Question

Use the image to answer the question.

A coordinate plane shows a parabola that opens downward. The x-axis ranges from 0 to 15 in increments of 5. The y-axis ranges from 0 to 15 in increments of 5.

The graph shows the trajectory of a firework. Which equation in vertex form matches this graph?

(1 point)
Responses

y=−12(x−5)2+10
y equals negative Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared plus 10

y=12(x−5)2+10
y equals Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared plus 10

y=12(x+5)2+10
y equals Start Fraction 1 over 2 End Fraction left parenthesis x plus 5 right parenthesis squared plus 10

y=−12(x−5)2−10

Answer 1

To identify the correct equation for the parabola that represents the trajectory of the firework in vertex form, we need to look for the following characteristics:

1. **Direction of Opening**: The parabola opens downward, which indicates that the coefficient of the squared term (in front of the parentheses) should be negative.

2. **Vertex**: The vertex form of a parabola is written as \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. Based on your description, the vertex seems to be at \((5, 10)\), given that it is the highest point of a downward-opening parabola.

Given these considerations, let’s analyze the given options:

- **Option 1**: \(y = -12(x - 5)^2 + 10\) (This opens downward and has vertex \((5, 10)\)).
- **Option 2**: \(y = \frac{1}{2}(x - 5)^2 + 10\) (This opens upward).
- **Option 3**: \(y = \frac{1}{2}(x + 5)^2 + 10\) (This opens upward and has a vertex at \((-5, 10)\)).
- **Option 4**: \(y = -12(x - 5)^2 - 10\) (This opens downward but has a vertex at \((5, -10)\)).

Based on this analysis, only **Option 1** \(y = -12(x - 5)^2 + 10\) fits the characteristics of the parabola showing downward opening with vertex at \((5, 10)\).

Therefore, the correct answer is:

**y = -12(x - 5)^2 + 10**.