Step 1 :Given the quadratic function \(f(x) = -0.01x^2 + 0.7x + 5.9\), we can find the maximum height, maximum horizontal distance, and the height from which the shot was released.
Step 2 :To find the maximum height, we first find the vertex of the parabola. The x-coordinate of the vertex is given by \(x_v = \frac{-b}{2a}\), where \(a = -0.01\) and \(b = 0.7\).
Step 3 :\(x_v = \frac{-0.7}{2(-0.01)} = 35\)
Step 4 :Substitute \(x_v\) into the function to find the maximum height: \(f(35) = -0.01(35)^2 + 0.7(35) + 5.9 = 18.15\)
Step 5 :\(\boxed{\text{a. The maximum height of the shot is 18.15 feet, which occurs 35 feet from the point of release.}}\)
Step 6 :To find the maximum horizontal distance, we need to find the x-intercepts of the function. Set \(f(x) = 0\) and solve for \(x\):
Step 7 :\(-0.01x^2 + 0.7x + 5.9 = 0\)
Step 8 :Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -0.01\), \(b = 0.7\), and \(c = 5.9\).
Step 9 :\(x = \frac{-0.7 \pm \sqrt{(0.7)^2 - 4(-0.01)(5.9)}}{2(-0.01)}\)
Step 10 :\(x \approx 0, 77.6\)
Step 11 :\(\boxed{\text{b. The shot's maximum horizontal distance is 77.6 feet (rounded to the nearest tenth).}}\)
Step 12 :The height from which the shot was released is the value of the function at \(x = 0\): \(f(0) = -0.01(0)^2 + 0.7(0) + 5.9 = 5.9\)
Step 13 :\(\boxed{\text{c. The height from which the shot was released is 5.9 feet.}}\)