Look at this table:

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-8 & -576.64 \\
\hline
-7 & -441.49 \\
\hline
-6 & -324.36 \\
\hline
-5 & -225.25 \\
\hline
-4 & -144.16 \\
\hline
\end{array}
\][/tex]

Write a linear [tex]\([y = mx + b]\)[/tex], quadratic [tex]\(\left[y = ax^2 \right]\)[/tex], or exponential [tex]\(\left[y = a(b)^x\right]\)[/tex] function that models the data.

[tex]\[ y = \][/tex]
[tex]\[ \square \][/tex]



Answer :

Given the data points:

| [tex]\( x \)[/tex] | [tex]\( y \)[/tex] |
|---------|-------------|
| -8 | -576.64 |
| -7 | -441.49 |
| -6 | -324.36 |
| -5 | -225.25 |
| -4 | -144.16 |

We are looking for a function in the form [tex]\( y = a x^2 \)[/tex] that fits this data.

To determine the value of [tex]\( a \)[/tex] in the quadratic model:

1. Take any data point from the table. For example, let's use the data point [tex]\((-8, -576.64)\)[/tex].
2. Since we assume the model [tex]\( y = a x^2 \)[/tex], substitute [tex]\( x = -8 \)[/tex] and [tex]\( y = -576.64 \)[/tex] into the equation:

[tex]\[ -576.64 = a \cdot (-8)^2 \][/tex]

3. Simplify the equation by calculating [tex]\((-8)^2 = 64\)[/tex]:

[tex]\[ -576.64 = a \cdot 64 \][/tex]

4. Solve for [tex]\( a \)[/tex]:

[tex]\[ a = \frac{-576.64}{64} \][/tex]
[tex]\[ a \approx -9.01 \][/tex]

Therefore, the quadratic function that models the given data is:

[tex]\[ y = -9.01 x^2 \][/tex]