William W. answered 10/27/22
Top Pre-Calc Tutor
h(x) = -0.01x2 + 1.18x + 2 is an inverted parabola so the max height will occur at the vertex. To find the x-value of the vertex of a parabola, we can use x = -b/(2a) for f(x) = ax2 + bx + c.
x = -b/(2a)
b = 1.18 and a = -0.01 so x = -1.18/(2(-0.01)) = 1.18/.02 = 59.00
x = 59.00 feet is then the answer to the second question of how far from the kicker's foot does the maximum height occur.
To find the maximum height, plug in x = 59:
h(59) = -0.01(59)2 + 1.18(59) + 2 = 36.81 feet high
To answer the question about how far the ball goes before hitting the ground, we would set the height equal to zero, and solve for "x":
0 = -0.01x2 + 1.18x + 2
Use the quadratic equation:
x = [-b ± √(b2 - 4ac)]/(2a) where a = -0.01, b = 1.18, and c = 2:
x = [-1.18 ± √(1.182 - 4(-0.01)(2))]/(2(-0.01))
x = [-1.18 ± √(1.3924 + 0.08)]/(-0.02)
x = [-1.18 ± √1.4724]/(-0.02)
x = [-1.18 ± 1.2134]/(-0.02)
So x = [-1.18 + 1.2134]/(-0.02) or [-1.18 - 1.2134]/(-0.02)
x = -1.67 or x = 119.67
We would ignore the negative answer and go with x = 119.67 feet
Regarding the y-intercept, the y-intercept occurs when x = 0
h(0) = -0.01(0)2 + 1.18(0) + 2 = 2 meaning that the height of the ball is initially at 2 feet when the distance away from the kickers foot is zero. This would be the height the ball when the kicker kicks it.