Question

Use the table to answer the question.

This table shows the corresponding y-values for a given set of x-values.
x-values y-values
−1
6
0 5
1 6
2 9
3 14
Create a quadratic model for the dataset in the xy
-chart.

(1 point)
Responses

f(x)=2x+5
f left parenthesis x right parenthesis equals 2 x plus 5

f(x)=x+5
f left parenthesis x right parenthesis equals x plus 5

f(x)=x^2+5
f left parenthesis x right parenthesis equals x squared plus 5

f(x)=x^2+7
f left parenthesis x right parenthesis equals x squared plus 7
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Answer 1

To create a quadratic model for the dataset, we need to find a quadratic function of the form \( f(x) = ax^2 + bx + c \) that fits the given x-values and y-values.

The x-values are:
-1, 0, 1, 2, 3

And the corresponding y-values are:
6, 5, 6, 9, 14

We'll plug in these points into the general form \( f(x) = ax^2 + bx + c \) to create a system of equations.

1. For \( x = 0 \):
\( f(0) = c = 5 \)

2. For \( x = -1 \):
\( f(-1) = a(-1)^2 + b(-1) + c = 6 \)
\( a - b + 5 = 6 \)
\( a - b = 1 \) → (1)

3. For \( x = 1 \):
\( f(1) = a(1)^2 + b(1) + c = 6 \)
\( a + b + 5 = 6 \)
\( a + b = 1 \) → (2)

4. For \( x = 2 \):
\( f(2) = a(2)^2 + b(2) + c = 9 \)
\( 4a + 2b + 5 = 9 \)
\( 4a + 2b = 4 \)
\( 2a + b = 2 \) → (3)

Now, we have a system of three equations:
1. \( a - b = 1 \)
2. \( a + b = 1 \)
3. \( 2a + b = 2 \)

From (1) and (2):
Solving \( a - b = 1 \) and \( a + b = 1 \):
Adding both equations:
\( (a - b) + (a + b) = 1 + 1 \)
\( 2a = 2 \)
\( a = 1 \)

Using \( a = 1 \) in (2):
\( 1 + b = 1 \)
\( b = 0 \)

Using \( a = 1 \) and \( b = 0 \) in (1):
\( 1 - 0 = 1 \) (checks out)

Now we have:
\( a = 1 \), \( b = 0 \), \( c = 5 \)

Thus, the quadratic model is:
\[ f(x) = 1x^2 + 0x + 5 = x^2 + 5 \]

The answer from the provided options is:
**f(x) = x^2 + 5**.