8. marleen and candice are both 6 ft tall, and they play on the same college volleyball team. in a game, candice set up marleen with an outside high ball for an attack hit. using a video of the game, their coach determined that the height of the ball above the court, in feet, on its path from candice to marleen could be defined by the function ( h(x) = -0.03(x - 9)^2 + 8 ) where ( x ) is the horizontal distance, measured in feet, from one edge of the court.

a) determine the axis of symmetry of the parabola.

b) marleen hit the ball at its highest point. how high above the court was the ball when she hit it?

c) how high was the ball when candice set it, if she was 2 ft from the edge of the court?

d) state the range for the ball’s path between candice and marleen. justify your answer.

Question

8. marleen and candice are both 6 ft tall, and they play on the same college volleyball team. in a game, candice set up marleen with an outside high ball for an attack hit. using a video of the game, their coach determined that the height of the ball above the court, in feet, on its path from candice to marleen could be defined by the function ( h(x) = -0.03(x - 9)^2 + 8 ) where ( x ) is the horizontal distance, measured in feet, from one edge of the court.

a) determine the axis of symmetry of the parabola.

b) marleen hit the ball at its highest point. how high above the court was the ball when she hit it?

c) how high was the ball when candice set it, if she was 2 ft from the edge of the court?

d) state the range for the ball’s path between candice and marleen. justify your answer.

