Question about Casio FX-9750GPlus Calculator

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.

- Turn calculator ON.
- If you do not see icons, press Menu.
- Use navigational arrows to highlight [GRAPH] and press [EXE]
- Type the function.

- 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)
- Use the caret [^] 3 to raise X to the third power and complete the function. and press [EXE]
- At the bottom of the screen you see some tabs.
- 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
- Press the Function key to above which you see [DRAW]
- Adjust the Window dimensions if necessary

- 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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Posted on Jan 02, 2017

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

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

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

-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

No, it is a flat line with one solution, 18

May 07, 2014 | Computers & Internet

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.

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

Jan 10, 2014 | Office Equipment & Supplies

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

2. step: left or lower bound

3. step: right or upper bound

4.step: extreme value of the function

2. step: left or lower bound

3. step: right or upper bound

4.step: extreme value of the function

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

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.

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

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.

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

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

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.

At this stage, use the Arrow keys to move the center of the graph to another point as in the next screen capture.

Notice the new center as a sign on the screen and its coordinates at the bottom. By pressing [ENTER] you get this

If you keep Zooming in you may end up having something that does not look like a parabola anymore.

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.

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

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.

At this stage, use the Arrow keys to move the center of the graph to another point as in the next screen capture.

Notice the new center as a sign on the screen and its coordinates at the bottom. By pressing [ENTER] you get this

If you keep Zooming in you may end up having something that does not look like a parabola anymore.

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

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]

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

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]

3. Supply the right limit of x (to the right of the minimum). Press [ENTER]

3. Supply an estimate of the X-value of the minimum.

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.

Hope this will help you set your problem correctly.

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]

3. Supply the right limit of x (to the right of the minimum). Press [ENTER]

3. Supply an estimate of the X-value of the minimum.

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.

Hope this will help you set your problem correctly.

Oct 28, 2009 | Texas Instruments TI-84 Plus Silver...

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