Answer :
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]
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]
