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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 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 negative 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 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 negative Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared plus 10
y=−12(x−5)2−10
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine which equation in vertex form matches the trajectory of the firework shown in the graph, we need to analyze the features of the parabola:
1. **Direction of Opening**: Since it is stated that the parabola opens downward, we know that the coefficient of the squared term in vertex form must be negative.
2. **Vertex of the Parabola**: The vertex form of a parabola is given as \( y = a(x - h)^2 + k \), where (h, k) is the vertex point. The graph will reflect the maximum point of the firework’s trajectory.
3. **Values from Options**:
- The options that include a negative coefficient are:
- \( y = -\frac{1}{2}(x - 5)^2 + 10 \)
- \( y = -\frac{1}{2}(x - 5)^2 - 10 \)
4. **Vertex Analysis**:
- If we consider the potential vertex of the graph indicated by the equations, we find that:
- The equation \( y = -\frac{1}{2}(x - 5)^2 + 10 \) has a vertex at (5, 10).
- The equation \( y = -\frac{1}{2}(x - 5)^2 - 10 \) has its vertex at (5, -10), which is not an appropriate value based on the provided y-axis range.
Therefore, given that the parabola opens downward and assuming the vertex is at (5, 10), the appropriate equation that matches this description is:
**\( y = -\frac{1}{2}(x - 5)^2 + 10 \)**.
1. **Direction of Opening**: Since it is stated that the parabola opens downward, we know that the coefficient of the squared term in vertex form must be negative.
2. **Vertex of the Parabola**: The vertex form of a parabola is given as \( y = a(x - h)^2 + k \), where (h, k) is the vertex point. The graph will reflect the maximum point of the firework’s trajectory.
3. **Values from Options**:
- The options that include a negative coefficient are:
- \( y = -\frac{1}{2}(x - 5)^2 + 10 \)
- \( y = -\frac{1}{2}(x - 5)^2 - 10 \)
4. **Vertex Analysis**:
- If we consider the potential vertex of the graph indicated by the equations, we find that:
- The equation \( y = -\frac{1}{2}(x - 5)^2 + 10 \) has a vertex at (5, 10).
- The equation \( y = -\frac{1}{2}(x - 5)^2 - 10 \) has its vertex at (5, -10), which is not an appropriate value based on the provided y-axis range.
Therefore, given that the parabola opens downward and assuming the vertex is at (5, 10), the appropriate equation that matches this description is:
**\( y = -\frac{1}{2}(x - 5)^2 + 10 \)**.
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