U3.L4.HW: Quadratic Functions: Question 21

Question

Answer first, then explain after.

The height y (in feet) of a ball thrown by a child is 

y = -1/12x² + 2x + 5 where x is the horizontal distance in feet from the point at which the ball is thrown.



What is the maximum height of the ball?

____________ feet

How far from the child does the ball strike the ground?

____________ feet

Answer 1

The maximum height of the ball will be where the height, y, is 0:

y = (-1/12)x² + 2x + 5

0 = (-1/12)x² + 2x + 5

0 = x² - 24x - 60

x = [-b±√(b²-4ac)]/2a = [-24±√(24²-4(1)(-60))]/(2(1)) ≈ 26.283 feet (taking only the positive solution)

The ball will be furthest from the child where the y-coordinate of the vertex is:

x = -b/2a = -2/(2(-1/12)) = -2/(-1/6) = -2(-6) = 12

y = (-1/12)(12)² + 2(12) + 5 = -12 + 24 + 5 = 12 + 5 = 17 feet

Hope these explanations helped!

Answer 2

To find the maximum height of the ball, we need to find the vertex of the parabolic function represented by the equation y = -1/12x² + 2x + 5. The x-coordinate of the vertex can be found using the formula -b/2a, where a = -1/12 and b = 2:

x = -b/2a = -(2)/(2*(-1/12)) = 12

So the maximum height occurs at x = 12 feet. To find the maximum height, we substitute x = 12 into the equation:

y = -1/12(12)² + 2(12) + 5 = 17 feet

Therefore, the maximum height of the ball is 17 feet.

To find the distance from the child where the ball strikes the ground, we need to find the value of x when y = 0. This represents the point where the ball hits the ground. We can solve the equation -1/12x² + 2x + 5 = 0 for x using the quadratic formula:

x = (-b ± sqrt(b² - 4ac))/2a

Plugging in a = -1/12, b = 2, and c = 5, we get:

x = (-2 ± sqrt(2² - 4(-1/12)(5)))/(2*(-1/12))

Simplifying this expression, we get:

x = 24 ± 2sqrt(29)

Since the distance cannot be negative, we take the positive value:

x = 24 + 2sqrt(29)

Therefore, the ball strikes the ground approximately 24 + 2sqrt(29) feet from the child.