To determine the quadratic regression equation that best fits the given data set, we need to find the equation of the form [tex]\( y = Ax^2 + Bx + C \)[/tex] where [tex]\( y \)[/tex] is the height in feet and [tex]\( x \)[/tex] is the number of seconds. Here are the steps involved in solving this:
1. Define the Data Points:
We begin with the following data points:
[tex]\[
\begin{array}{|c|c|}
\hline \text{Number of seconds } (x) & \text{Height (in feet) } (y) \\
\hline 0 & 12 \\
\hline 1 & 14 \\
\hline 2 & 15 \\
\hline 3 & 14 \\
\hline 4 & 10 \\
\hline 5 & 6 \\
\hline
\end{array}
\][/tex]
2. Fit a Quadratic Regression Model:
Utilizing regression techniques, we can determine the coefficients [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] that minimize the difference between the observed values [tex]\( y \)[/tex] and the values predicted by the quadratic model. The coefficients from this fitting are approximately [tex]\( A = -0.89285714 \)[/tex], [tex]\( B = 3.23571429 \)[/tex], and [tex]\( C = 11.92857143 \)[/tex].
3. Formulate the Quadratic Equation:
Substituting the values of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] into the quadratic model [tex]\( y = Ax^2 + Bx + C \)[/tex], we get the following equation:
[tex]\[
y = -0.89285714x^2 + 3.23571429x + 11.92857143
\][/tex]
Rounding these values to two decimal places, the equation becomes approximately:
[tex]\[
y = -0.89x^2 + 3.24x + 11.93
\][/tex]
4. Verify the Options:
Comparing this derived equation to the provided options:
A. [tex]\( y = -0.89x^2 + 3.24x + 11.93 \)[/tex]
B. [tex]\( y = -1.94x^2 + 0.62x + 9.62 \)[/tex]
C. [tex]\( y = 15.68(0.88^x) \)[/tex]
D. [tex]\( y = -3.69x^2 + 2.08x + 8.07 \)[/tex]
The closest match is option A.
Thus, the correct equation that fits the data is:
[tex]\[ \boxed{y = -0.89x^2 + 3.24x + 11.93} \][/tex]
