Q:

A fountain shoots a water arc that can be modeled by the graph of the equation y=-0.006x^2+1.2x+10, where x is the horizontal distance (in feet) from the river's north shore and y is the height (in feet) above the river. Does the water arc reach a height of 50 feet? If so, about how far from the north shore is the water arc 50 feet above the water?​

Accepted Solution

A:
ANSWERThe water arc is 50 ft above the water approximately 42ft and 158ft from the north shore.EXPLANATIONThe water arc is modeled by the function:[tex]y = - 0.006 {x}^{2} + 1.2x + 10[/tex]We write this function in vertex form:[tex]y = - 0.006 ({x}^{2} - 200x )+ 10[/tex][tex]y = - 0.006 ({x}^{2} - 200x + ( - 100)^{2} ) - - 0.006( - 100)^{2} + 10[/tex][tex]y = - 0.006 ( x- 100)^{2} ) + 60+ 10[/tex][tex]y = - 0.006 ( x- 100)^{2} ) +70[/tex]The vertex of this function is at(100,70).This means that the water arc reached a height of 50ft.We put y=50 and solve for x.[tex]- 0.006 ( x- 100)^{2}+70 = 50[/tex][tex]- 0.006 ( x- 100)^{2} = - 20[/tex][tex]( x- 100)^{2} = \frac{10000}{3} [/tex][tex]x = 100 \pm \frac{100 \sqrt{3} }{3} [/tex]x=42.3 or x=157.7Hence the water arc is 50 ft above the water approximately 42ft and 158ft from the north shore.