ProblemWhen a sprinkler is installed in the ground, the spray of water goes up and falls in the pattern of a parabola. The height, in inches, of a spray of water is given by the equation h(x)=160x−16x2ℎ(𝑥)=160𝑥−16𝑥2 where x𝑥 is the number of feet away from the sprinkler head the spray is. What is the height of the spray 22 feet away from the sprinkler head? Solutionh(2)=ℎ(2)= After 22 seconds, the height of the spray is inches.How many feet along the ground away from the sprinkler head will the spray reach a maximum height? feetWhat is the maximum height of the water spray? inchesHow many feet away from the sprinkler head will the water hit the ground again? feet

Instructions: Use the given information to answer the questions and interpret key features. Use any method of graphing or solving.  Round to one decimal place, if necessary.An irrigation system (sprinkler) has a parabolic pattern. The height, in feet, of the spray of water is given by the equation h(x)=−x2+10x+7.5ℎ(𝑥)=−𝑥2+10𝑥+7.5, where x𝑥 is the number of feet away from the sprinkler head (along the ground) the spray is.The irrigation system is positioned Answer 1 Question 6 feet above the ground to start.The spray reaches a Answer 2 Question 6 height of Answer 3 Question 6 feet at a horizontal distance of Answer 4 Question 6 feet away from the sprinkler head.The spray reaches all the way to the ground at about Answer 5 Question 6 feet away.

A water fountain is designed to shoot a stream of water in the shape of a parabolic arc. The equation of the parabola is given by ℎ(𝑡)=−0.5𝑡2+4𝑡+1h(t)=−0.5t 2 +4t+1, where ℎ(𝑡)h(t) represents the height of the water stream in meters and t  represents the time in seconds since the water was shot. Answer the following questions.Determine the maximum height reached (in meters).

ProblemEarlier, you were told about a toy rocket fired into the air from the top of a barn. Its height (hℎ) above the ground in yards after x𝑥 seconds is given by the function:h(x)=−5x2+10x+20ℎ(𝑥)=−5𝑥2+10𝑥+20What was the maximum height of the rocket? SolutionThe maximum height was reached by the rocket at one second as you found in part b from the previous example. It takes second to reach the maximum height. We will substitute that value in for x𝑥 in our function and simplify.The maximum height reached by the rocket was yards. What is the time it takes for the rocket to hit the ground? (Use a graph or any other method to solve.)It takes approximately seconds for the rocket to hit the ground. (Round to the nearest tenth.)CheckQuestion 9

ProblemSuppose Jeremiah is a diver for his summer swim team. The function h(x)=−4.9x2+8x+5ℎ(𝑥)=−4.9𝑥2+8𝑥+5 represents Jeremiah's height (hℎ) in meters above the water x𝑥 seconds after he leaves the diving board.What is the initial height of the diving board?At what time did Jeremiah reach his maximum height?What was Jeremiah’s maximum height?Sketch a graph of the function. (You can use your calculator for this or create a table of values.) SolutionThe initial height of the diving board is when the time is zero.h(0)=−4.9x2+8x+5ℎ(0)=−4.9𝑥2+8𝑥+5h(0)=−4.9(0)2+8(0)+5ℎ(0)=−4.9(0)2+8(0)+5h(0)=0+0+5ℎ(0)=0+0+5h(0)=5ℎ(0)=5The initial height of the diving board is 55 m.The time at which Jeremiah reaches his maximum height is the x𝑥-coordinate of the vertex.x=−b2a𝑥=−𝑏2𝑎x=𝑥=2(2( ))x=−8−9.8𝑥=−8−9.8x=0.82𝑥=0.82 secIt took Jeremiah seconds to reach his maximum height.The maximum height was reached Jeremiah at seconds. The maximum height is the y𝑦-coordinate of the vertex.h(t)=−4.9x2+8x+5ℎ(𝑡)=−4.9𝑥2+8𝑥+5h(0.82)=−4.9(0.82)2+8(0.82)+5ℎ(0.82)=−4.9(0.82)2+8(0.82)+5h(0.82)=−3.29+6.56+5ℎ(0.82)=−3.29+6.56+5h(0.82)=8.27ℎ(0.82)=8.27 mThe maximum height reached by Jeremiah was m.CheckQuestion 8

4. A student throws a bag of chips to her friend. Unfortunately, herfriend does not catch the chips, and the bag hits the ground. Thedistance from the ground (height) for the bag of chips is modeledby the function h(t) 5 216(t 2 1)2 1 20, where h is the height(distance from the ground in feet) of the chips, and t is the numberof seconds the chips are in the air.A. Graph h(t) 5 216(t 2 1)2 1 20.B. From what height are the chips being thrown? Explain howyou know.C. What is the maximum height the bag ofchips reaches while airborne? Explain howyou know.D. About how many seconds a$er the bagwas thrown did it hit the ground?E. What is the average rate of change ofheight for the interval from 0 to 12 second?What does that number represent in termsof the context?F. Based on your answer to Part E, what is the average rate of change for the interval from1.5 to 2 sec.?© Tribalium/Shutterstock.com