Use the image to answer the question.

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
y equals negative Start Fraction 1 over 2 End Fraction left parenthesis x minus 5 right parenthesis squared minus 10
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Answer 1

To determine the correct equation in vertex form for the parabola showing the trajectory of the firework, we need to identify the vertex and the direction the parabola opens.

1. Since the parabola opens downward, the value of "a" in vertex form \(y = a(x - h)^2 + k\) must be negative. This eliminates any option with a positive coefficient.

2. The vertex form of a parabola is characterized by its vertex \((h, k)\). We can use the coordinates of the vertex provided in the description.

If we assume the vertex is at \((5, 10)\):

• This implies \(h = 5\) and \(k = 10\).

3. The equations provided:

   • \(y = 12(x + 5)^2 + 10\) (positive, opens upward)
   • \(y = -\frac{1}{2}(x - 5)^2 + 10\) (negative, opens downward)
   • \(y = \frac{1}{2}(x - 5)^2 + 10\) (positive, opens upward)
   • \(y = -\frac{1}{2}(x - 5)^2 - 10\) (negative, opens downward)

4. Since we established the vertex is at \((5, 10)\) and the parabola opens downward, we can focus on the equations that meet these criteria:

   The correct equation is:

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

Thus, the correct response is: y = -\frac{1}{2}(x - 5)^2 + 10.