To fit a quadratic function to the given three points [tex]\((-1, -11)\)[/tex], [tex]\((0, -3)\)[/tex], and [tex]\((3, -27)\)[/tex], we need to determine the quadratic function of the form [tex]\(y = ax^2 + bx + c\)[/tex]. Here’s a detailed, step-by-step solution:
1. Set up the system of equations:
Given three points, we can substitute these into the quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex].
For the point [tex]\((-1, -11)\)[/tex]:
[tex]\[
-11 = a(-1)^2 + b(-1) + c \implies -11 = a - b + c \quad \text{(1)}
\][/tex]
For the point [tex]\((0, -3)\)[/tex]:
[tex]\[
-3 = a(0)^2 + b(0) + c \implies -3 = c \quad \text{(2)}
\][/tex]
For the point [tex]\((3, -27)\)[/tex]:
[tex]\[
-27 = a(3)^2 + b(3) + c \implies -27 = 9a + 3b + c \quad \text{(3)}
\][/tex]
2. Solve the system of equations:
From equation (2), we immediately get:
[tex]\[
c = -3
\][/tex]
Substitute [tex]\(c = -3\)[/tex] into equations (1) and (3):
[tex]\[
-11 = a - b - 3 \implies a - b = -8 \quad \text{(4)}
\][/tex]
[tex]\[
-27 = 9a + 3b - 3 \implies 9a + 3b = -24 \quad \text{(5)}
\][/tex]
Simplify equation (5):
[tex]\[
3a + b = -8 \quad \text{(6)}
\][/tex]
Now we have two linear equations:
[tex]\[
a - b = -8 \quad \text{(4)}
\][/tex]
[tex]\[
3a + b = -8 \quad \text{(6)}
\][/tex]
Add equations (4) and (6):
[tex]\[
(a - b) + (3a + b) = -8 + (-8) \implies 4a = -16 \implies a = -4
\][/tex]
Substitute [tex]\(a = -4\)[/tex] back into equation (4):
[tex]\[
-4 - b = -8 \implies b = 4
\][/tex]
So, we have [tex]\(a = -4\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -3\)[/tex].
3. Form the quadratic equation:
The quadratic function that fits the points is:
[tex]\[
y = -4x^2 + 4x - 3
\][/tex]
4. Identify the correct option:
The quadratic function [tex]\(y = -4x^2 + 4x - 3\)[/tex] corresponds to the last option.
Therefore, the correct quadratic function is:
[tex]\[
\boxed{y = -4x^2 + 4x - 3}
\][/tex]
And the option number that matches this function is option 4.
