Question

Write the function given the input-output table.

x f(x)
-2 -11
-1 -2
0 1

Answer 1

To find a function \( f(x) \) that fits the given input-output table, we can look for a polynomial function. Since we have three points, we can use a quadratic function in the form:

\[
f(x) = ax^2 + bx + c
\]

We will use the given points to create a system of equations.

### Given points:

1. \( f(-2) = -11 \)
2. \( f(-1) = -2 \)
3. \( f(0) = 1 \)

### Create equations:

From the first point:
\[
f(-2) = a(-2)^2 + b(-2) + c = -11 \implies 4a - 2b + c = -11 \quad \text{(1)}
\]

From the second point:
\[
f(-1) = a(-1)^2 + b(-1) + c = -2 \implies a - b + c = -2 \quad \text{(2)}
\]

From the third point:
\[
f(0) = c = 1 \quad \text{(3)}
\]

### Substitute \( c \) from Equation (3) into Equations (1) and (2):

Substituting \( c = 1 \) into Equation (1):
\[
4a - 2b + 1 = -11 \implies 4a - 2b = -12 \implies 2a - b = -6 \quad \text{(4)}
\]

Substituting \( c = 1 \) into Equation (2):
\[
a - b + 1 = -2 \implies a - b = -3 \quad \text{(5)}
\]

### Now, we have two equations:

1. From Equation (4): \( 2a - b = -6 \)
2. From Equation (5): \( a - b = -3 \)

### Solve this system of equations:

Subtract Equation (5) from Equation (4):
\[
(2a - b) - (a - b) = -6 - (-3)
\]
\[
2a - b - a + b = -6 + 3
\]
\[
a = -3
\]

Substituting \( a = -3 \) back into Equation (5):
\[
-3 - b = -3 \implies -b = 0 \implies b = 0
\]

### Now we have \( a \), \( b \), and \( c \):
- \( a = -3 \)
- \( b = 0 \)
- \( c = 1 \)

### Therefore, the function is:
\[
f(x) = -3x^2 + 1
\]

### Verification:

1. \( f(-2) = -3(-2)^2 + 1 = -3(4) + 1 = -12 + 1 = -11 \) ✅
2. \( f(-1) = -3(-1)^2 + 1 = -3(1) + 1 = -3 + 1 = -2 \) ✅
3. \( f(0) = -3(0)^2 + 1 = 0 + 1 = 1 \) ✅

Thus, the function \( f(x) \) that fits the input-output table is:

\[
\boxed{f(x) = -3x^2 + 1}
\]