To find the value of \( a \) in the quadratic function \( y = ax^2 + bx + c \), we'll use the given table of values.
We have these 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]
First, we use the point \((0, -3)\):
[tex]\[
y = ax^2 + bx + c \\
-3 = a(0)^2 + b(0) + c \Rightarrow c = -3
\][/tex]
Now, let's use the points \((1, -3.75)\) and \((2, -4)\).
For \((1, -3.75)\):
[tex]\[
-3.75 = a(1)^2 + b(1) + c \\
-3.75 = a + b - 3 \\
a + b - 3 = -3.75 \Rightarrow a + b = -0.75
\][/tex]
For \((2, -4)\):
[tex]\[
-4 = a(2)^2 + b(2) + c \\
-4 = 4a + 2b - 3 \\
4a + 2b - 3 = -4 \Rightarrow 4a + 2b = -1
\][/tex]
Now we have the system of linear equations:
[tex]\[
1) \quad a + b = -0.75 \\
2) \quad 4a + 2b = -1
\][/tex]
To solve this system, we can use substitution. From the first equation:
[tex]\[
a + b = -0.75 \Rightarrow b = -0.75 - a
\][/tex]
Substitute this into the second equation:
[tex]\[
4a + 2(-0.75 - a) = -1 \\
4a - 1.5 - 2a = -1 \\
2a - 1.5 = -1 \\
2a = 0.5 \\
a = 0.25
\][/tex]
Hence, the value of \( a \) in the function's equation is:
[tex]\[
\boxed{\frac{1}{4}}
\][/tex]
So, the correct answer is:
D. [tex]\(\frac{1}{4}\)[/tex]
