that does not look like a parabola anymore.
4ea7132.jpg

To summarize: Use the Zoom In or Zoom Out functions or find where the minimum/maximum is, center the graph on that point and modify the windows dimensions (Xmin, Xmax, Ymin, and Ymax) so as to get as much of the interesting part of the graph as possible.

Sep 02, 2010 | Texas Instruments TI-83 Plus Calculator

1 Answer

Graphing Problems


In home folder (main calculator screen) press [Diamond][F2:Window]] to access Window configuration screen.
Press [F1] ( not mecessary now because the F1 page is open by default). In the F1 page of configuration screen set Xmin=-10,
Xmax=10
Ymin=-10
Ymax=10
Xscale=1.
Yscale=1.

OR
Press [Diamond][F2:Window][F2][6:ZoomStd] (Zoom Standard) to set all values as above.

If a graph is active (axes displayed, or part of a curve is showing) press [F2:Zoom][6:ZoomStd]

Feb 15, 2010 | Texas Instruments TI-89 Calculator

1 Answer

The maximum and minimum are not working, when i put in the problem and it will give me the graph but when i use maximum or minimum it will give me the same answer on every problem


Hello,
When you are use the calculator to find a maximum or a minimum, you are supposed to give the calculator a range of values between which to search for i (interval where the max or min lies)
1.Look at the graph to locate approximately the position (the x coordinate) of the minimum.
2. Supply the left limit of x ( to the left of the minimum) Press [ENTER]
3498c8f.jpg 3. Supply the right limit of x (to the right of the minimum). Press [ENTER]
4a5a77d.jpg 3. Supply an estimate of the X-value of the minimum.

afc61ed.jpg
After a moment the calculator displays the X and Y values of the maximum. The better your estimate, the faster the maximum will be found.

d8247ba.jpg
Hope this will help you set your problem correctly.

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