You live near a bridge that goes over a river

Asked by maham237 @ 25/12/2020 in Mathematics viewed by 275 People

You live near a bridge that goes over a river. The underneath side of the bridge is an arch that can be modeled with the function y= -0.000475x^2 + 0.851x, where x and y in feet. How high above is the river bridge (the top of the arch)? How long is section of bridge above the arch?
A)The bridge is about 1,791.58 ft. above the river, and the length of the bridge above the arch is about 381.16 ft.

B)The bridge is about 1,791.58 ft. above the river, and the length of the bridge above the arch is about 895.79 ft.

C)The bridge is about 381.16 ft. above the river, and the length of the bridge above the arch is about 895.79 ft.

D)The bridge is about 381.16 ft. above the river, and the length of the bridge above the arch is about 1,791.58 ft You live near a bridge that goes over a river. The underside of the bridge is an arch that can be modeled with the function y = –0.000475x^2 + 0.851x, where x and y are in feet. How high above the river is the bridge (the top of the arch)? How long is the section of bridge above the arch?

Answered by maham237 @ 25/12/2020

Answer:

D)The bridge is about 381.16 ft. above the river, and the length of the bridge above the arch is about 1,791.58 ft

Step-by-step explanation:

Finding the top of the arch means finding the maximum of the quadratic function given. The maximum will be the vertex; to find the vertex, we first find the axis of symmetry. We do this using the formula

x = -b/2a

In this function, a = -0.000475 and b = 0.851. This gives us:

$x=\frac{-0.851}{2(-0.000475}=\frac{-0.851}{-0.00095}=895.7894737$

This is the x-coordinate of the vertex. To find the y-coordinate, substitute this value in for x in the function:

$y = -0.000475x^2+0.851(x)\\\\=-0.000475(895.7894737)^2+0.851(895.7894737)\\\\=-0.000475(802438.7812)+0.851(895.7894737)\\\\=-381.1584211+762.3168421\\\\=381.158421\approx 381.16$

This is the height of the arch. The x-coordinate of the vertex tells us how far horizontally the vertex, or high point, of the arch is. This will be halfway across the arch; multiply this by 2 to get the width of the arch:

895.7894737(2) = 1791.578947 ≈ 1791.58

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