Answer :
The question is incomplete. The full question is below.
The path of a particular punt follows the quadratic function
[tex]h(x)=\frac{-1}{8}(x-25)^{2}+50[/tex]
where h(x) is the height of the ball in yards and x corresponds to the horizontal distance in yards. Assume x=0 corresponds to midfield (the 50 yards line). For example, x=20 corresponds to the punter's own 30 yard line, whereas corresponds to the other team's 30 yard line.
a. Find the maximum height the ball achieves.
b. Find the horizontal distance the ball covers. Assume the height is zero when the ball is kicked and when the ball is caught.
Answer: a. Maximum height = 50
b. Horizontal distance = 40
Step-by-step explanation: A quadratic function has its maximum or minimum point on its vertex.
When the function is negative, function has a maximum point.
The quadratic function h(x) is written in vertex form, meaning the coordinate of vertex is explicit in the function.
General vertex form: [tex]f(x)=A(x-h)^{2}+k[/tex]
A is vertical scaling parameter
(h,k) are the coordinates of vertex
So, the quadratic function for the height for the ball:
[tex]h(x)=\frac{-1}{8}(x-25)^{2}+50[/tex]
shows its vertex:
(h,k) = (25,50)
a. The maximum height the ball achieves the maximum value corresponds to the y-axis, which means, maximum height is 50 yards.
b. Horizontal distance is when h(x)=0, so:
[tex]\frac{-1}{8}(x-25)^{2}+50=0[/tex]
[tex]\frac{-1}{8}(x-25)^{2}=-50[/tex]
[tex](x-25)^{2}=400[/tex]
[tex]x^{2}-50x+625=400[/tex]
[tex]x^{2}-50x+225=0[/tex]
Solving quadratic equation:
[tex]x_{1}=\frac{50+\sqrt{1600} }{2}[/tex] = 45
[tex]x_{2}=\frac{50-\sqrt{1600} }{2}[/tex] = 5
When ball is kicked, it is in position 5. When is caught, is position 45. So, distance the balls covers is 40 yards.