Question about Texas Instruments TI-84 Plus Calculator

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When I use my intersect function to find the intersection of y = 4 and the square root of the quantity 2x-3, it will not recognize the square root part. It will only recognize the y = 4 part.

Posted by Anonymous on

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See captured image below



when I use my intersect function to find the inter - when-use-intersect-function-find-gvce23bt1fji13zwokxrc5ja-2-0.jpg

Posted on Jan 07, 2016

6 Suggested Answers

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Posted on Jan 02, 2017

  • 2 Answers

SOURCE: ti-89 titanium SIN!!!!!

The answer that you have received in actuality is correct. However, it is in exact form when the answer that you were looking for was an approximation (decimal form) answer. To achieve this all you have to do is hit the alpha button (the yellow button with a diamond in the center) and then enter which will give you your desired answer :-)

Posted on Jan 15, 2008

  • 11 Answers

SOURCE: error :invalid DIM

This is my first post on this site, but I've had friends at school ask me about this. Apparently I'm like a calculator genius or something...

One solution I know of is clearing the RAM. Sounds strange, doesn't it? But it fixed three different calcs.

Be sure to back up or archive important data before doing this.

Posted on Oct 21, 2008

  • 11 Answers

SOURCE: ERR: No Sign change...

-0.171x+43.43 and -0.146x+39.83
i tried that and i got the same thing, but i noticed that if you change your window so it fits more of the graph on the right, were you can actually see the lines intersect, then you can get the intersection.
oh, and by the way, the intersection is (144, 18.806)

Posted on Mar 01, 2009

  • 1 Answer

SOURCE: Graphing radical functions

I have a TI-89 Titanium and I have experienced the same problem. HERE IS HOW TO FIX IT:

Goto " MODE " and scroll down the options until you get to
" COMPLEX FORMAT " change that option to " RECTANGULAR " and press ENTER until you exit out of the " MODE " screen. (very important that you press ' enter ' until you exit because if you don't, it will not save your changes. )

Posted on Mar 15, 2009

  • 7993 Answers

SOURCE: TI-15 square root Function

Hello,
You should enter it as follows [SQRT]36 ) [ENTER/=].
If you define some function f of a variable x you write that f(x), where the parentheses enclose the so-called argument (objet on wich the function acts). It seems that on this calculator the opening parenthesis is implicit: the calculator supplies it when you press [SQRT] but does not display it (a design flaw?). However the closing parenthesis must be entered by you to signifie to the function [SQRT] that you have finished entering the argument. Weird but one can live with it.
Hope it helps.

Posted on Sep 07, 2009

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1 Answer

Is it possible to find the intersection of two graphs on a casio fx-500ES ?


In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions, if it exists, is a point (x_i,y_i) such that f(x_i)=g(x_i).
Graphically this means that (x_i, f(x_i))=(x_i, g(x_i)): The two curves pass through the same point in the Cartesian plane.
Now consider f(x_i)=g(x_i). That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or f(x)-g(x)=0.
If you can feed it to the SOLVER in EQN, the equation mode, and the calculator gives you the roots of the equation, then for each root x_i found you have an intersection point (x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to calculate the y-value since the two functions are supposed to be equal at the root x_i.
However it is much simpler to do with a graphing calculator.

Oct 05, 2013 | Casio Office Equipment & Supplies

1 Answer

Find the intersection between two graphs


In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions, if it exists, is a point (x_i,y_i) such that f(x_i)=g(x_i).
Graphically this means that (x_i, f(x_i))=(x_i, g(x_i)): The two curves pass through the same point in the Cartesian plane.
Now consider f(x_i)=g(x_i). That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or f(x)-g(x)=0.
If you can feed it to the SOLVER in EQN, the equation mode, and the calculator gives you the roots of the equation, then for each root x_i found you have an intersection point (x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to calculate the y-value since the two functions are supposed to be equal at the root x_i.
However it is much simpler to do with a graphing calculator.

Oct 05, 2013 | Casio FX-115ES Scientific Calculator

1 Answer

HELP WITH TI INSPIRE CX calculator


Draw the graphs of the functions you are interested in. While the graphs are displayed Press the Menu key. In the drop down window locate the submenu Settings (number 8 or 9) depending on your calculator
6d8395f7-b10e-4d9e-a7e6-45beadb303a3.jpg76f72147-9dcf-457f-a122-ae70490e2815.jpgThe following screen is displayed with at the top the setting Float.
eb304039-5513-442e-886f-d435de8aa653.jpgFollow the right pointing arrow on the same line as float and select Float 3 or 4 as on the screen capture above.
0c4e79fb-d2eb-4461-821f-87adbd63b387.jpgPoint to the graph of one of the functions select it then press the menu select option Analyze graph then intersection. You will be asked to select the second function, then the lower bound and upper bound. As you see on the previous screen capture, the coordinates of the intersection point are given with 3 decimal digits.
3acdb05f-f573-4e5e-a467-03db35a80ea2.jpg

Apr 27, 2013 | Texas Instruments TI-Nspire Graphic...

