Chapter 4: Problem 7
The graph of a quadratic function is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of f. (c) Find the domain and range of f. $$ f(x)=2 x^{2}-4 x-1 $$
Short Answer
Expert verified
(a) Vertex: (1, -3); (b) Minimum value: -3; (c) Domain: (-∞, ∞), Range: [-3, ∞).
Step by step solution
01
Identify the form of the quadratic function
The given quadratic function \( f(x) = 2x^2 - 4x - 1 \) is in the standard form \( ax^2 + bx + c \), where \( a=2 \), \( b=-4 \), and \( c=-1 \).
02
Find the vertex using the vertex formula
The vertex of a quadratic function \( ax^2 + bx + c \) is given by \( x = \frac{-b}{2a} \). Substitute \( b = -4 \) and \( a = 2 \) into the formula to find \( x \):\[ x = \frac{-(-4)}{2 \times 2} = \frac{4}{4} = 1 \]Then, substitute \( x = 1 \) back into the function to find \( y \):\[ f(1) = 2(1)^2 - 4(1) - 1 = 2 - 4 - 1 = -3 \]Thus, the vertex is \((1, -3)\).
03
Determine if the vertex is a maximum or a minimum
Since the coefficient \( a = 2 \) is positive, the parabola opens upwards, indicating that the vertex represents the minimum point of the function. Therefore, the minimum value of \( f(x) \) is \(-3\) at \( x = 1 \).
04
Find the domain of the function
The domain of any quadratic function is all real numbers, since it is a polynomial function. Thus, the domain is \((-\infty, \infty)\).
05
Find the range of the function
Since the vertex is a minimum at \( y = -3 \) and the parabola opens upwards, the range of the function is all real numbers greater than or equal to \(-3\). Therefore, the range is \([-3, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex of a Parabola
To understand a quadratic function, it's important to identify its vertex. The vertex of a parabola is a point that represents the peak or valley of the curve. In the context of the quadratic function, this either means the highest point for a downward-opening parabola or the lowest point for an upward-opening one. The standard formula for a quadratic function is given by:- \(f(x) = ax^2 + bx + c\).To find the vertex, you use the vertex formula:- \(x = \frac{-b}{2a}\).
Substitute \(b = -4\) and \(a = 2\) into the formula to find the x-coordinate of the vertex. For the function \(f(x) = 2x^2 - 4x - 1\), the x-coordinate of the vertex is:- \(x = \frac{-(-4)}{2 \times 2} = 1\).
Once you have the x-coordinate, substitute it back into the function to find the y-coordinate. When \(x = 1\), the function evaluates to:- \(f(1) = 2(1)^2 - 4(1) - 1 = -3\).Thus, the vertex is located at the coordinates \((1, -3)\). Identifying the vertex is crucial for understanding the behavior of the parabola and predicting where it changes direction.
Substitute \(b = -4\) and \(a = 2\) into the formula to find the x-coordinate of the vertex. For the function \(f(x) = 2x^2 - 4x - 1\), the x-coordinate of the vertex is:- \(x = \frac{-(-4)}{2 \times 2} = 1\).
Once you have the x-coordinate, substitute it back into the function to find the y-coordinate. When \(x = 1\), the function evaluates to:- \(f(1) = 2(1)^2 - 4(1) - 1 = -3\).Thus, the vertex is located at the coordinates \((1, -3)\). Identifying the vertex is crucial for understanding the behavior of the parabola and predicting where it changes direction.
Domain and Range
The domain and range are essential elements in describing a function. They tell you where the function lives in terms of input and output values. For any quadratic function, which is a type of polynomial function, the domain is always all real numbers. This means that you can input any real number into a quadratic function and it will yield a result. Hence, the domain is:- \((-\infty, \infty)\).The range of a quadratic function, however, depends on the direction in which the parabola opens. In our function \(f(x) = 2x^2 - 4x - 1\), the parabola opens upwards (since \(a = 2\) is positive). This means the vertex provides the minimum point of the parabola. For our specific function, the vertex is \(-3\), and because the parabola opens upwards, the range includes every number larger than or equal to \(-3\). Thus, the range is:- \([-3, \infty)\).Understanding the domain and range gives a complete picture of where the parabola lies and how it behaves with respect to the x and y axes.
Maximum and Minimum Value
A quadratic function can have either a maximum or minimum value, and this depends on the orientation of its parabola. If the parabola opens upwards, like in the function \(f(x) = 2x^2 - 4x - 1\), it means that the function has a minimum value.As determined through calculations, the vertex of \(f(x)\) provides the coordinates \((1, -3)\). Therefore, the minimum value of the function is \(-3\). This occurs at the x-coordinate where the vertex is located, which is \(x = 1\). It's important to remember that when the coefficient \(a\) in the quadratic formula is positive, this indicates an upward opening, leading to a minimum value. Conversely, a downward opening signal would mean \(a\) is negative, resulting in a maximum value. By noting the orientation of the parabola, one can immediately identify whether the vertex signifies a minimum or maximum value, which is a key aspect in analyzing quadratic functions.