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

To determine which equation in vertex form matches the downward-opening parabola shown in your description, we can analyze the general form of the vertex equation:

\[ y = a(x - h)^2 + k \]

Where:
- \( (h, k) \) is the vertex of the parabola,
- \( a \) determines the direction and width of the parabola (if \( a < 0 \), the parabola opens downward).

Since the graph describes a parabola that opens downward, we are looking for an equation where \( a \) is negative.

Assuming the vertex is at a point like (5, 10) based on your description (10 could be the maximum height), we can start narrowing down the equations:

1. \( y = 12(x + 5)^2 + 10 \) — This opens upwards (not a match).
2. \( y = -\frac{1}{2}(x - 5)^2 + 10 \) — This opens downwards with vertex (5, 10).
3. \( y = -\frac{1}{2}(x - 5)^2 - 10 \) — This also opens downwards, but vertex is at (5, -10).
4. \( y = \frac{1}{2}(x - 5)^2 + 10 \) — This opens upwards (not a match).

Thus, the equation that correctly represents a downward-opening parabola with vertex (5, 10) is:

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

So the correct response is:

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