1 Answer

How to use ti 84 plus calculator to graph linear equations


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












2_26_2012_7_15_54_am.jpg

2_26_2012_7_16_12_am.jpg

Feb 25, 2012 | Texas Instruments TI-89 Calculator

1 Answer

Give the polynomial function whose roots are -2 1 and 3


We can write this polynomial as:
  • (x-(-2))*(x-1)*(x-3)=
  • (x+2)(x-1)(x-3)=
  • (x+2)[x*(x-3)-1*(x-3)]=
  • (x+2)*(x^2-3x-x+3)=
  • (x+2)(x^2-4x+3)=
  • x*(x^2-4x+3)+2*(x^2-4x+3)=
  • x^3-4x^2+3x+2x^2-8x+6=
  • x^3-2x^2-5x+6
x^3-2x^2-5x+6 is polynomial with roots -2, 1, 3.

You can see this polynomial in following picture:

elessaelle_2.png

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

Oct 03, 2011 | Office Equipment & Supplies

1 Answer

Given a length of 30 cm on the x axis and a length of 10 cm on the y axis and a angle of 90 at the x/y axis intersection is there a short cut to calculate interior angles and hypot. of the triangle?


Yes, there is shortcut because this is right triangle, so you can use Pythagorean theorem (see picture).
  1. Length of hypotenuse is square root of sum of squares of lengths of other two sides of triangle, which is equal to square root of 30^2+10^2=31.6 cm.
  2. Sin(a)=longer cathetus/hypotenuse=0.949 so a=arcsin(0.949)=71.6 degrees
  3. Finally b=90-a=18.4 degrees.
elessaelle.png

If this was helpful please rate 4 thumbs :)

Sep 05, 2011 | Texas Instruments TI-30XA Calculator

1 Answer

WHen I use my intersect function to find intersection points, it asks "first curve" then "second curve" and it wont say anything after that. It just repeats "second curve" continuously. PLEASE HELP!


the curve is refering to the function. So you have two different "lines". You choose intersect then one line, then it will ask for the other line, then a guess. your guess may or may not matter if there is only 1 intersection it will only calculate faster. If there is more than one intersection it will let you choose which one you want.

Basically you can graph 1,2,3,4,5 or how ever many lines it will take, but you may only want the intersect of 2 of them, thats why it asks.

hope this helps

Apr 24, 2010 | Texas Instruments TI-84 Plus Calculator

1 Answer

When you have a problem like that ex: 5-x^2 = x -9, how do you know what to set the ymin, ymax at in order to get ur answers to solve


Half of the work involved in solving a problem is being able to formulate it so that it can be solved. I am afraid your formulation leaves too many details in the dark.
But I think I figured out what you want.
"You want to solve the equation graphically, ie find the zeros of a function"

FIRST METHOD
  1. Create a single expression from the equation : gather all terms on one side so as to make "expression"=0
  2. One such expression is X^2 +X -14=0
  3. Draw the function y=X^2+X-14
  4. Find the X-coordinates of the points where y=0
If that is the method you had in mind, it does not matter what specific values you set for Ymin and Ymax, because you are interested in Y values near Zero. You have to make sure the X-axis appears on the screen.
I assume you know how to use the [2nd][TRACE] (CALC) [2:Zero] function to find the zeros.

Here is the negative one Here is the positive one
13c7e20.jpgafd8447.jpg

SECOND METHOD
The second method entails
  1. defining two functions, Y1=-X^2+5 and Y2=X-9,
  2. Graphing the two functions.
  3. Finding their intersections.
  4. The X-Values of the two points of intersection of Y1 and Y2 are the solutions of the equation.
Here is the graph in standard window dimensions
c51df7b.jpg91c2f1a.jpg
As you see, one root does not show and you have either to Zoom out or move the Yrange downward (as seen on the right picture for which I set Ymin=-15, Ymax=5

I assume you know how to find the intersection of the two curves, and I will show you only one point.
To find the intersection you use the [2nd][TRACE] (CALC) [5:Intersect] command


Jan 29, 2010 | Texas Instruments TI-84 Plus Calculator

1 Answer

I graph two rational equations (both the positive and negative versions) which ceate circles. I then try to find the intersection of the two circles (2 points of intersections) using the intersect function...


Hi,
You should check your understanding of what a function is. You are drawing two functions the ranges of which do not overlap, since one branch is positive and the other is negative. You know that the only two points where there could be overlapping are the points where y=0 for both functions. Why would you need the calculator to confirm to you what you already know.
To define the two branches you had to take the square root of some expression say y= SQRT(5-x^2). That is a circle centered on O(0,0) with radius SQRT(5). The two points where the positive branch intersects the negative one are for y=0, meaning x1= SQRT(5) or x2= -SQRT(5).
What do you think is the exact value of SQRT(5): 2.236067977....? No, because SQRT(5) is an irrational number that has an infinite number of digits and no matter how many additional digits you may align to determine it will not make that representation the EXACT value of SQRT(5).

That does not mean you will never be able to find an intersection of the two curves. Maybe, if you take y=SQRT(4-x^ 2) the calculator will be able to find the intersections but that will remain one case.In general the calculator will not find the intersection.

I hope that I convinced that it is futile to seek, with the help of the calculator, the intersection of two irrational functions ( for they are irrational not rational as you claim) that share only two points.

Hope it helps.

Dec 03, 2009 | Texas Instruments TI-84 Plus Calculator

1 Answer

How you use a stanley center square #46-101


No need for a manual. Just hold it so that the two edges of the "vee" line up with the circumference of the circle and scribe a line along the steel rule. Then turn approx. 90 degrees and scribe another line along the rule that intersects the first. The point of intersection is exactly the center of the circle. This tools only function is to find the center of circles.

Mar 05, 2009 | Measuring Tools & Sensors

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