# Math Help - develope a linear, extonential and quadratic model using the following table

1. ## develope a linear, extonential and quadratic model using the following table

So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
Age Value
1 14000
2 9100
3 6200
4 4000
3 3000

I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
Thank you

2. Originally Posted by surffan
So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
Age Value
1 14000
2 9100
3 6200
4 4000
3 3000

I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
Thank you
Graphing Calculator Help

3. i'm giving it a try but the problem is that i cant get it to match the graph, to get the exponential function to start at around 17500 and end at 0 for y.

4. I developed the exponential, its just the quadratic I have to do now. I'm pretty lost I must admit

5. Originally Posted by surffan
I developed the exponential, its just the quadratic I have to do now. I'm pretty lost I must admit
did you perform a quadratic regression using the data in your table?

6. Originally Posted by surffan
So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
Age Value
1 14000
2 9100
3 6200
4 4000
3 3000

I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
Thank you
Hi Surffan,

If you're trying to do this manually, good luck with that.

I used a TI-84+ and got a linear regression equation of y = -2710x + 15390, which is a little better than your calculation. The correlation coefficient is .9659370337.

Using the same calculator data, the quadratic regression equation was $607.1428571x^2-6352.857143x+19640$ which is pretty darn good. The correlation coefficient is .9985988704.

My exponential regression equation was equally good at $y=20130.14771(.676862578)^x$. The correlation coefficient is .9979550889

7. i used that link given in the first response to create a table and use my data. Wow much easier than trying to figure it all out manually