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Select the correct answer.

This table represents a quadratic function.
[tex]\[
\begin{tabular}{|c|c|}
\hline $x$ & $y$ \\
\hline 0 & -3 \\
\hline 1 & -3.75 \\
\hline 2 & -4 \\
\hline 3 & -3.75 \\
\hline 4 & -3 \\
\hline 5 & -1.75 \\
\hline
\end{tabular}
\][/tex]

What is the value of [tex]$a$[/tex] in the function's equation?

A. [tex]$\frac{1}{4}$[/tex]

B. [tex]$-\frac{1}{2}$[/tex]

C. [tex]$-\frac{1}{4}$[/tex]

D. [tex]$\frac{1}{2}$[/tex]


Sagot :

To determine the value of [tex]\(a\)[/tex] in the quadratic function that fits the given data points, follow these steps:

1. Identify the data points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -3 \\ \hline 1 & -3.75 \\ \hline 2 & -4 \\ \hline 3 & -3.75 \\ \hline 4 & -3 \\ \hline 5 & -1.75 \\ \hline \end{array} \][/tex]

2. Fit a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] to these data points.

3. Determine the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].

From the problem's calculations, we have:

- The quadratic polynomial that fits the data points is represented with coefficients approximately found.

- The coefficient for [tex]\(x^2\)[/tex], denoted as [tex]\(a\)[/tex], is approximately [tex]\(0.25\)[/tex].

4. Match the calculated value of [tex]\(a\)[/tex] with the provided options:

[tex]\[ a \approx 0.25 \][/tex]

5. Identify the correct option:

[tex]\[ \begin{aligned} \text{A. } & \frac{1}{4} \\ \text{B. } & -\frac{1}{2} \\ \text{C. } & -\frac{1}{4} \\ \text{D. } & \frac{1}{2} \end{aligned} \][/tex]

Since [tex]\(a \approx 0.25\)[/tex], it corresponds to [tex]\(\frac{1}{4}\)[/tex].

Therefore, the correct answer is [tex]\(\boxed{A}\)[/tex].
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