To represent the scenario of the golf ball hit by Lorraine, we can derive a quadratic function based on the given information.
Information Given:
- The ball lands 60 feet away: x-intercepts = (0, 0) and (60, 0)
- The highest point of the ball is 30 feet: y-coordinate of the vertex = 30 feet
- The axis of symmetry is at half the distance to the landing point: axis of symmetry = 30 feet (at x = 30)
Step-by-step Breakdown:
-
Vertex Form of the Quadratic Equation: The general form of a quadratic equation can be presented as: \[ f(x) = a(x - h)^2 + k \] where (h, k) is the vertex.
Based on the highest point (vertex) at (30, 30): \[ f(x) = a(x - 30)^2 + 30 \]
-
Finding 'a': Since we know the ball lands on the x-axis at (0, 0), we can use this point to find the value of 'a'. \[ 0 = a(0 - 30)^2 + 30 \] \[ 0 = a(900) + 30 \] \[ a(900) = -30 \] \[ a = -\frac{30}{900} = -\frac{1}{30} \]
-
Final Equation: Substituting 'a' back into the vertex form: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \] For a standard form or simplified version (using vertex coordinates): \[ f(x) = -\frac{1}{30}(x^2 - 60x + 900) + 30 \] Expanding it gives: \[ f(x) = -\frac{1}{30}x^2 + 2x \]
-
Finding the Height when x = 15: Substitute x = 15 into the equation: \[ f(15) = -\frac{1}{30}(15)^2 + 2(15) \] \[ = -\frac{1}{30}(225) + 30 \] \[ = -7.5 + 30 \] \[ = 22.5 \]
Summary Answers:
- Equation of the scenario: \[ f(x) = -\frac{1}{30}x^2 + 2x \]
- Height of the ball when it is 15 feet away from Lorraine: \[ \text{Height} \approx 23 \text{ feet (rounded to the nearest whole number)} \]
You may write the equation and height in the format you need. If you need further assistance, please let me know!
1 answer