Question

use the graphing tool at the right to create a scatter plot for the data in the table. use the shape of the scatter plot to determine if the data are best modeled by a linear function, an exponential function, a logarithmic function, or a quadratic function.

x	y
-	-
0	-3
1	-2
2	0
3	4
4	12

plot each point, one at a time. using the graphing tool, start by plotting (0, - 3).
use the graphing tool to plot (1, - 2).
use the graphing tool to plot (2, 0).
use the graphing tool to plot (3, 4).
use the graphing tool to plot (4,12).

Explanation:

Step1: Plot the points

Plot the points \((0, - 3)\), \((1,-2)\), \((2,0)\), \((3,4)\), \((4,12)\) on a graph.

Step2: Analyze the shape

Observe that as \(x\) increases, the differences in \(y - values\) between consecutive \(x - values\) are not constant (so not linear). The rate of change of \(y\) with respect to \(x\) is increasing. For a logarithmic function, the rate of change decreases as \(x\) increases. For an exponential function, the ratio of consecutive \(y - values\) for equally - spaced \(x - values\) would be constant. Here, the differences in \(y\) are not following an exponential pattern. For a quadratic function \(y=ax^{2}+bx + c\), the second - differences (the differences of the differences of \(y\) values) are constant.
Let's find the first - differences:
When \(x = 0\) to \(x = 1\), \(\Delta y=-2-(-3)=1\); when \(x = 1\) to \(x = 2\), \(\Delta y=0 - (-2)=2\); when \(x = 2\) to \(x = 3\), \(\Delta y=4 - 0=4\); when \(x = 3\) to \(x = 4\), \(\Delta y=12 - 4 = 8\).
The second - differences: \(2 - 1=1\), \(4 - 2=2\), \(8 - 4=4\) (not constant for a quadratic). But if we consider the general behavior of the curve formed by the points, it has a non - linear, upward - curving shape which is more indicative of a quadratic function.

Answer:

The data are best modeled by a quadratic function.