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Magazine Chapter 3: Problem 20
In Exercises \(5-36,\) graph the given functions. $$h=20 t-5 t^{2}$$
Short Answer
Expert verified
The function is a downward-opening parabola with vertex (2, 20) and intercepts at (0, 0) and (4, 0).
Step by step solution
01
Identify the Function Type
The function provided is \( h = 20t - 5t^2 \). This is a quadratic function, which typically has the form \( at^2 + bt + c \). The graph of a quadratic function is a parabola.
02
Determine the Vertex Form
The function can be rewritten in vertex form by completing the square. Start by rearranging terms: \( h = -5(t^2 - 4t) \). Next, complete the square inside the parentheses: \( h = -5((t-2)^2 -4) \). Thus, \( h = -5(t-2)^2 + 20 \). The vertex form is \( h = a(t-h)^2 + k \), where \( (h, k) = (2, 20) \) is the vertex of the parabola.
03
Identify Key Features
From the vertex form \( h = -5(t-2)^2 + 20 \), we identify that the parabola opens downwards since the coefficient of the squared term is negative (-5). The vertex of the parabola is at \( (2, 20) \). This is the maximum point of the parabola.
04
Calculate Intercepts
To find the \( t \)-intercepts, set \( h = 0 \): \( 0 = 20t - 5t^2 \). Factoring gives \( 0 = -5t(t-4) \), resulting in \( t = 0 \) and \( t = 4 \). These are the \( t \)-intercepts. For the \( h \)-intercept, use \( t = 0 \): \( h = 20(0) - 5(0)^2 = 0 \). The \( h \)-intercept is also at \( (0, 0) \).
05
Sketch the Graph
Plot the vertex \( (2, 20) \) and the \( t \)-intercepts \( (0, 0) \) and \( (4, 0) \) on a coordinate plane. Since the parabola opens downward, draw a symmetrical curve through these points, making sure the vertex is the highest point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola Graphing
Graphing a parabola involves plotting a quadratic function on a coordinate plane. A quadratic function takes the form \( y = ax^2 + bx + c \), where the graph will look like the U-shape or an upside-down U-shape. This shape is known as a parabola. In this exercise, we look at the quadratic function \( h = 20t - 5t^2 \).
The key parts of a parabola include the vertex, axis of symmetry, direction in which it opens (either upwards or downwards), and its intercepts. To draw the graph correctly:
The key parts of a parabola include the vertex, axis of symmetry, direction in which it opens (either upwards or downwards), and its intercepts. To draw the graph correctly:
- First, identify the vertex, as it is the most crucial point on a parabola. It is either its highest or lowest point.
- Determine the intercepts, where the graph crosses the axes.
- Look at the direction the parabola opens based on the coefficient of the squared term.
Vertex Form
The vertex form of a quadratic function provides a convenient way to determine the vertex of a parabola. The vertex form is expressed as \( h = a(t-h)^2 + k \), where \((h, k)\) is the vertex. Converting a quadratic equation into vertex form involves completing the square.
In this exercise, the given function \( h = 20t - 5t^2 \) is converted into vertex form through a series of steps:
In this exercise, the given function \( h = 20t - 5t^2 \) is converted into vertex form through a series of steps:
- Rearrange terms to focus on the \( t^2 \) and \( t \) parts, resulting in \( h = -5(t^2 - 4t) \).
- Complete the square within the parentheses to get \( h = -5((t-2)^2 - 4) \).
- Simplify to find the function as \( h = -5(t-2)^2 + 20 \).
t-Intercepts
The \( t \)-intercepts of a quadratic function are the points where the graph crosses the \( t \)-axis. These are found by setting \( h = 0 \) in the equation and solving for \( t \). In this problem, set the equation \( 0 = 20t - 5t^2 \); this is equivalent to:
- Factoring out common terms, we get \( 0 = -5t(t-4) \).
- Solve for \( t \) to find the intercepts as \( t = 0 \) and \( t = 4 \).
Completing the Square
Completing the square is a technique used to convert a quadratic expression from its standard form \( ax^2 + bx + c \) to vertex form \( a(t-h)^2 + k \). This process helps to easily identify the vertex and other graph features of a quadratic function. Here鈥檚 how it is done in the provided exercise:
- Start with \( h = 20t - 5t^2 \) and rearrange it to focus on the quadratic part as \( h = -5(t^2 - 4t) \).
- Inside the parentheses, add and subtract the square of half the coefficient of \( t \), turning \( t^2 - 4t \) into \( (t-2)^2 - 4 \).
- The function then becomes \( h = -5((t-2)^2 - 4) \) and simplifies further to the vertex form \( h = -5(t-2)^2 + 20 \).
