I need help with this

Question

The height of a swimmers dive off a 10-foot platform into a diving pool is modeled by the equation y=2x²-12x+10 , where x represents the number of seconds since the swimmer left the diving board and y represents the number of feet above or below the waters surface . What is the farthest depth below the waters surface that the swimmer will reach ?

Answer 1

Method 1: Rewrite the equation in vertex form to solve for the vertex. The vertex of a parabola will give you the maximum or minimum value. Since the x² term in this equation is positive, the vertex will be the minimum value of the function, which represents the maximum depth below the surface.

y = 2x² - 12x + 10

y = 2(x−3)² − 8

→ vertex = (3,-8)

→ max depth = 8 feet



Method 2: For maximization problems, you can also find the derivative, set it equal to 0, and then solve for x. This will be the time that the minimum or maximum depth occurs. Plug x into the original problem to determine the actual depth below the surface.

y = 2x² - 12x + 10

y' = 4x -12 = 0

→ x = 12 / 4 = 3 s

→ y(3) = 2(3)² - 12(3) + 10 = -8 ft

→ max depth = 8 feet

Answer 2

At lovest point velocity wiil be 0, so y' = 4x - 12 = 0 and x = 3sec. At this time his hite y = 2·3² - 12·3 + 10 = - 8 foot. So his farthers depth below waters level will be 8 foot.