Question A rocket is launched from a tower. The height of the rocket, \( y \) in feet, is related to the time after launch, \( x \) in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. \[ y=-16 x^{2}+121 x+135 \] Answer Attempt 2 out of 2

To find the maximum height reached by the rocket, we need to determine the vertex of the quadratic equation given by: \[ y = -16x^2 + 121x + 135 \] The vertex of a quadratic equation in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] In this case, \( a = -16 \) and \( b = 121 \). Plugging in these values: \[ x = -\frac{121}{2 \times -16} = \frac{121}{32} \] Now, we calculate \( \frac{121}{32} \): \[ \frac{121}{32} \approx 3.78125 \] Next, we substitute this value back into the original equation to find the maximum height \( y \): \[ y = -16\left(\frac{121}{32}\right)^2 + 121\left(\frac{121}{32}\right) + 135 \] Calculating \( \left(\frac{121}{32}\right)^2 \): \[ \left(\frac{121}{32}\right)^2 = \frac{14641}{1024} \] Now substituting this into the equation: \[ y = -16\left(\frac{14641}{1024}\right) + 121\left(\frac{121}{32}\right) + 135 \] Calculating each term: 1. For the first term: \[ -16 \times \frac{14641}{1024} = -\frac{234256}{1024} = -229.0 \] 2. For the second term: \[ 121 \times \frac{121}{32} = \frac{14641}{32} = 457.53125 \] 3. The constant term is: \[ 135 \] Now, combining these results: \[ y = -229.0 + 457.53125 + 135 \] Calculating this step-by-step: \[ y = 457.53125 - 229.0 = 228.53125 \] \[ y = 228.53125 + 135 = 363.53125 \] Rounding to the nearest tenth: \[ y \approx 363.5 \] Thus, the maximum height reached by the rocket is approximately: \[ \boxed{363.5} \text{ feet} \]