Using Data Ponts, Write in Point-Slope form

[babybells96]

Using the first and last data points, write an equation of a line in point-slope form. the data points are (-3, 15) (-2,-5) (-1,1) (0,3) (1,1) (2, -5) and (3,-15). And a graph of the data points, a parabola, has been included as well. My problem is if you use the first and last data points of the graph, and plug them into the point-slope form, you get y=-15 and if you repeat it for the first and last data points you keep getting y= whatever y-value they share. I am aware that I am doing something wrong, but I can't figure out how to do this correctly.

 

[Deleted member 4993]

Using the first and last data points, write an equation of a line in point-slope form. the data points are (-3, 15) (-2,-5) (-1,1) (0,3) (1,1) (2, -5) and (3,-15). And a graph of the data points, a parabola, has been included as well. My problem is if you use the first and last data points of the graph, and plug them into the point-slope form, you get y=-15 and if you repeat it for the first and last data points you keep getting y= whatever y-value they share. I am aware that I am doing something wrong, but I can't figure out how to do this correctly.

Why do think your work is incorrect?

Since you are drawing a straight-line using only two points - the line must go through those points.

 

[babybells96]

The problem is my graph is a parabola, not a straight line.

 

[Bob Brown MSEE]

If you draw a line through the first and last data points you will get a horizontal line,
That line is a new graph, who's formula is y=-15 in point-slope form.

 

[wjm11]

Using the first and last data points, write an equation of a line in point-slope form. the data points are (-3, 15) (-2,-5) (-1,1) (0,3) (1,1) (2, -5) and (3,-15). And a graph of the data points, a parabola, has been included as well. My problem is if you use the first and last data points of the graph, and plug them into the point-slope form, you get y=-15 and if you repeat it for the first and last data points you keep getting y= whatever y-value they share. I am aware that I am doing something wrong, but I can't figure out how to do this correctly.

babybells 96, you should congratulate yourself on your sense of unease. It shows you are thinking about the validity of your answer. You have not done anything wrong however. Your unease comes from recognizing that the line you calculated does not describe that set of data points very well.

Certainly we can use linear equations in attempts to analyze data or make guesses about where other points may lie, but this example shows us that lines may not always be a good choice to represent our data. (Quadratics, cubics, sinusoidals, exponentials, etc., might be other choices to consider.)

Perhaps that is the point of this exercise???

 

[Deleted member 4993]

Using the first and last data points, write an equation of a line in point-slope form. the data points are (-3, 15) (-2,-5) (-1,1) (0,3) (1,1) (2, -5) and (3,-15). And a graph of the data points, a parabola, has been included as well. My problem is if you use the first and last data points of the graph, and plug them into the point-slope form, you get y=-15 and if you repeat it for the first and last data points you keep getting y= whatever y-value they share. I am aware that I am doing something wrong, but I can't figure out how to do this correctly.

Are those correct?

 

[babybells96]

babybells 96, you should congratulate yourself on your sense of unease. It shows you are thinking about the validity of your answer. You have not done anything wrong however. Your unease comes from recognizing that the line you calculated does not describe that set of data points very well.

Certainly we can use linear equations in attempts to analyze data or make guesses about where other points may lie, but this example shows us that lines may not always be a good choice to represent our data. (Quadratics, cubics, sinusoidals, exponentials, etc., might be other choices to consider.)

Perhaps that is the point of this exercise???

I think you a right that the point of the exercise has to do with proving that lines are not always the best choice. I think my teacher might have been trying to make us overthink this a bit, as well. thank you.

 

[lookagain]

the data points are (-3, 15) (-2,-5) (-1,1) (0,3) (1,1) (2, -5) and (3,-15). And a graph of the data points, > > a parabola, < < has been included as well.

The first co-ordinate is wrong. It must be (-3, -15) instead.

With that change, all of the seven points fit perfectly on the graph of \(\displaystyle \ \ y \ = \ -2x^2 + 3.\)