Question about Texas Instruments TI-84 Plus Calculator

Ad

In all probability, someone used the Transfrm Application before. The problem is that it lingers in memory and has to be unloaded from RAM.

To do so

- Press the APPS button.
- Scroll Up or down to reach the Transfrm APP.
- Select it and press ENTER.
- In the subsequent screen, press the number that corresponds to the UNINSTALL option.
- Validate the selection by pressing the ENTER button.
- Go back to your Y= Editor and enter up to 10 functions, then press the GRAPH key.
- Graphs should know display correctly.

Posted on Jun 01, 2011

Ad

What are the question marks in the two equations? Get rid of them or replace them with numbers and you'll get rid of the error.

Apr 28, 2012 | Texas Instruments TI-83 Plus Graphing...

You have to use **Y Editor** to define functions on the both sides of the equation and then graph it. Finally option is **F5: Intersection**.

See captured images for equation**5x+10=-2x-7**, **y1=5x+10**, **y2=-2x-7** and **x=-17/7**

See captured images for equation

Feb 25, 2012 | Texas Instruments TI-89 Calculator

This is problem with graphing differential equation. Invalid Axes error occurs if Axes=TIME or if t(time) is set as a CUSTOM axis. This is important to noticed: if Fields=DIRFLD you cannot plot a time axis. See captured image:

This graph represented differential system of equations y1'(x)=y2(x), y2'(x)=-y1(x) for initial conditions y1(0)=0, y2(0)=-1.Finally, x axes=y1(x) and y axes=y2(x)

This graph represented differential system of equations y1'(x)=y2(x), y2'(x)=-y1(x) for initial conditions y1(0)=0, y2(0)=-1.Finally, x axes=y1(x) and y axes=y2(x)

Feb 08, 2012 | Texas Instruments TI-89 Calculator

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

1.) make sure the two lines are not the same line

2.) under the y= button, when you enter the two equations make sure for both y1 and y2 the icon on the far left for both is a solid \ and the = on the right of y1 and y2 should be highlighted.

3.) clear memory if all else fails as a last resort.

hope this helps

2.) under the y= button, when you enter the two equations make sure for both y1 and y2 the icon on the far left for both is a solid \ and the = on the right of y1 and y2 should be highlighted.

3.) clear memory if all else fails as a last resort.

hope this helps

Apr 27, 2010 | Texas Instruments TI-84 Plus Calculator

This is my last post. It seems that I squanderd my time for zilch. There is a simple way to reset the defaults or the RAM. I stopped suggesting it because ..(what is the point?).

Press [2nd][MEM][7:Reset][1:All Ram][2:Yes] [ENTER]. Message RAM cleared is displayed on screen. Press [CLEAR] key. Screen should clear the message.

If that does not solve your problem..

Press [2nd][MEM][7:Reset][1:All Ram][2:Yes] [ENTER]. Message RAM cleared is displayed on screen. Press [CLEAR] key. Screen should clear the message.

If that does not solve your problem..

Oct 06, 2009 | Texas Instruments TI-83 Plus Silver...

Hello,

Suppose you want to draw the circle the equation of which is X^2+Y^2=4, a circle centered at the origin with radius 2.

When you isolate Y^2 and then extarct the squre root you obtain two solutions, call them Y1 and Y2

Y1=SQRT(4-X^2), Y2= -SQRT(4-X^2)

You press[Y=] and type in the right hand side of the equations above, one in Y1= and the other in Y2. Press [GRAPH].

Ih a graphic does not appear within the scree window adjust the window parameters ( [WINDOW]) or Press [ZOOM] and select ZoomIn, ZoomOut, ZoomSTD, etc.

Explore the program Draw accessible by pressing [2nd][DRAW].

Hope it helps

Suppose you want to draw the circle the equation of which is X^2+Y^2=4, a circle centered at the origin with radius 2.

When you isolate Y^2 and then extarct the squre root you obtain two solutions, call them Y1 and Y2

Y1=SQRT(4-X^2), Y2= -SQRT(4-X^2)

You press[Y=] and type in the right hand side of the equations above, one in Y1= and the other in Y2. Press [GRAPH].

Ih a graphic does not appear within the scree window adjust the window parameters ( [WINDOW]) or Press [ZOOM] and select ZoomIn, ZoomOut, ZoomSTD, etc.

Explore the program Draw accessible by pressing [2nd][DRAW].

Hope it helps

Sep 04, 2009 | Texas Instruments TI-84 Plus Calculator

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.

Aug 12, 2009 | Texas Instruments TI-84 Plus Calculator

At first you have to slving this equations in explicit form** y=f(x)** and the graphing it. See captured images

Jan 23, 2009 | Texas Instruments TI-83 Plus Calculator

just try resetting your calc. Hit 2nd then + then 7 then 1 then 2

Jan 28, 2008 | Texas Instruments TI-84 Plus Calculator

41 people viewed this question

Usually answered in minutes!

×