Accepted Answer

### Part (a) # Explanation: ## Step1: Recall vertex form of parabola The vertex form of a parabola is $ y = a(x - h)^2 + k $, where the axis of symmetry is $ x = h $. ## Step2: Identify $ h $ from given function Given $ h(x)=- 0.03(x - 9)^2+8 $, comparing with $ y=a(x - h)^2 + k $, we have $ h = 9 $. So the axis of symmetry is $ x = 9 $. # Answer: The axis of symmetry is $ x = 9 $ ### Part (b) # Explanation: ## Step1: Analyze the vertex form For the function $ h(x)=-0.03(x - 9)^2 + 8 $, in the vertex form $ y=a(x - h)^2 + k $, the vertex is $ (h,k) $. Since $ a=- 0.03<0 $, the parabola opens downwards, and the vertex is the maximum point. ## Step2: Find the maximum height The vertex of the parabola $ h(x) $ is $ (9,8) $. So the maximum height of the ball (when Marleen hits it at the highest point) is 8 feet above the court. # Answer: The ball was 8 feet above the court when Marleen hit it at its highest point. ### Part (c) # Explanation: ## Step1: Substitute $ x = 2 $ into the function We need to find $ h(2) $ where $ h(x)=-0.03(x - 9)^2+8 $. First, calculate $ (2 - 9)=- 7 $. Then $ (-7)^2 = 49 $. ## Step2: Compute the value $ h(2)=-0.03\times49 + 8=-1.47 + 8 = 6.53 $ # Answer: The ball was approximately $ 6.53 $ feet high when Candice set it (2 ft from the edge of the court). ### Part (d) # Explanation: ## Step1: Find the domain first The function $ h(x)=-0.03(x - 9)^2 + 8 $ models the height of the ball from Candice to Marleen. To find the domain, we need to find the $ x $-intercepts (where the ball hits the court, $ h(x) = 0$) and also consider the starting point. Set $ h(x)=0 $: $ 0=-0.03(x - 9)^2+8 $ $ 0.03(x - 9)^2=8 $ $ (x - 9)^2=\frac{8}{0.03}=\frac{800}{3}\approx266.67 $ $ x - 9=\pm\sqrt{\frac{800}{3}}\approx\pm16.33 $ $ x = 9\pm16.33 $ We get $ x_1=9 + 16.33\approx25.33 $ and $ x_2=9-16.33\approx - 7.33 $. But since $ x $ represents the horizontal distance from Candice (we can assume the starting point is around $ x = 0$ (Candice's position) and the end point is around $ x\approx25.33 $ (where the ball hits the court). But from the context, the ball is moving from Candice to Marleen, and we know the axis of symmetry is at $ x = 9 $. Also, when $ x = 0 $ (Candice's position, we can check $ h(0)=-0.03(0 - 9)^2+8=-0.03\times81 + 8=-2.43+8 = 5.57$ (positive, so the ball is above the court at $ x = 0$). The ball hits the court at $ x\approx25.33 $ and starts from $ x = 0$ (approximate starting point of Candice's position). But we also know that the vertex is at $ x = 9 $. ## Step2: Find the range Since the parabola opens downwards with vertex at $ (9,8) $ and the minimum height (when it hits the court) is $ 0 $ (but in reality, when moving from Candice to Marleen, the height starts from around $ h(0) = 5.57$ (or we can consider the practical domain from $ x = 0$ to $ x\approx25.33 $). But the range of the function $ h(x) $ for the domain $ x\in[0,25.33] $ (approximate domain between Candice and Marleen) will have a minimum value of $ 0 $ (when it hits the court) and a maximum value of $ 8 $ (at the vertex). But also, when $ x = 0 $, $ h(0)=5.57 $, but since the ball is hit by Candice (height of Candice is 6 ft, and the height at $ x = 0$ is 5.57 ft which is less than 6 ft? Wait, maybe the domain is from $ x = 0$ (Candice's position) to $ x = 18$ (assuming the court length is such that Marleen is at $ x = 18$? Wait, no, from the axis of symmetry at $ x = 9 $, maybe the distance between Candice and Marleen is such that $ x $ ranges from $ 0 $ to $ 18 $ (symmetric around $ x = 9$). Let's re - evaluate. The function $ h(x)=-0.03(x - 9)^2 + 8 $. The vertex is at $ (9,8) $. The ball is above the court when $ h(x)>0 $. We found the roots at $ x\approx9\pm16.33 $, so $ x\in[- 7.33,25.33] $. But in the context of the game, the ball is moving from Candice to Marleen, so the domain is from $ x = 0$ (Candice's position) to $ x = 18$ (assuming the distance between Candice and Marleen is 18 ft, symmetric around $ x = 9$). The range of a downward - opening parabola with vertex $ (9,8) $ and domain $ x\in[0,18] $: When $ x = 0 $ or $ x = 18 $: $ h(0)=h(18)=-0.03(0 - 9)^2+8=-0.03\times81 + 8=-2.43 + 8 = 5.57$ The minimum value in the domain $ [0,18] $ occurs at the endpoints (since the vertex is in the middle) and the maximum at $ x = 9 $ (8). But if we consider the entire path from when the ball is hit by Candice (height 6 ft, so when $ x = 0 $, $ h(0)=5.57$ which is less than 6, maybe the actual domain is from $ x = 0$ (Candice hits the ball at height 6 ft, so we can find $ x $ when $ h(x)=6 $): Set $ h(x)=6 $: $ 6=-0.03(x - 9)^2+8 $ $ - 0.03(x - 9)^2=-2 $ $ (x - 9)^2=\frac{2}{0.03}=\frac{200}{3}\approx66.67 $ $ x - 9=\pm\sqrt{\frac{200}{3}}\approx\pm8.16 $ $ x = 9\pm8.16 $, so $ x_1\approx0.84 $ and $ x_2\approx17.16 $ So the domain is approximately $ [0.84,17.16] $ (when the ball is between Candice and Marleen, from the time Candice hits it (height 6 ft) to Marleen hits it or the ball hits the court). The range: since the parabola opens downwards, the maximum value is 8 (at $ x = 9$), and the minimum value occurs at the endpoints of the domain. At $ x = 0.84 $ or $ x = 17.16 $, $ h(x)=6 $. Also, when the ball hits the court, $ h(x)=0 $, but in the interval between Candice and Marleen (from $ x\approx0.84 $ to $ x\approx17.16 $), the height starts at 6 ft, goes up to 8 ft, and comes back down to 6 ft (or maybe to the court). But more accurately, from the function $ h(x)=-0.03(x - 9)^2 + 8 $, the range of $ h(x) $ for $ x $ such that the ball is in the air between Candice and Marleen: The minimum height is $ 0 $ (when it hits the court) and the maximum is $ 8 $, but considering the practical situation where the ball is hit by Candice at height 6 ft, the range is from $ 0 $ (when it hits the court) up to $ 8 $ (the maximum height), but also including the height at the start (6 ft). But from the function's vertex form, since $ a=-0.03<0 $, the range of $ h(x) $ for the domain where the ball is moving from Candice to Marleen (from $ x = 0$ to $ x$ where $ h(x)=0 $) is $ 0\leq h(x)\leq8 $. But we can also see that when $ x = 0 $, $ h(0)=5.57\approx5.5 $, and when $ x = 18 $, $ h(18)=5.57 $. So the range of the ball's path between Candice and Marleen is from $ 0 $ (when it hits the court) to $ 8 $ (the maximum height), but in the context of the game (the ball is in play between them), the height ranges from approximately $ 0 $ (if it hits the court) up to $ 8 $ feet. But more precisely, since the vertex is at $ (9,8) $ and the parabola opens down, the range of $ h(x) $ for $ x $ in the domain (distance from Candice to Marleen) is $ 0\leq h(x)\leq8 $. We can justify this because the ball starts above the court (height > 0), reaches a maximum of 8 ft, and then comes back down to the court (height = 0). # Answer: The range of the ball's path between Candice and Marleen is $ 0\leq h(x)\leq8 $ (or in practical terms, from approximately $ 0 $ feet (when it hits the court) up to $ 8 $ feet (the maximum height)). The justification is that the function $ h(x)=-0.03(x - 9)^2 + 8 $ is a downward - opening parabola with a maximum value of $ 8 $ (at $ x = 9 $) and the minimum value (when the ball hits the court) is $ 0 $, and the ball's height between Candice and Marleen is between these two values.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Recall vertex form of parabola

Step2: Identify \( h \) from given function

Step1: Analyze the vertex form

Step2: Find the maximum height

Step1: Substitute \( x = 2 \) into the function

Step2: Compute the value

Answer:

Part (b)