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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

1 answer

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:

\( y = -\frac{1}{2}(x - 5)^2 + 10 \).