Problem 7
Question
Graph each function by plotting points, and identify the domain and range. $$f(x)=-x^{2}-1$$
Step-by-Step Solution
Verified Answer
The vertex of the given quadratic function $$f(x) = -x^2 - 1$$ is at point \((0,-1)\). After plotting additional points, we form a symmetric, downward-opening parabola. The domain of the function is all real numbers: \(\text{Domain} = (-\infty, \infty)\). The range of the function is all real numbers less than or equal to -1: \(\text{Range} = (-\infty, -1]\).
1Step 1: 1. Identify the vertex
The vertex form of a quadratic function is given by: \(f(x)=a(x-h)^{2}+k\), where (h, k) is the vertex of the parabola. The given quadratic function is already in vertex form, with \(a=-1\), \(h=0\), and \(k=-1\). Thus, the vertex of the given parabola is \((0, -1)\).
2Step 2: 2. Choose and plot some points around the vertex
Now, let's pick a few points around the vertex, plug in their x-coordinates into the given function, and find their corresponding y-coordinates:
- When \(x= -2\), \(f(x)= -(-2)^{2}-1 = -4 - 1 = -5\), so we get the point \((-2, -5)\).
- When \(x= -1\), \(f(x)= -(-1)^{2}-1 = -1 - 1 = -2\), so we get the point \((-1, -2)\).
- When \(x= 0\), \(f(x)= -(0)^{2}-1 = -1\), which we already got as the vertex \((0, -1)\).
- When \(x= 1\), \(f(x)= -(1)^{2}-1 = -1 - 1 = -2\), so we get the point \((1, -2)\).
- When \(x= 2\), \(f(x)= -(2)^{2}-1 = -4 - 1 = -5\), so we get the point \((2, -5)\).
3Step 3: 3. Create the graph of the function
Using the points above, along with the knowledge that it is a quadratic function with \(a=-1\), which indicates a downward-opening parabola, we can create the graph of the function. The points determined above, including the vertex, create a symmetric parabola.
4Step 4: 4. Determine the domain and range
The domain of a function consists of all possible input values (x-values) for which the function is defined. Since a quadratic function is defined for all real numbers, the domain of the given function is: \(\text{Domain} = (-\infty, \infty)\).
The range of a function consists of all possible output values (y-values) for which the function is defined. Since our parabola is downward-opening with a vertex of \((0, -1)\), its maximum value is -1. The parabola will continue downward, so there is no minimum value. Therefore, the range of the given function is: \(\text{Range} = (-\infty ,-1] \).
Key Concepts
Graphing Quadratic FunctionsVertex Form of a Quadratic FunctionDomain and Range of Quadratic Functions
Graphing Quadratic Functions
Graphing a quadratic function helps us understand its shape and behavior. In this exercise, we graphed the function \( f(x) = -x^2 - 1 \). To start, we identified the vertex, which is a key point that helps in sketching the parabola.
The vertex is at the highest or lowest point, depending on whether the parabola opens downward or upward. Since the coefficient \(a = -1\) is negative, our parabola opens downward, creating an upside-down 'U' shape.
We then plotted a series of points around the vertex. By choosing various \(x\) values like -2, -1, 0, 1, and 2, and substituting them into the quadratic equation, we obtained the points:
The vertex is at the highest or lowest point, depending on whether the parabola opens downward or upward. Since the coefficient \(a = -1\) is negative, our parabola opens downward, creating an upside-down 'U' shape.
We then plotted a series of points around the vertex. By choosing various \(x\) values like -2, -1, 0, 1, and 2, and substituting them into the quadratic equation, we obtained the points:
- \((-2, -5)\)
- \((-1, -2)\)
- \((0, -1)\) - the vertex
- \((1, -2)\)
- \((2, -5)\)
Vertex Form of a Quadratic Function
The vertex form of a quadratic function is essential for easily identifying a parabola's vertex. It is represented as \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex.
For the function \(f(x) = -x^2 - 1\), we can see it in the form \( f(x) = -1(x-0)^2 - 1 \), showing:
If \(a\) is positive, the parabola opens upwards, resembling a 'U' shape. Conversely, if \(a\) is negative like in our case, the parabola opens downwards. The vertex form is particularly convenient when converting standard form equations and understanding the parabola's key features more intuitively.
For the function \(f(x) = -x^2 - 1\), we can see it in the form \( f(x) = -1(x-0)^2 - 1 \), showing:
- \(a = -1\)
- \(h = 0\)
- \(k = -1\)
If \(a\) is positive, the parabola opens upwards, resembling a 'U' shape. Conversely, if \(a\) is negative like in our case, the parabola opens downwards. The vertex form is particularly convenient when converting standard form equations and understanding the parabola's key features more intuitively.
Domain and Range of Quadratic Functions
Understanding the domain and range of a quadratic function is crucial for analyzing its behavior. The domain of quadratic functions is always all real numbers, \((-\infty, \infty)\), because you can input any \(x\) value into the quadratic and still calculate \(f(x)\).
For the function \(f(x) = -x^2 - 1\), this is true as well.
When it comes to the range, it depends on the orientation of the parabola. Since our parabola opens downward, it has a maximum point at the vertex, \( (0, -1) \).
This means that \(f(x)\) cannot exceed -1. As the parabola continues indefinitely downwards, the range of the function is \((-\infty, -1] \).
Graphically, this tells us the lowest \(y\)-values that \(f(x)\) can take, with -1 being the highest point it can reach.
For the function \(f(x) = -x^2 - 1\), this is true as well.
When it comes to the range, it depends on the orientation of the parabola. Since our parabola opens downward, it has a maximum point at the vertex, \( (0, -1) \).
This means that \(f(x)\) cannot exceed -1. As the parabola continues indefinitely downwards, the range of the function is \((-\infty, -1] \).
Graphically, this tells us the lowest \(y\)-values that \(f(x)\) can take, with -1 being the highest point it can reach.
Other exercises in this chapter
Problem 7
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