To create a quadratic model for the provided dataset, we have to analyze the given x-values and corresponding y-values.
Let's represent the function as \( f(x) = ax^2 + bx + c \).
Using the given points, we have:
- For \( x = 0 \), \( f(0) = 5 \) → \( c = 5 \)
- For \( x = 1 \), \( f(1) = 6 \) → \( a(1^2) + b(1) + c = 6 \)
- For \( x = 2 \), \( f(2) = 9 \) → \( a(2^2) + b(2) + c = 9 \)
- For \( x = 3 \), \( f(3) = 14 \) → \( a(3^2) + b(3) + c = 14 \)
Since \( c = 5 \), we can substitute this into the equations:
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From \( f(1) = 6 \): \[ a(1^2) + b(1) + 5 = 6 \implies a + b + 5 = 6 \implies a + b = 1 \tag{1} \]
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From \( f(2) = 9 \): \[ a(2^2) + b(2) + 5 = 9 \implies 4a + 2b + 5 = 9 \implies 4a + 2b = 4 \implies 2a + b = 2 \tag{2} \]
Now, we have a system of two equations: \[ \begin{align*} a + b &= 1 \quad \text{(1)}\ 2a + b &= 2 \quad \text{(2)} \end{align*} \]
Subtract equation (1) from equation (2): \[ (2a + b) - (a + b) = 2 - 1 \implies 2a - a = 1 \implies a = 1 \]
Now substitute \( a = 1 \) back into equation (1): \[ 1 + b = 1 \implies b = 0 \]
Thus, we have \( a = 1 \), \( b = 0 \), and \( c = 5. \)
The quadratic model is: \[ f(x) = 1x^2 + 0x + 5 = x^2 + 5. \]
Therefore, the correct response is:
f(x) = x^2 + 5