5.7.3 Quiz: Linear, Quadratic, and Exponential

These tables represent a quadratic function with a vertex at [tex]$(0, -1)$[/tex]. What is the average rate of change for the interval from [tex]$x=9$[/tex] to [tex]$x=10$[/tex]?

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
0 & -1 \\
\hline
1 & -2 \\
\hline
2 & -5 \\
\hline
3 & -10 \\
\hline
4 & -17 \\
\hline
5 & -26 \\
\hline
6 & -37 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline
Interval & \begin{tabular}{c}
Average rate \\
of change
\end{tabular} \\
\hline
0 to 1 & -1 \\
\hline
1 to 2 & -3 \\
\hline
2 to 3 & -5 \\
\hline
3 to 4 & -7 \\
\hline
4 to 5 & -9 \\
\hline
5 to 6 & -11 \\
\hline
\end{tabular}



Answer :

To determine the average rate of change of the quadratic function for the interval from [tex]\( x = 9 \)[/tex] to [tex]\( x = 10 \)[/tex], follow these steps:

1. Identify the [tex]$y$[/tex]-values for the given [tex]$x$[/tex]-values:
- Given: [tex]\( x = 9 \)[/tex] and [tex]\( x = 10 \)[/tex].
- From the table, the [tex]$y$[/tex]-values corresponding to the given [tex]$x$[/tex]-values:
[tex]\[ (9, -82) \quad \text{and} \quad (10, -101) \][/tex]

2. Calculate the change in [tex]$y$[/tex] ([tex]\(\Delta y\)[/tex]) and the change in [tex]$x$[/tex] ([tex]\(\Delta x\)[/tex]):
[tex]\[ \Delta y = y_{final} - y_{initial} = -101 - (-82) \][/tex]
Simplifying:
[tex]\[ \Delta y = -101 + 82 = -19 \][/tex]
[tex]\[ \Delta x = x_{final} - x_{initial} = 10 - 9 = 1 \][/tex]

3. Calculate the average rate of change:
[tex]\[ \text{Average rate of change} = \frac{\Delta y}{\Delta x} = \frac{-19}{1} = -19 \][/tex]

So, the average rate of change for the interval from [tex]\( x = 9 \)[/tex] to [tex]\( x = 10 \)[/tex] is [tex]\(\boxed{-19}\)[/tex].