To create a quadratic model for the dataset, we can use the given x-values and y-values to find a quadratic equation of the form \( f(x) = ax^2 + bx + c \).
We have the following points from the table:
- (-1, 6)
- (0, 5)
- (1, 6)
- (2, 9)
- (3, 14)
We can start by observing how the y-values change with respect to the x-values.
At \( x = 0 \), the y-value is 5, which suggests that when \( x = 0 \), \( f(0) = c = 5 \).
Next, we can calculate the first differences (the differences between consecutive y-values):
- From \( 5 \) to \( 6 \) (at \( x = 1 \)): \( 6 - 5 = 1 \)
- From \( 6 \) to \( 9 \) (at \( x = 2 \)): \( 9 - 6 = 3 \)
- From \( 9 \) to \( 14 \) (at \( x = 3 \)): \( 14 - 9 = 5 \)
Now, let's calculate the second differences (the differences of the first differences):
- From \( 1 \) to \( 3 \): \( 3 - 1 = 2 \)
- From \( 3 \) to \( 5 \): \( 5 - 3 = 2 \)
The second differences are constant and equal to \( 2 \), which confirms that a quadratic function is appropriate for this data set.
From the value of the second differences, we can conclude that the coefficient \( a \) in our quadratic equation \( ax^2 + bx + c \) can be determined as follows:
Since the second difference is \( 2 \), we have \( 2a = 2 \) => \( a = 1 \).
So we can start building our quadratic formula with \( a = 1 \):
Now, we already know \( c = 5 \):
Thus our equation starts as:
\[ f(x) = x^2 + bx + 5 \]
To find \( b \), we can use any of the given points. Let's use the point (1, 6):
\[ f(1) = 1^2 + b(1) + 5 = 6 \] \[ 1 + b + 5 = 6 \] \[ b + 6 = 6 \] \[ b = 0 \]
So we have determined the quadratic function:
\[ f(x) = x^2 + 0x + 5 = x^2 + 5 \]
So, the model we arrived at is:
f(x) = x² + 5
Thus, the correct response from your options provided would be:
f(x)=x²+5