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A volleyball is served by a 6-foot player at an initial upward velocity of 33 feet per second. The situation is modeled by the equation h=−16t2+33t+6 h representing the height in feet and t representing the time in seconds. Using this equation, define the domain of the ball when it reaches its maximum height.

While hiking, Marek throws a rock off a cliff that is roughly 10 meters above a lake. This situation is modeled by the equation h=−4.9t2+10
, with h representing the height in meters and t the time in seconds. If you were to set up a table to determine when the rock hits the water, what would be a reasonable scale for the range?

An owl swoops down from its perch in a tree at 20 feet to catch field mice below. It follows the function h=−6t2−24t+20
. Plot points to graph the equation. Which ordered pair represents the vertex?

Luca is building a garden in his backyard. He needs to erect a fence to keep his dog Luna from digging in it. He has enough materials for a fence with 120 feet around the perimeter. This function is represented by the formula A=−w2+60w
, where A
is the area and w is the width. Which of the following correctly identifies the x-intercept(s)?

An owl swoops down from its perch in a tree at 30 feet to catch field mice. It follows the function h=6t2 −24t+30
, where t is the time in seconds and h is the height in feet. Which of the following best interprets the vertex?(1 point)

3 answers

Let's tackle each situation one by one:

  1. Volleyball Height Equation: The equation for the volleyball is \( h = -16t^2 + 33t + 6 \). This is a quadratic function that models a parabola opening downwards. To find the domain when the ball reaches its maximum height, we need to find the time at which that occurs. The vertex of a parabola given by the equation \( ax^2 + bx + c \) is found at \( t = -\frac{b}{2a} \).

    Here, \( a = -16 \), \( b = 33 \): \[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]

    The ball reaches its maximum height at approximately 1.031 seconds and will drop afterwards. The domain of the function for the height up until it reaches that maximum is therefore \( t \in [0, \frac{33}{32}] \).

  2. Rock Thrown off a Cliff: The rock's height is modeled by the equation \( h = -4.9t^2 + 10 \). To find when the rock hits the water, we need to find when \( h = 0 \): \[ 0 = -4.9t^2 + 10 \ 4.9t^2 = 10 \ t^2 = \frac{10}{4.9} \approx 2.0408 \ t \approx 1.428 \text{ seconds} \]

    To set up a table that reasonably captures this event, a scale for the range can be set from \( h = 0 \) to \( h = 10 \). A reasonable scale could be increments of 1 meter to capture the rock's height as it falls.

  3. Owl's Descent Equation: The function given is \( h = -6t^2 - 24t + 20 \). To plot points and find the vertex, we calculate the vertex using \( t = -\frac{b}{2a} \): Here, \( a = -6 \), \( b = -24 \): \[ t = -\frac{-24}{2 \times -6} = \frac{24}{-12} = -2 \text{ seconds} \] To find corresponding height: \[ h = -6(-2)^2 - 24(-2) + 20 = -6(4) + 48 + 20 = -24 + 48 + 20 = 44 \text{ feet} \] The vertex is at the ordered pair \( (-2, 44) \).

    In practice, the vertex does not make physical sense in terms of time since negative time generally is non-physical; for practical purposes, we'd want to compute the maximum height reached on the path.

  4. Garden Fence Area Equation: The area \( A = -w^2 + 60w \) is a quadratic function. To find the x-intercepts, we set \( A = 0 \): \[ 0 = -w^2 + 60w \ 0 = w(-w + 60) \] The x-intercepts are at \( w = 0 \) and \( w = 60 \) feet. These values indicate the widths at which the area of the garden is zero (either not using any width or using all the available material).

  5. Owl's Swoop Interpretation: For the equation \( h = 6t^2 - 24t + 30 \), the vertex can be found as follows: \[ t = -\frac{-24}{2 \times 6} = \frac{24}{12} = 2 \text{ seconds} \] To find the height: \[ h = 6(2)^2 - 24(2) + 30 = 6(4) - 48 + 30 = 24 - 48 + 30 = 6 \text{ feet} \] The vertex represents the maximum height reached by the owl at 2 seconds after it swoops down, at a height of 6 feet.

This is for Algebra Quadratic Function from Situations

Certainly! Here’s a breakdown of each situation using the context of quadratic functions as they relate to algebra:

  1. Volleyball Height Equation:

    • Equation: \( h = -16t^2 + 33t + 6 \)
    • Finding the vertex: The time \( t \) at which the maximum height is achieved can be found using the vertex formula \( t = -\frac{b}{2a} \).
      • Here, \( a = -16 \) and \( b = 33 \):
        \[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
    • Domain of the ball until maximum height: Since a volleyball is served at \( t=0 \) and they only care about the positive time until maximum height:
      • Domain: \( [0, \frac{33}{32}] \) or approximately \( [0, 1.031] \) seconds.

  2. Rock Thrown off a Cliff:

    • Equation: \( h = -4.9t^2 + 10 \)
    • Finding when it hits the water (when \( h = 0 \)): \[ 0 = -4.9t^2 + 10 \ 4.9t^2 = 10 \ t^2 = \frac{10}{4.9} \approx 2.0408 \ t \approx 1.428 \text{ seconds} \]
    • Reasonable scale for the range: Initially, the rock is at 10 meters and eventually goes to 0 meters when it hits the water. A suitable range scale would be from 0 meters to 10 meters, with increments (like 1 or 2 meters) to observe the descent.

  3. Owl's Descent Equation:

    • Equation: \( h = -6t^2 - 24t + 20 \)
    • Finding the vertex:
      • Using \( t = -\frac{b}{2a} \) with \( a = -6 \) and \( b = -24 \): \[ t = -\frac{-24}{2 \times -6} = \frac{24}{-12} = -2 \text{ seconds (not physically sensible)} \]
    • Find height at the point of interest (e.g., t = 0): \[ h(0) = -6(0)^2 - 24(0) + 20 = 20 \text{ feet} \]
    • Finding max height effectively requires knowing how high it goes at t=0, but the vertex time is negative.

  4. Garden Fence Area Equation:

    • Equation: \( A = -w^2 + 60w \)
    • Finding the x-intercepts (A = 0): \[ 0 = -w^2 + 60w \ 0 = w(60 - w) \]
      • X-intercepts are at \( w = 0 \) and \( w = 60 \) feet. This means if no width is used, the area is 0, and if the total material is dedicated to width, the area is also zero at that width.

  5. Owl's Swoop Interpretation:

    • Equation: \( h = 6t^2 - 24t + 30 \)
    • Finding the vertex:
      • Using \( t = -\frac{b}{2a} \): \[ t = -\frac{-24}{2 \times 6} = \frac{24}{12} = 2 \text{ seconds} \]
    • Finding height at that time: \[ h = 6(2)^2 - 24(2) + 30 = 6(4) - 48 + 30 = 24 - 48 + 30 = 6 \text{ feet} \]
    • Interpretation: At 2 seconds after the owl begins swooping, it reaches its maximum height of 6 feet.

This breakdown aligns with principles of quadratic functions in algebra, examining height equations in specific real-world contexts. Would you like any further explanations or additional details on any of these situations?