To find the value of [tex]\( a \)[/tex] in the quadratic function, we start by recalling the general form of a quadratic equation:
[tex]\[ y = ax^2 + bx + c \][/tex]
Given the table of values, we have six points: [tex]\((0, -3)\)[/tex], [tex]\((1, -3.75)\)[/tex], [tex]\((2, -4)\)[/tex], [tex]\((3, -3.75)\)[/tex], [tex]\((4, -3)\)[/tex], and [tex]\((5, -1.75)\)[/tex].
We can use three of these points to set up a system of equations to solve for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
First, we use the point [tex]\((0, -3)\)[/tex]:
[tex]\[ -3 = a(0)^2 + b(0) + c \][/tex]
This simplifies to:
[tex]\[ c = -3 \][/tex]
Next, we use the point [tex]\((1, -3.75)\)[/tex]:
[tex]\[ -3.75 = a(1)^2 + b(1) - 3 \][/tex]
[tex]\[ -3.75 = a + b - 3 \][/tex]
[tex]\[ a + b = -0.75 \][/tex]
Then, we use the point [tex]\((2, -4)\)[/tex]:
[tex]\[ -4 = a(2)^2 + b(2) - 3 \][/tex]
[tex]\[ -4 = 4a + 2b - 3 \][/tex]
[tex]\[ 4a + 2b = -1 \][/tex]
We now have the system of equations:
1. [tex]\( c = -3 \)[/tex]
2. [tex]\( a + b = -0.75 \)[/tex]
3. [tex]\( 4a + 2b = -1 \)[/tex]
Since [tex]\( c \)[/tex] has already been determined to be [tex]\(-3\)[/tex], we solve for [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
We can simplify the third equation:
[tex]\[ 4a + 2b = -1 \][/tex]
[tex]\[ 2a + b = -0.5 \][/tex]
We have two equations now:
1. [tex]\( a + b = -0.75 \)[/tex]
2. [tex]\( 2a + b = -0.5 \)[/tex]
Subtract the first equation from the second:
[tex]\[ (2a + b) - (a + b) = -0.5 - (-0.75) \][/tex]
[tex]\[ 2a + b - a - b = -0.5 + 0.75 \][/tex]
[tex]\[ a = 0.25 \][/tex]
Thus, the value of [tex]\( a \)[/tex] is [tex]\( 0.25 \)[/tex].
Therefore, the correct answer is:
D. [tex]\(\frac{1}{4}\)[/tex]
