Question
The center of the parabola is (4,2) and opens downward and goes through (2,0) and (6,0)
A sports analyst graphed the trajectory of a soccer ball that was kicked by a goalkeeper. In the graph, x is the ball’s distance from the goalpost, and y is the height of the ball in meters. What was the goalkeeper’s distance from the goalpost when the ball was kicked?
(1 point)
Responses
6 meters
6 meters
2 meters
2 meters
0 meters
0 meters
4 meters
4 meters
A sports analyst graphed the trajectory of a soccer ball that was kicked by a goalkeeper. In the graph, x is the ball’s distance from the goalpost, and y is the height of the ball in meters. What was the goalkeeper’s distance from the goalpost when the ball was kicked?
(1 point)
Responses
6 meters
6 meters
2 meters
2 meters
0 meters
0 meters
4 meters
4 meters
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the goalkeeper’s distance from the goalpost when the ball was kicked, we need to analyze the information given about the parabola.
The vertex (center) of the parabola is at (4, 2). The parabola opens downward and passes through the points (2, 0) and (6, 0). This indicates that these points are the x-intercepts of the parabola.
In the standard form of a downward-opening parabola, the vertex form is given by:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola. Here, \(h = 4\) and \(k = 2\), so the equation becomes:
\[ y = a(x - 4)^2 + 2 \]
Since the parabola passes through the points (2, 0) and (6, 0), we can use one of these points to find the coefficient \(a\). Let's use the point (2, 0):
\[ 0 = a(2 - 4)^2 + 2 \]
\[ 0 = a(2)^2 + 2 \]
\[ 0 = 4a + 2 \]
\[ 4a = -2 \]
\[ a = -\frac{1}{2} \]
Now we can write the equation of the parabola:
\[ y = -\frac{1}{2}(x - 4)^2 + 2 \]
To determine the point where the ball was kicked, we look for the maximum height of the parabola (the vertex), which occurs at \(x = 4\). This corresponds to the height of the ball when it was kicked.
Since the parabola opens downward, the maximum height occurs at the vertex point:
- The goalkeeper’s distance from the goalpost when the ball was kicked is **4 meters**.
Thus, the answer is **4 meters**.
The vertex (center) of the parabola is at (4, 2). The parabola opens downward and passes through the points (2, 0) and (6, 0). This indicates that these points are the x-intercepts of the parabola.
In the standard form of a downward-opening parabola, the vertex form is given by:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola. Here, \(h = 4\) and \(k = 2\), so the equation becomes:
\[ y = a(x - 4)^2 + 2 \]
Since the parabola passes through the points (2, 0) and (6, 0), we can use one of these points to find the coefficient \(a\). Let's use the point (2, 0):
\[ 0 = a(2 - 4)^2 + 2 \]
\[ 0 = a(2)^2 + 2 \]
\[ 0 = 4a + 2 \]
\[ 4a = -2 \]
\[ a = -\frac{1}{2} \]
Now we can write the equation of the parabola:
\[ y = -\frac{1}{2}(x - 4)^2 + 2 \]
To determine the point where the ball was kicked, we look for the maximum height of the parabola (the vertex), which occurs at \(x = 4\). This corresponds to the height of the ball when it was kicked.
Since the parabola opens downward, the maximum height occurs at the vertex point:
- The goalkeeper’s distance from the goalpost when the ball was kicked is **4 meters**.
Thus, the answer is **4 meters**.
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