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 Start Fraction 1 over 2 End Fraction left parenthesis x plus 5 right parenthesis squared plus 10
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
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 Start Fraction 1 over 2 End Fraction left parenthesis x plus 5 right parenthesis squared plus 10
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
Answers
GPT-4o mini
To determine which equation in vertex form matches the graph, we need to analyze the characteristics of the parabola portrayed in the graph.
1. **Direction of the parabola**: Since the parabola opens downward, the coefficient in front of the squared term must be negative.
2. **Vertex of the parabola**: The vertex form is given by \( y = a(x - h)^2 + k \), where (h, k) is the vertex of the parabola. According to the responses you provided, we can identify the vertex points from the equation.
- If the vertex is at (5, 10), the vertex form would look something like \( y = a(x - 5)^2 + 10 \).
- If the vertex is at (-5, 10), the vertex form would then be \( y = a(x + 5)^2 + 10 \).
Since the correct vertex will guide us towards the right equation, we will check the provided options:
- **y=12(x+5)²+10**: This opens upwards.
- **y=−12(x−5)²+10**: This opens downwards with a vertex at (5, 10).
- **y=12(x−5)²+10**: This opens upwards.
- **y=−12(x−5)²−10**: This opens downwards but has a vertex at (5, -10).
Given that the parabola opens downwards, the only equation that fits all these criteria with a vertex at (5, 10) is:
**y=−12(x−5)²+10**
Therefore, the correct answer is:
**y = −12(x - 5)² + 10**.
1. **Direction of the parabola**: Since the parabola opens downward, the coefficient in front of the squared term must be negative.
2. **Vertex of the parabola**: The vertex form is given by \( y = a(x - h)^2 + k \), where (h, k) is the vertex of the parabola. According to the responses you provided, we can identify the vertex points from the equation.
- If the vertex is at (5, 10), the vertex form would look something like \( y = a(x - 5)^2 + 10 \).
- If the vertex is at (-5, 10), the vertex form would then be \( y = a(x + 5)^2 + 10 \).
Since the correct vertex will guide us towards the right equation, we will check the provided options:
- **y=12(x+5)²+10**: This opens upwards.
- **y=−12(x−5)²+10**: This opens downwards with a vertex at (5, 10).
- **y=12(x−5)²+10**: This opens upwards.
- **y=−12(x−5)²−10**: This opens downwards but has a vertex at (5, -10).
Given that the parabola opens downwards, the only equation that fits all these criteria with a vertex at (5, 10) is:
**y=−12(x−5)²+10**
Therefore, the correct answer is:
**y = −12(x - 5)² + 10**.