Let's walk through the process of estimating [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] based on the provided data points [tex]\((x_i, y_i)\)[/tex]:
1. Fit a Polynomial to the Data:
The data suggests a nonlinear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Given the values provided, we use a quadratic polynomial (a second-degree polynomial) to model this relationship.
2. Determine the Polynomial Coefficients:
The quadratic polynomial can be expressed as:
[tex]\[
y = ax^2 + bx + c
\][/tex]
Using the provided data points:
[tex]\[
\begin{align*}
(2.5, 6.25) \\
(9.4, 88.36) \\
(15.6, 243.63) \\
(19.5, 380.25) \\
(25.8, 665.64)
\end{align*}
\][/tex]
the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] can be found through polynomial regression. These coefficients are approximately:
[tex]\[
a \approx 0.9990, \quad b \approx 0.0286, \quad c \approx -0.0888
\][/tex]
3. Form the Polynomial Equation:
Substituting the coefficients, the quadratic polynomial becomes:
[tex]\[
y \approx 0.9990x^2 + 0.0286x - 0.0888
\][/tex]
4. Calculate [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex]:
Plugging [tex]\( x = 4 \)[/tex] into the polynomial equation gives:
[tex]\[
\begin{align*}
y & \approx 0.9990(4)^2 + 0.0286(4) - 0.0888 \\
y & \approx 0.9990 \cdot 16 + 0.0286 \cdot 4 - 0.0888 \\
y & \approx 15.984 + 0.1144 - 0.0888 \\
y & \approx 16.0098
\end{align*}
\][/tex]
5. Select the Closest Approximate Value:
The calculated value [tex]\( y \approx 16.0098 \)[/tex] is closest to option B.
Therefore, the approximate value of [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] is:
[tex]\[
\boxed{16}
\][/tex]
