To create a quadratic model for the given dataset, we need to find a quadratic function of the form \( f(x) = ax^2 + bx + c \) that fits the data points.
Let's analyze the data:
- For \( x = -1 \), \( y = 6 \)
- For \( x = 0 \), \( y = 5 \)
- For \( x = 1 \), \( y = 6 \)
- For \( x = 2 \), \( y = 9 \)
- For \( x = 3 \), \( y = 14 \)
We can start by substituting the known points into the quadratic equation and solving for the coefficients \(a\), \(b\), and \(c\).
Using points:
- \( (-1, 6) \): \( 6 = a(-1)^2 + b(-1) + c \) → \( 6 = a - b + c \) → (Equation 1)
- \( (0, 5) \): \( 5 = a(0)^2 + b(0) + c \) → \( 5 = c \) → (Equation 2)
- \( (1, 6) \): \( 6 = a(1)^2 + b(1) + c \) → \( 6 = a + b + c \) → (Equation 3)
From Equation 2, we have \( c = 5 \). Substituting this into Equations 1 and 3 gives us:
-
For Equation 1: \( 6 = a - b + 5 \) → \( a - b = 1 \) → (Equation 4)
-
For Equation 3: \( 6 = a + b + 5 \) → \( a + b = 1 \) → (Equation 5)
Now we can solve Equations 4 and 5: From Equation 4 \( a - b = 1 \) and from Equation 5 \( a + b = 1 \).
Adding these two equations: \( (a - b) + (a + b) = 1 + 1 \) → \( 2a = 2 \) → \( a = 1 \).
Substituting \( a = 1 \) back into Equation 5: \( 1 + b = 1 \) → \( b = 0 \).
Now we have \( a = 1, b = 0, c = 5 \). Thus, the quadratic model is: \[ f(x) = x^2 + 5 \]
Looking at the responses provided:
- \( f(x) = x^2 + 5 \)
This is the correct quadratic model for the dataset.
The correct response is: f(x) = x² + 5