Question about Texas Instruments TI-83 Plus Calculator

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TI-83 graph problems-help please. Hi, sorry to sound so stupid, but trying for hours to try to solve these using my calculator,but doing something wrong as can't get the right answers. I wonder if anyone can tell me how to do these on my TI-83. (1)Parabola y=2+3x-2x^2, I need to see the gradient, the vertex, the equation of the symmetry, the intercept on the y axis, if the graph is the same of y=6+9x-6x^2 and if it does it cross the point at the same point as the original parabola? (2) Input data into calculator to get best fit regression function for data. X 1,2,3,4,5,6 Y 2.46, 8.96, 22.30,41.66, 64.85,98.15 To find equation y= What Y is when x is 3.5? The value of the correlation coefficient? (3) Sunset at 4 weekly intervals, with times in GMT in hours to 2 decs. input data to get a scatterplot and best fit sine regression y in terms of x for data. week (x) 0, 4,8,12,16,20,24,28,32,36,40,44,48,52 sunset (y) 16.04, 16.75, 17.62,18.44,19.24,19.99,20.45,20.28, 19.55,18.54, 17.45, 16.49, 15.92,16.02 Does the scatterplot indicate curve is a good fit or poor one? The equation of the best fit curve? Predition of sunset in wk 38? (4) Temp in degrees from 6am to 10pm, with 6 with t time in hrs on 24 hr clock. H(t)=18+6sin(pi t divided by 8 - 5pi divided by 4) for 6 t 22 To get temp at 12 noon and highest and lowest temps in day? A big thankyou to anyone who can help me with these, as head aching now and abusing calculator and guide for my lack of skill.

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  • abjustso May 03, 2010

    I have after hours managed to work out how to do question 4 and think that I have the correct answers.

    The other 3 I keep still doing something wronge as get error messages. I have viewed lots of you tube videos, but none seem to give me a simple step by step way to to these probems.



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Open Y= editor and type in the two functions
TI-83 graph problems-help please.  - 6447ca5.jpgbbeaa23.jpg
The calculaus functions are accessible by pressing [2nd][TRACE] to open the CALCulate menu options. For the gradient (I think you mean the derivative) use option 6:dy/dx. But first choose the point where you want it calculated (use cursor to move along the curve) and press ENTER. The value of the deivative will be calculated at the chosen poTI-83 graph problems-help please. - 50e3257.jpg>50e3257.jpg005508b.jpg
TI-83 graph problems-help please. - 24d4168.jpgertex of the parabolas are maxima. Thus yoTI-83 graph problems-help please. - b57bf8d.jpgoption 4:maximum
You will be prompted for a left bound. Move cursor to the left of the maximum (not too far) and press [ENTER]. A fat arrow is displayed on the graph that shows the left limit of the interval. You will be prompted for a right bound. Move cursor along the curve to the right of the the vertex. Press ENTER. A seconf fat arrow will be displayed to show the right limiTI-83 graph problems-help please. - 3b702ee.jpgterval.

You will be prompted foTI-83 graph problems-help please. - 22c11ef.jpgf the maximum. Move cursor newar the max or enter a numerical value and press ENTER.


The location of the vertex is displayed (X and Y values)
I have no idea what you mean by the equation of symmetry


Posted on May 18, 2010

  • k24674 May 18, 2010

    For the Y-intercept , use the Trace function (4th function key). Move the cursor near the point of intersection and read the Y-value. -1E-9 is practically zero. The intercept is at Y=2.

    I do not know what you mean by the two graphs are the same because they are not. The second function is 3 times the first. They do intersect the X-axis at the same points. Because if equation 2+3x-2x^2=0 for a certain value of x, then 6+9x-6x^2= 3(2+3x-2x^2) is also equal to zero for the same value.

    For the points 2 and 3 regression reread my earlier post answering you question about an exponential regression. Point C shows you how to configure calculator to calculate the correlation coefficient, if one is defined for the regression
    Press the [STAT] key then Right arrow to highlight CALC and select the option C:SineReg

    Once you have the equation of best fit stored in Y1 you can draw it and use the trace to move the cursor to X=3.5 and read off the value or substitute the value x=3.5 in the equation to calculate the corresponding Y

    Concerning the goodness of a fit, only YOU can decide if it is good enough for you: It may help to
    draw the scatter plot and the equation of best fit stored in Y1 on the same graph. See my post on exponential regression to review how to draw them on same screen.


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Try contacting the company that manufactured the product.

Posted on Oct 04, 2010


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