Problem 18
Question
Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=x^{2}-1 $$
Step-by-Step Solution
Verified Answer
The vertex is \((0, -1)\), parabola opens upward, y-intercept is \((0, -1)\), x-intercepts are \((1, 0)\) and \((-1, 0)\).
1Step 1: Identify the standard form
The given quadratic function is \( f(x) = x^2 - 1 \). It is already in the standard form \( ax^2 + bx + c \) where \( a = 1 \), \( b = 0 \), and \( c = -1 \).
2Step 2: Determine the direction of opening
Since \( a = 1 \) is positive, the graph of the quadratic function opens upward.
3Step 3: Find the vertex
The vertex of a parabola in the form \( ax^2 + bx + c \) is given by the formula \( x = \frac{-b}{2a} \). Here, \( b = 0 \) and \( a = 1 \), so \( x = \frac{-0}{2 \times 1} = 0 \). Substitute \( x = 0 \) into the function to find \( y \): \( f(0) = 0^2 - 1 = -1 \). Thus, the vertex is \((0, -1)\).
4Step 4: Find the y-intercept
The y-intercept can be found by evaluating \( f(x) \) at \( x = 0 \). We have already determined this in Step 3, where \( f(0) = -1 \). Therefore, the y-intercept is \((0, -1)\).
5Step 5: Find the x-intercepts
The x-intercepts occur where \( f(x) = 0 \). Solve \( x^2 - 1 = 0 \):\[ x^2 = 1 \]which simplifies to \( x = \pm 1 \). Thus, the x-intercepts are \((1, 0)\) and \((-1, 0)\).
6Step 6: Graph the function
Plot the vertex at \((0, -1)\) on the coordinate plane. Plot the y-intercept at \((0, -1)\) and the x-intercepts at \((1, 0)\) and \((-1, 0)\). Draw the parabola opening upward through these points.
Key Concepts
Parabola VertexInterceptsGraphing Quadratics
Parabola Vertex
The vertex of a parabola is a fundamental feature of a quadratic function. It's the point where the parabola reaches its minimum or maximum value. For a quadratic function in standard form, \[ ax^2 + bx + c \],the vertex can be found using the formula \( x = \frac{-b}{2a} \).
In our example, the function is \( f(x) = x^2 - 1 \).Here, \( a = 1 \)and \( b = 0 \).Plugging these values into the formula gives us:\[ x = \frac{-0}{2 \times 1} = 0 \].
Now, substitute \( x = 0 \) back into the function to find the \( y \)-value:\[ f(0) = 0^2 - 1 = -1 \].
Thus, the vertex of the parabola is at the point \((0, -1)\).This means the parabola's lowest point (since it opens upward) is at \((0, -1)\).It's important to note this vertex is the turning point of the graph.
In our example, the function is \( f(x) = x^2 - 1 \).Here, \( a = 1 \)and \( b = 0 \).Plugging these values into the formula gives us:\[ x = \frac{-0}{2 \times 1} = 0 \].
Now, substitute \( x = 0 \) back into the function to find the \( y \)-value:\[ f(0) = 0^2 - 1 = -1 \].
Thus, the vertex of the parabola is at the point \((0, -1)\).This means the parabola's lowest point (since it opens upward) is at \((0, -1)\).It's important to note this vertex is the turning point of the graph.
Intercepts
Intercepts provide key points that show where a graph crosses the axes. There are two types of intercepts: the y-intercept and the x-intercepts. For our quadratic function, \( f(x) = x^2 - 1 \), let's find these intercepts.
- **Y-Intercept**: The y-intercept is the point where the graph crosses the y-axis. This occurs when \( x = 0 \).By evaluating the function at \( x = 0 \),we find \( f(0) = 0^2 - 1 = -1 \).Thus, the y-intercept is \((0, -1)\).
- **X-Intercepts**: The x-intercepts are the points where the graph crosses the x-axis. These points occur where \( f(x) = 0 \).Set \( x^2 - 1 = 0 \)and solve:\[ x^2 = 1 \]\[ x = \pm 1 \].
Hence, the x-intercepts are \((1, 0)\) and \((-1, 0)\).
Graphing Quadratics
Graphing a quadratic function may initially seem challenging, but it becomes simpler once you understand its components. Let's graph the given quadratic function \( f(x) = x^2 - 1 \), using the vertex and intercepts.
To start, plot the vertex \((0, -1)\).This is the lowest point of the parabola because it opens upwards. Next, plot the intercepts. Place the y-intercept at \((0, -1)\)(the graph will pass through this point as well, making it a critical location).
Then, plot the x-intercepts at \((1, 0)\) and \((-1, 0)\).The parabola will cross these points because these are where the graph touches the x-axis.
Finally, draw a smooth curve that passes through these key points, forming a 'U' shape that opens upward from the vertex. This curve visually represents the input-output relationship set by the quadratic equation \( f(x) = x^2 - 1 \).
Graphing allows us to not only visualize the function but also understand its real-world implications, such as finding maximum and minimum values or roots.
To start, plot the vertex \((0, -1)\).This is the lowest point of the parabola because it opens upwards. Next, plot the intercepts. Place the y-intercept at \((0, -1)\)(the graph will pass through this point as well, making it a critical location).
Then, plot the x-intercepts at \((1, 0)\) and \((-1, 0)\).The parabola will cross these points because these are where the graph touches the x-axis.
Finally, draw a smooth curve that passes through these key points, forming a 'U' shape that opens upward from the vertex. This curve visually represents the input-output relationship set by the quadratic equation \( f(x) = x^2 - 1 \).
Graphing allows us to not only visualize the function but also understand its real-world implications, such as finding maximum and minimum values or roots.
Other exercises in this chapter
Problem 18
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Graph each quadratic function. Label the vertex and sketch and label the axis of symmetry. \(h(x)=(x+4)^{2}-6\)
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