How do you find a linear model using a graphing calculator?

dannyboi2006tk
2022-09-28
Answered

How do you find a linear model using a graphing calculator?

You can still ask an expert for help

Sanaa Hudson

Answered 2022-09-29
Author has **7** answers

To find a linear model for a scatterplot (which is what I assume you want), you just need to do a couple of things.

Firstly, you need to enter your data into the calculator. To do this, hit your "STAT" key, and select "EDIT". You should see a table with lists. Enter all your x values into one list, and all your y values into the other.

Once you have done this, hit 2ND and QUIT (normally mode) to return to your home screen.

Then, go back to your Stat key, but this time scroll left to the "CALC" option. Once you're there, scroll down to "LinReg (ax+b)", and select it. It will ask you to verify which list you're using for which variable, and you can change this if needed by pressing 2nd $\to$ Number of the list you want. Once you have finished with that, hit enter.

You should see a screen with all the information necessary for you to have your linear model.

Firstly, you need to enter your data into the calculator. To do this, hit your "STAT" key, and select "EDIT". You should see a table with lists. Enter all your x values into one list, and all your y values into the other.

Once you have done this, hit 2ND and QUIT (normally mode) to return to your home screen.

Then, go back to your Stat key, but this time scroll left to the "CALC" option. Once you're there, scroll down to "LinReg (ax+b)", and select it. It will ask you to verify which list you're using for which variable, and you can change this if needed by pressing 2nd $\to$ Number of the list you want. Once you have finished with that, hit enter.

You should see a screen with all the information necessary for you to have your linear model.

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I realised that the second derivative of any cubic equation in the form of ${x}^{3}+a{x}^{2}+bx+c=0$ is ${y}^{\u2033}=6x+2a$, and when we equalize $y$ to $0$, $x$ is equal to $-\frac{a}{3}$. This had me thinking about any possible relation between the quadratic term and the second derivative.

It is also sort of the same in quadratic functions in the form of ${x}^{2}+ax+b=0$. When we replace $x$ with $-\frac{a}{2}$, which is the first derivative of a quadratic function and the vertex point of it, we obtain the vertex form of the equation which does not have the linear term, ${x}^{1}$.

Is there a relation between the quadratic term and the second derivative of a cubic equation? If the answer is yes, what is it and how is it observed on graphs, in equations?

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