These tables represent a quadratic function with a vertex at [tex]\((0,3)\)[/tex]. What is the average rate of change for the interval from [tex]\(x=7\)[/tex] to [tex]\(x=8\)[/tex]?

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 3 \\
\hline
1 & 2 \\
\hline
2 & -1 \\
\hline
3 & -6 \\
\hline
4 & -13 \\
\hline
5 & -22 \\
\hline
6 & -33 \\
\hline
\end{array}
\][/tex]

[tex]\[
\begin{array}{|c|c|}
\hline
\text{Interval} & \text{Average rate of change} \\
\hline
0 \text{ to } 1 & -1 \\
\hline
1 \text{ to } 2 & -3 \\
\hline
2 \text{ to } 3 & -5 \\
\hline
3 \text{ to } 4 & -7 \\
\hline
4 \text{ to } 5 & -9 \\
\hline
5 \text{ to } 6 & -11 \\
\hline
\end{array}
\][/tex]

A. [tex]\(-61\)[/tex]

B. [tex]\(-2\)[/tex]

C. [tex]\(-15\)[/tex]

D. [tex]\(-46\)[/tex]



Answer :

To find the average rate of change for the interval from [tex]\(x = 7\)[/tex] to [tex]\(x = 8\)[/tex] in a quadratic function with a given vertex at [tex]\((0, 3)\)[/tex], we can follow these steps:

1. Calculate [tex]\( y \)[/tex] values at [tex]\( x = 7 \)[/tex] and [tex]\( x = 8 \)[/tex]:
- A quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] will have specific [tex]\( y \)[/tex]-values at these points. For this problem, they are:

- When [tex]\( x = 7 \)[/tex]:
[tex]\[ y_1 = -(7^2) + 3 = -49 + 3 = -46 \][/tex]

- When [tex]\( x = 8 \)[/tex]:
[tex]\[ y_2 = -(8^2) + 3 = -64 + 3 = -61 \][/tex]

2. Apply the average rate of change formula:
The formula to calculate the average rate of change between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] on the function is:
[tex]\[ \text{Average rate of change} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

Plugging in the values we have:
[tex]\[ \frac{y_2 - y_1}{x_2 - x_1} = \frac{-61 - (-46)}{8 - 7} = \frac{-61 + 46}{1} = \frac{-15}{1} = -15 \][/tex]

Thus, the average rate of change for the interval from [tex]\( x = 7 \)[/tex] to [tex]\( x = 8 \)[/tex] is:
[tex]\[ \boxed{-15} \][/tex]

Option C. -15 is the correct answer.