To create a quadratic model for the given dataset in the xy-chart, we can use the data points provided and identify a quadratic function of the form \( f(x) = ax^2 + bx + c \).
Based on the data:
- At \( x = 0 \), \( y = 5 \)
- At \( x = 1 \), \( y = 6 \)
- At \( x = 2 \), \( y = 9 \)
- At \( x = 3 \), \( y = 14 \)
We can see that the y-values are not consistent with a linear function and suggest a quadratic relation.
To find a quadratic function, we can use the points to set up a system of equations. Let's take three points and plug them into the general form:
- For \( x = 0 \), \( f(0) = c = 5 \)
- For \( x = 1 \), \( f(1) = a(1)^2 + b(1) + c = 6 \)
- For \( x = 2 \), \( f(2) = a(2^2) + b(2) + c = 9 \)
This leads to the following equations:
- \( c = 5 \)
- \( a + b + 5 = 6 \) → \( a + b = 1 \) (Equation 1)
- \( 4a + 2b + 5 = 9 \) → \( 4a + 2b = 4 \) → \( 2a + b = 2 \) (Equation 2)
Now we have two equations:
- Equation 1: \( a + b = 1 \)
- Equation 2: \( 2a + b = 2 \)
We can solve these equations. First, subtract Equation 1 from Equation 2:
\[ (2a + b) - (a + b) = 2 - 1 \ 2a - a = 1 \ a = 1 \]
Now that we have \( a \), we can find \( b \):
Using Equation 1: \[ 1 + b = 1 \ b = 0 \]
Thus, we have \( a = 1 \), \( b = 0 \), and \( c = 5 \). The quadratic model is:
\[ f(x) = 1x^2 + 0x + 5 \ f(x) = x^2 + 5 \]
So, the final answer is:
\( f(x) = x^2 + 5 \)
This matches one of the options given:
f(x) = x² + 5