SOLUTION: The arch under the kemp channel bridge is in the shape of a parabola lake archway. The arch is modeled by the equation h(x)=16x-x^2, where h(x) is the distance, in feet, the archer

Algebra ->  Exponents -> SOLUTION: The arch under the kemp channel bridge is in the shape of a parabola lake archway. The arch is modeled by the equation h(x)=16x-x^2, where h(x) is the distance, in feet, the archer      Log On


   

Question 999197: The arch under the kemp channel bridge is in the shape of a parabola lake archway. The arch is modeled by the equation h(x)=16x-x^2, where h(x) is the distance, in feet, the archer is above the water from the value of x.
A. For what value of X will the Archway be 28 feet above the water?
B. How wide is the Archway At the water?
C. What is the maximum height of the arch?
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
h(x) = 16x - x^2.
A) For what value of X will the Archway be 28 feet above the water?
28 = 16x - x^2
Multiply through by -1
x^2 - 16x + 28 = 0
(x - 2)(x - 14) = 0
x = 2 and x = 14
B) How wide is the Archway At the water?
16x - x^2 = 0
x(16 - x) = 0
x = 0 and x = 16
Width = 16 ft
C)What is the maximum height of the arch?
Mid point x = 8
h(x) = 16(8) - 8^2
h(x) = 128 - 64
Height = 64 ft.
Hope this helps :-)