Question

Points on the grid/ graph: ( -1.5, 6.75) (0,6) (3,0) and (-6,0).

The graph of a quadratic function is shown on the grid. Which function is best represented by the graph?

A: f(x) = -x^2 + 12

B: f(x) = -1/3x^2 + 12

C: f(x) = -x^2 -x +6

D: f(x) = -1/3x^2 -x + 6

Answer 1

To determine which quadratic function best matches the provided points, we can evaluate each candidate function at the x-values of the points and compare the respective outputs.

The points to evaluate against are:
1. (-1.5, 6.75)
2. (0, 6)
3. (3, 0)
4. (-6, 0)

Let's evaluate each function step-by-step for each of the candidate functions.

### A: \( f(x) = -x^2 + 12 \)

1. For \( x = -1.5 \):
\[
f(-1.5) = -(-1.5)^2 + 12 = -2.25 + 12 = 9.75 \quad \text{(not 6.75)}
\]
2. For \( x = 0 \):
\[
f(0) = -0^2 + 12 = 12 \quad \text{(not 6)}
\]
3. For \( x = 3 \):
\[
f(3) = -3^2 + 12 = -9 + 12 = 3 \quad \text{(not 0)}
\]
4. For \( x = -6 \):
\[
f(-6) = -(-6)^2 + 12 = -36 + 12 = -24 \quad \text{(not 0)}
\]

### B: \( f(x) = -\frac{1}{3}x^2 + 12 \)

1. For \( x = -1.5 \):
\[
f(-1.5) = -\frac{1}{3}(-1.5)^2 + 12 = -\frac{1}{3}(2.25) + 12 = -0.75 + 12 = 11.25 \quad \text{(not 6.75)}
\]
2. For \( x = 0 \):
\[
f(0) = -\frac{1}{3}(0)^2 + 12 = 12 \quad \text{(not 6)}
\]
3. For \( x = 3 \):
\[
f(3) = -\frac{1}{3}(3)^2 + 12 = -\frac{1}{3}(9) + 12 = -3 + 12 = 9 \quad \text{(not 0)}
\]
4. For \( x = -6 \):
\[
f(-6) = -\frac{1}{3}(-6)^2 + 12 = -\frac{1}{3}(36) + 12 = -12 + 12 = 0 \quad \text{(works)}
\]

### C: \( f(x) = -x^2 - x + 6 \)

1. For \( x = -1.5 \):
\[
f(-1.5) = -(-1.5)^2 - (-1.5) + 6 = -2.25 + 1.5 + 6 = 5.25 \quad \text{(not 6.75)}
\]
2. For \( x = 0 \):
\[
f(0) = -0^2 - 0 + 6 = 6 \quad \text{(works)}
\]
3. For \( x = 3 \):
\[
f(3) = -3^2 - 3 + 6 = -9 - 3 + 6 = -6 \quad \text{(not 0)}
\]
4. For \( x = -6 \):
\[
f(-6) = -(-6)^2 - (-6) + 6 = -36 + 6 + 6 = -24 \quad \text{(not 0)}
\]

### D: \( f(x) = -\frac{1}{3}x^2 - x + 6 \)

1. For \( x = -1.5 \):
\[
f(-1.5) = -\frac{1}{3}(-1.5)^2 - (-1.5) + 6 = -\frac{1}{3}(2.25) + 1.5 + 6 = -0.75 + 1.5 + 6 = 6.75 \quad \text{(works)}
\]
2. For \( x = 0 \):
\[
f(0) = -\frac{1}{3}(0)^2 - 0 + 6 = 6 \quad \text{(works)}
\]
3. For \( x = 3 \):
\[
f(3) = -\frac{1}{3}(3)^2 - 3 + 6 = -\frac{1}{3}(9) - 3 + 6 = -3 - 3 + 6 = 0 \quad \text{(works)}
\]
4. For \( x = -6 \):
\[
f(-6) = -\frac{1}{3}(-6)^2 - (-6) + 6 = -\frac{1}{3}(36) + 6 + 6 = -12 + 6 + 6 = 0 \quad \text{(works)}
\]

### Conclusion
The function that matches all the points is:
**D: \( f(x) = -\frac{1}{3}x^2 - x + 6 \)**.