Question

The Gateway Arch in St. Louis, Missouri is not a parabola but a shape known as a catenary. The name is given to the shaoe formed by the Graph of the hyperbolic cosine (cosh). The arch has a height of 625 feet andna span of 600 feet. The hyperbolic cosine is defined as:
Cosh x= 1/2 e^x + 1/2 e^-x
•on the same set of axis graph f(x)=1/2e^x, g(x)=1/2e^-x,and h(x)=(f+g)(x)
•create a parabolic model that matches the arch at the vertex and ground level
•the equation that gives shape of the arch is c(x)=693.8597-68.7672cosh(0.0100333x). Define x and c(x) in terms of the situation
•rewrite the gateway arch equation into exponential function form.

Thanks ahead of time!

Answer 1

you can do all the graphing stuff. Lots of online places for that. Given the definition of cosh(x), you have the arch as

c(x)=693.8597-68.7672(e^0.0100333x+e^-0.0100333x)/2.

The vertex of the arch is at
c(0) = 693.8597-65.7672 = 828.0925
the arch meets the ground where c(x) = 0, or x = ±299.226

So, the parabola whose vertex is where the arch's is is

y = -0.007750x^2 + 693.8957

Answer 2

oops. wrong vertex. Should be (0,628.0925)

y = -0.007015x^2+628.0925