Question

consider the data points (-4,-14), (-2,-7), (0,-3), (2,-2), and (6,-6). which type of function would best model the data set? exponential linear quadratic

Explanation:

Step1: Recall characteristics of functions

• Exponential function: Has a constant ratio between consecutive \(y\) - values for equally - spaced \(x\) - values. For example, if \(y = a\cdot b^{x}\), then \(\frac{y_{n + 1}}{y_{n}}=b\) (constant) for \(x_{n+1}-x_{n}=\text{constant}\).
• Linear function: Has a constant slope \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\).
• Quadratic function: The second - differences of \(y\) - values for equally - spaced \(x\) - values are constant.

Step2: Calculate slopes for linearity check

Let \((x_1,y_1)=(-4,-14)\), \((x_2,y_2)=(-2,-7)\), \((x_3,y_3)=(0, - 3)\), \((x_4,y_4)=(2,-2)\), \((x_5,y_5)=(6,-6)\)
The slope formula is \(m=\frac{y_{i + 1}-y_{i}}{x_{i+1}-x_{i}}\)
For \(i = 1\): \(m_1=\frac{-7-(-14)}{-2-(-4)}=\frac{-7 + 14}{-2 + 4}=\frac{7}{2}=3.5\)
For \(i = 2\): \(m_2=\frac{-3-(-7)}{0-(-2)}=\frac{-3 + 7}{0 + 2}=\frac{4}{2}=2\)
Since \(m_1
eq m_2\), it is not a linear function.

Step3: Check for exponentiality

For \(x=-4,y=-14\); \(x=-2,y = - 7\). If \(y=a\cdot b^{x}\), then \(\frac{y(-2)}{y(-4)}=\frac{-7}{-14}=\frac{1}{2}\). For \(x=-2,y=-7\) and \(x = 0,y=-3\), \(\frac{y(0)}{y(-2)}=\frac{-3}{-7}=\frac{3}{7}
eq\frac{1}{2}\). So, it is not an exponential function.

Step4: Check for quadraticity

First, list the \(x\) and \(y\) values:

\(x\)	\(y\)	First - differences (\(\Delta y\))	Second - differences (\(\Delta^2y\))
\(-2\)	\(-7\)	\(-7-(-14)=7\)
\(0\)	\(-3\)	\(-3-(-7) = 4\)	\(4 - 7=-3\)
\(2\)	\(-2\)	\(-2-(-3)=1\)	\(1 - 4=-3\)
\(6\)	\(-6\)	\(-6-(-2)=-4\)	\(-4 - 1=-5\) (Note: the \(x\) - values from \(2\) to \(6\) have a step of \(4\) instead of \(2\), but if we consider the general trend of non - constant first differences and the fact that for equally - spaced \(x\) (with step \(h = 2\)) the first two second - differences are constant (\(-3\)) before the non - standard \(x\) - step. A quadratic function \(y=ax^{2}+bx + c\) has second - differences \(2ah^{2}\) (constant for constant \(h\)).

Answer:

quadratic