How do I do this?
It is usually a good idea to use the procedures outlined in the lesson and the problem statement.
1a) You can plot the points on a graph to see if the relation looks linear.
Or, you can examine the change in output for a chang in input as described in the lesson.
The input change from 9 to 12 (a change of 3) corresponds to an output change from 17 to 21 (a change of 4).
The input change from 12 to 15 (a change of 3) corresponds to an output change from 21 to 25 (a change of 4).
The input change from 3 to 9 (a change of 6) coresponds to an output change from 9 to 17 (a change of 8). The rates of change in all these cases is
.. 4/3 = 4/3 = 8/6
Since they are equal, the lesson tells us the function is linear.
1b) For the purpose of writing the equation, you can use the point-slope form of the equation of a line, then rearrange it.
.. for a line through (h, k) with slope m, the equation can be written as
.. y = m(x -h) +k
We have a relation with a slope of 4/3 through point (3, 9), so we can write the equation as
.. y = (4/3)*(x -3) +9
.. y = (4/3)x +5 . . . . . . . . combine constant terms
As in the attachment, you can also have your graphing calculator do this.
1c) See the attachment for a graph. It looks like a straight line with a slope of 4/3 and a y-intercept of 5.
2a) A graph of the points shows this is a non-linear function.
2b) It looks a lot like a parabola, so we try using a 2nd-degree fit to the data points* and find that they are described by the equation
.. y = 2x^2
___
* If you want to do this analytically, you start with the equation
.. y = ax^2 +bx +c
Each pair of values (x, y) will give you an equation in a, b, c. Using three (x, y) pairs, you can get three equations in those unknowns. They can be solved in any of the usual ways that systems of linear equations are solved. The point (0, 0) tells you that c=0, so you have a head start there.
Your calculator's regression function capability is something worth learning about if you have to do much of this.
It is usually a good idea to use the procedures outlined in the lesson and the problem statement.
1a) You can plot the points on a graph to see if the relation looks linear.
Or, you can examine the change in output for a chang in input as described in the lesson.
The input change from 9 to 12 (a change of 3) corresponds to an output change from 17 to 21 (a change of 4).
The input change from 12 to 15 (a change of 3) corresponds to an output change from 21 to 25 (a change of 4).
The input change from 3 to 9 (a change of 6) coresponds to an output change from 9 to 17 (a change of 8). The rates of change in all these cases is
.. 4/3 = 4/3 = 8/6
Since they are equal, the lesson tells us the function is linear.
1b) For the purpose of writing the equation, you can use the point-slope form of the equation of a line, then rearrange it.
.. for a line through (h, k) with slope m, the equation can be written as
.. y = m(x -h) +k
We have a relation with a slope of 4/3 through point (3, 9), so we can write the equation as
.. y = (4/3)*(x -3) +9
.. y = (4/3)x +5 . . . . . . . . combine constant terms
As in the attachment, you can also have your graphing calculator do this.
1c) See the attachment for a graph. It looks like a straight line with a slope of 4/3 and a y-intercept of 5.
2a) A graph of the points shows this is a non-linear function.
2b) It looks a lot like a parabola, so we try using a 2nd-degree fit to the data points* and find that they are described by the equation
.. y = 2x^2
___
* If you want to do this analytically, you start with the equation
.. y = ax^2 +bx +c
Each pair of values (x, y) will give you an equation in a, b, c. Using three (x, y) pairs, you can get three equations in those unknowns. They can be solved in any of the usual ways that systems of linear equations are solved. The point (0, 0) tells you that c=0, so you have a head start there.
Your calculator's regression function capability is something worth learning about if you have to do much of this.