Finding Y1 on the TI-83 Plus Calculator for Linear Regression

AI Thread Summary
To find Y1 on the TI-83 Plus calculator for linear regression, simply enter the values directly as you input data points; there is no dedicated "Y1" key. The calculator automatically assigns the first equation to Y1 when performing linear regression. You enter a series of points (xi, yi), and the calculator calculates the best-fit line using the least squares method. Understanding this process is key to effectively using the TI-83 Plus for linear regression tasks. This approach allows for accurate data analysis and graphing of linear relationships.
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Homework Statement


Where can I find Y(subscript: 1) on the TI-83 plus calculator? I was working on Linear Regressions in the book and I came across a question that requires me to input Y1 on the screen but I can't find the 'y'.. how can I find it?


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The Attempt at a Solution

 
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You are being prompted to input a value which the calculator will assign to "Y1". There is no "Y1" key, just enter a number. Basically, with linear regression, you are entering a series of points, (xi, yi), and the calculator responds with the "closest" line in the sense of least squares.
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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