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Chapter 9: Problem 9

The function \(h(t)=-4.9(t-0.4)^{2}+2.5\) describes the height of a softball thrown by a pitcher, where \(h(t)\) is in meters and \(t\) is in seconds. a. How high does the ball go? (T b. What is an equivalent function in general form? c. At what height did the pitcher release the ball when \(t\) was \(0 \mathrm{~s}\) ? (A) d. What domain and range values make sense in this situation?

Short Answer

Expert verified
a. The ball reaches a height of 2.5 meters. b. Equivalent general form is \(-4.9t^2 + 3.92t + 1.716\). c. The release height is 1.716 meters. d. Domain for time \(t\) is \([0, \text{when it lands}]\), range for height \(h(t)\) is \([0, 2.5]\) meters.

Step by step solution

01

Determine the Vertex

The function is in vertex form \( h(t) = a(t-h)^2 + k \). In this case, \( a = -4.9 \), \( h = 0.4 \), and \( k = 2.5 \). The vertex \((h, k)\) represents the maximum point of the parabola, so the height of the ball is \( 2.5 \) meters.
02

Convert to General Form

Expand \( h(t) = -4.9(t-0.4)^2 + 2.5 \). First, calculate \((t-0.4)^2 = t^2 - 0.8t + 0.16\). Then substitute it back into the function: \( h(t) = -4.9(t^2 - 0.8t + 0.16) + 2.5 \). Distributing \(-4.9\) gives \(-4.9t^2 + 3.92t - 0.784 + 2.5\). Simplify to get the general form \( h(t) = -4.9t^2 + 3.92t + 1.716 \).
03

Find Initial Height at t=0

Substitute \( t = 0 \) into the function \( h(t) = -(4.9)(0 - 0.4)^2 + 2.5 \). Calculating, we get \( h(0) = -(4.9)(0.16) + 2.5 = -0.784 + 2.5 = 1.716 \) meters. Thus, the initial height is \( 1.716 \) meters.
04

Determine Domain and Range

The domain of the function represents the time interval the ball is in motion. Since a negative time makes no real-world sense, the domain starts at \( t = 0 \) until the ball hits the ground. The ball hits the ground when \( h(t) = 0 \), solving \(-4.9t^2 + 3.92t + 1.716 = 0 \) can determine this. By calculating, you'll find roots, but the context should verify positive time values. For range, as the ball cannot go lower than the ground \( h(t) \geq 0 \) and the maximum is at the vertex \( 2.5 \) meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex Form
In mathematics, the vertex form of a quadratic function highlights the vertex of its graph, which is a parabola. This form is expressed as \( h(t) = a(t-h)^2 + k \). Here, \((h, k)\) represents the vertex point. In our exercise, the quadratic function for the height of a softball is given in vertex form: \[ h(t) = -4.9(t-0.4)^2 + 2.5 \] Breaking it down:
  • \(a = -4.9\): This coefficient affects the direction and width of the parabola. Since it is negative, the parabola opens downward.
  • \(h = 0.4\) and \(k = 2.5\): These are the coordinates of the vertex. The vertex tells us the highest point the ball reaches, which is \(2.5\) meters above the ground.
Understanding the vertex form helps to quickly locate the maximum or minimum value of the function, which is essential for determining the ball's peak height.
General Form
Quadratic functions can also be written in the general form: \( ax^2 + bx + c \). Converting the function in vertex form to general form involves expanding and simplifying the expression. Starting from the vertex form: \[ h(t) = -4.9(t-0.4)^2 + 2.5 \] First, expand \((t-0.4)^2\) to get \(t^2 - 0.8t + 0.16\). Then, substitute this back into the equation: \[ h(t) = -4.9(t^2 - 0.8t + 0.16) + 2.5 \] Distribute \(-4.9\) across the terms:
  • \(-4.9t^2\)
  • \(+3.92t\)
  • \(-0.784 + 2.5\)
This simplifies to the general form: \[ h(t) = -4.9t^2 + 3.92t + 1.716 \] The general form is useful for solving quadratic equations and finding various points on the graph.
Domain and Range
Domain and range are fundamental concepts of functions, indicating where they exist and the limits of their outputs. The domain refers to all possible input values, whereas the range concerns all potential output values. For this quadratic function, the domain is rooted in the physical context of the problem:
  • The start of the motion, \( t = 0 \), represents when the ball leaves the pitcher's hand.
  • The end, when \( h(t) = 0 \), signifies when the ball hits the ground. Calculating this involves finding the positive root of the equation \(-4.9t^2 + 3.92t + 1.716 = 0\).
Thus, the domain can be expressed in terms of time \(t\):
  • \([0, t_{ground}]\) where \(t_{ground}\) is the time when \(h(t) = 0\).
As for the range:
  • Begins at the initial height, \(1.716\) meters, when \(t = 0\).
  • Reaches up to the maximum height, which is \(2.5\) meters at the vertex.
By examining the domain and range, one can understand the real-world applicability of the quadratic function.
Maximum Height
In quadratic functions, the maximum or minimum height is found at the vertex, which is either the top or bottom of the parabola, respectively. This point represents a significant feature of the function's behavior in real-world scenarios. Since the function \( h(t) = -4.9(t-0.4)^2 + 2.5 \) is in vertex form, identifying the maximum height is straightforward. The vertex \( (0.4, 2.5) \) indicates the peak point, revealing that the ball reaches its highest point at \(0.4\) seconds after being thrown.To confirm, consider these details:
  • The coefficient \(a = -4.9\) is negative, indicating a parabola that opens downward, thus having a maximum value.
  • The value \(k = 2.5\) directly tells us the maximum height in meters.
Understanding the vertex and maximum height helps in analyzing and predicting the trajectory of the ball during its flight.

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Most popular questions from this chapter

Graph the equation \(y=x^{2}+3 x+5\). Use the graph and the quadratic formula to answer these questions. a. How many \(x\)-intercepts does the graph have? b. Use the quadratic formula to try to find the roots of \(0=x^{2}+3 x+5\). What happens when you take the square root? c. Without looking at a graph, how can you use the quadratic formula to tell if a quadratic equation has any real roots? (a)

Some numbers are both perfect squares and perfect cubes. a. Find at least three numbers that are both a perfect square and a perfect cube. b. Define a rule that you can use to find as many numbers as you like that are both perfect squares and perfect cubes.

In this exercise you will discover whether knowing the \(x\)-intercepts determines a unique quadratic equation. Work through the steps in 9 a-e to find an answer. Graph each equation to check your work. a. Write an equation of a parabola with \(x\)-intercepts at \(x=3\) and \(x=7\). b. Name the vertex of the parabola in 9 a. c. Modify your equation in 9 a so that the graph is reflected across the \(x\)-axis. Where are the \(x\)-intercepts? Where is the vertex? d. Modify your equation in 9 a to apply a vertical stretch with a factor of 2 . Where are the \(x\)-intercepts? Where is the vertex? (a) e. How many quadratic equations do you think there are with \(x\)-intercepts at \(x=3\) and \(x=7 ?\) How are they related to one another?

Consider the equation \(y=x^{2}+6 x+10\). a. Convert this equation to vertex form by completing the square. b. Find the vertex. Graph both equations. c. Find the roots of the equation \(0=x^{2}+6 x+10\). What happens and why?

Solve each equation symbolically and check your answer. a. \((x+3)^{2}=7\) b. \((x-2)^{2}-8=13\)
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