Question 726708
After careful analysis, he has found that the rainbow can be modeled by the quadratic equation;
h(x)= 0.73x - 0.12x^2 - 0.73
normally written
h(x) = -.12x^2 + .73x - .73
where h = height and x = horizontal distance in kilometers.
(a)What assumptions can be made?
That it is a parabola, opening downward.
:
(b)Find the total distance between the ends of the rainbow.
That would be the distance between the x intercepts, solve the equation using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
where: a=-.12; b=.73; c=.-73
{{{x = (-.73 +- sqrt(.73^2-4*-.12*-.73 ))/(2*-.12) }}}
{{{x = (-.73 +- sqrt(.5329-.3504 ))/-.24 }}}
{{{x = (-.73 +- sqrt(.1825 ))/-.24 }}}
Two solutions
{{{x = (-.73 + .4272)/-.24 }}}
x = {{{(-.3028)/(-.24)}}}
x = +1.26
and
{{{x = (-.73 - .4272)/-.24 }}}
x = {{{(-1.1572)/(-.24)}}}
x = +4.82
The distance between the ends of the rainbow
4.82 - 1.26 = 3.56 km
:
(c) Indiana knows that if clouds were to obstruct his view, he will be unable to follow the rainbow's path.
 What is the lowest that the clouds can be to not obstruct Indiana's quest.
That would be the maximum of the parabola, find the axis of symmetry
x = {{{(-.73)/(2*-.12)}}} 
x = 3.04 ~ 3 should be close enough
Substitute 3 for x in the original equation to find the max height
h(x) = -.12(3^2) + .73(3) - .73
h(x) = .38 km or about 380 meters max height of the rainbow
:
I am sure of the method but you should check the math here.
:
Looks like this
{{{ graph( 300, 200, -1, 6, -.5, 1, -.12x^2+.73x-.73) }}}