Question
You are attending a professional baseball game. The batter hits the ball into right field. You notice the ball, as it is hit, forms a parabolic path from the bat to where it lands in the field. The path the ball traveled can be expressed by the quadratic equation y = - 1/42 (x - 50)^2 + 60, where x is the distance from home plate, and y is the height in the air as the ball travels in the air. Interpret the following key features of the parabola in terms of the context.
Range: (0 ≤ y ≤ __
Range: (0 ≤ y ≤ __
Answers
GPT-4o mini
To interpret the range of the parabolic path of the baseball in the context of the game, we need to analyze the quadratic equation given:
\[ y = -\frac{1}{42} (x - 50)^2 + 60. \]
This is a downward-opening parabola (because the coefficient of the quadratic term is negative).
### Key Features
1. **Vertex of the Parabola**: The vertex form of a parabolic equation is in the format \( y = a(x - h)^2 + k \), where (h, k) is the vertex. In this case:
- The vertex occurs at \( x = 50 \) and \( y = 60 \).
- The vertex represents the maximum height the ball reaches, which is 60 feet.
2. **Intercepts and Range**:
- Since the parabola opens downwards, the maximum value of \( y \) (the height) is 60 feet at its vertex.
- Mathematically, the range of a downward-opening parabola is from the lowest point it can reach (which we find by determining when \( y = 0 \)) to the maximum height.
### Finding the Roots (x-intercepts):
To find where the ball hits the ground (y = 0), we set the equation to 0:
\[ 0 = -\frac{1}{42}(x - 50)^2 + 60. \]
Rearranging gives:
\[ \frac{1}{42}(x - 50)^2 = 60 \]
\[ (x - 50)^2 = 60 \times 42 \]
\[ (x - 50)^2 = 2520. \]
Taking the square root of both sides gives:
\[ x - 50 = \pm \sqrt{2520}. \]
\[ x = 50 \pm \sqrt{2520}. \]
Calculating \( \sqrt{2520} \) yields approximately 50.2. Thus, the x-values where the ball hits the ground are:
\[ x \approx 50 - 50.2 \quad \text{and} \quad x \approx 50 + 50.2. \]
This simplifies to approximately \( x \approx -0.2 \) and \( x \approx 100.2 \).
Since x represents distance from home plate, the negative value doesn't make sense in context, so we take the positive x-value. The ball lands roughly around **100.2 feet** from home plate.
### Conclusion on Range:
Therefore, for the quadratic equation representing the ball's trajectory, the height \( y \) can range from **0 feet** (when the ball hits the ground) to a maximum of **60 feet** (the highest point of the trajectory).
Thus, the range of the parabola in this context is:
**Range: (0 ≤ y ≤ 60)**.
\[ y = -\frac{1}{42} (x - 50)^2 + 60. \]
This is a downward-opening parabola (because the coefficient of the quadratic term is negative).
### Key Features
1. **Vertex of the Parabola**: The vertex form of a parabolic equation is in the format \( y = a(x - h)^2 + k \), where (h, k) is the vertex. In this case:
- The vertex occurs at \( x = 50 \) and \( y = 60 \).
- The vertex represents the maximum height the ball reaches, which is 60 feet.
2. **Intercepts and Range**:
- Since the parabola opens downwards, the maximum value of \( y \) (the height) is 60 feet at its vertex.
- Mathematically, the range of a downward-opening parabola is from the lowest point it can reach (which we find by determining when \( y = 0 \)) to the maximum height.
### Finding the Roots (x-intercepts):
To find where the ball hits the ground (y = 0), we set the equation to 0:
\[ 0 = -\frac{1}{42}(x - 50)^2 + 60. \]
Rearranging gives:
\[ \frac{1}{42}(x - 50)^2 = 60 \]
\[ (x - 50)^2 = 60 \times 42 \]
\[ (x - 50)^2 = 2520. \]
Taking the square root of both sides gives:
\[ x - 50 = \pm \sqrt{2520}. \]
\[ x = 50 \pm \sqrt{2520}. \]
Calculating \( \sqrt{2520} \) yields approximately 50.2. Thus, the x-values where the ball hits the ground are:
\[ x \approx 50 - 50.2 \quad \text{and} \quad x \approx 50 + 50.2. \]
This simplifies to approximately \( x \approx -0.2 \) and \( x \approx 100.2 \).
Since x represents distance from home plate, the negative value doesn't make sense in context, so we take the positive x-value. The ball lands roughly around **100.2 feet** from home plate.
### Conclusion on Range:
Therefore, for the quadratic equation representing the ball's trajectory, the height \( y \) can range from **0 feet** (when the ball hits the ground) to a maximum of **60 feet** (the highest point of the trajectory).
Thus, the range of the parabola in this context is:
**Range: (0 ≤ y ≤ 60)**.