Question

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?​

Answer 1

The water arc is 50 ft above the water approximately 42ft and 158ft from the north shore.

Step-by-step explanation

The water arc is modeled by the function:

`y = - 0.006 {x}^(2) + 1.2x + 10`

We write this function in vertex form:

`y = - 0.006 ({x}^(2) - 200x )+ 10`

`y = - 0.006 ({x}^(2) - 200x + ( - 100)^(2) ) - - 0.006( - 100)^(2) + 10`

`y = - 0.006 ( x- 100)^(2) ) + 60+ 10`

`y = - 0.006 ( x- 100)^(2) ) +70`

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.

`- 0.006 ( x- 100)^(2)+70 = 50`

`- 0.006 ( x- 100)^(2) = - 20`

`( x- 100)^(2) = (10000)/(3)`

`x = 100 \pm (100 √(3) )/(3)`

x=42.3 or x=157.7

Hence the water arc is 50 ft above the water approximately 42ft and 158ft from the north shore.