Problem 25
Question
Graph each of the following linear and quadratic functions. $$f(x)=-2 x^{2}-1$$
Step-by-Step Solution
Verified Answer
Graph the downward-opening parabola with vertex at (0, -1) and through (1, -3) and (-1, -3).
1Step 1: Identify the Type of Function
The given function is \( f(x) = -2x^2 - 1 \). This is a quadratic function because it has the form \( ax^2 + bx + c \), where \( a = -2 \), \( b = 0 \), and \( c = -1 \).
2Step 2: Determine the Direction of the Parabola
Look at the coefficient of \( x^2 \). Since \( a = -2 \) is negative, the parabola opens downward.
3Step 3: Find the Vertex of the Parabola
For a quadratic function in the form \( ax^2 + bx + c \), the x-coordinate of the vertex is calculated by \( x = -\frac{b}{2a} \). Here, \( b = 0 \), so \( x = 0 \). Substitute \( x = 0 \) into the function to find \( y = -2(0)^2 - 1 = -1 \). Thus, the vertex is at \( (0, -1) \).
4Step 4: Find the y-Intercept
The y-intercept of a function \( f(x) \) is found by evaluating \( f(0) \). In this case, \( f(0) = -2(0)^2 - 1 = -1 \), so the y-intercept is \( (0, -1) \).
5Step 5: Find Additional Points for the Graph
Choose x-values to find more points. For example, use \( x = 1 \) and \( x = -1 \):\[ f(1) = -2(1)^2 - 1 = -3 \] \[ f(-1) = -2(-1)^2 - 1 = -3 \]Thus, two more points are \( (1, -3) \) and \( (-1, -3) \).
6Step 6: Sketch the Graph
Plot the points \( (0, -1) \), \( (1, -3) \), \( (-1, -3) \), and draw a symmetric parabola opening downward with vertex \( (0, -1) \). Connect these points smoothly to complete the graph.
Key Concepts
ParabolaVertexy-interceptGraphing
Parabola
A parabola is the graph of a quadratic function, which is always shaped like a "U" or an upside-down "U" (also known as an "n"). It is defined by a quadratic equation, which takes the form \( ax^2 + bx + c \). The parabola's direction depends on the sign of the coefficient \( a \). If \( a \) is positive, the parabola opens upwards, resembling a valley. If \( a \) is negative, it opens downwards, appearing like a hill.
Key features of the parabola include:
Key features of the parabola include:
- The axis of symmetry, which is a vertical line passing through the vertex, dividing the parabola into two mirror-image halves.
- The direction in which it opens is determined by the sign of \( a \).
- The steepness or width of the parabola depends on the absolute value of \( a \). Smaller values create wider parabolas, while larger values create narrower ones.
Vertex
The vertex of a parabola is its highest or lowest point, dependent on the direction it opens. It serves as a critical turning point.
For a quadratic function in standard form \( ax^2 + bx + c \), the vertex can be calculated using the formula for the x-coordinate, \( x = -\frac{b}{2a} \).
Once the x-coordinate is found, substitute it back into the function to find the y-coordinate. Together, these form the vertex \( (x, y) \).
In our case with the function \( f(x) = -2x^2 - 1 \), where \( a = -2 \) and \( b = 0 \), using \( x = -\frac{0}{2(-2)} = 0 \), we determine that our vertex is at \( (0, -1) \). The vertex acts as a peak since the parabola opens downwards.
For a quadratic function in standard form \( ax^2 + bx + c \), the vertex can be calculated using the formula for the x-coordinate, \( x = -\frac{b}{2a} \).
Once the x-coordinate is found, substitute it back into the function to find the y-coordinate. Together, these form the vertex \( (x, y) \).
In our case with the function \( f(x) = -2x^2 - 1 \), where \( a = -2 \) and \( b = 0 \), using \( x = -\frac{0}{2(-2)} = 0 \), we determine that our vertex is at \( (0, -1) \). The vertex acts as a peak since the parabola opens downwards.
y-intercept
The y-intercept is where the graph of the function crosses the y-axis. For any function, it is found by substituting \( x = 0 \) into the function. This gives the point \( (0, c) \) assuming the quadratic equation \( ax^2 + bx + c \).
In the given function \( f(x) = -2x^2 - 1 \), plugging in \( x = 0 \) results in \( f(0) = -2(0)^2 - 1 = -1 \). Hence, the y-intercept is the point \( (0, -1) \), which coincidentally in this function coincides with the vertex.
Knowing the y-intercept aids in sketching the graph, providing a pivotal point for alignment.
In the given function \( f(x) = -2x^2 - 1 \), plugging in \( x = 0 \) results in \( f(0) = -2(0)^2 - 1 = -1 \). Hence, the y-intercept is the point \( (0, -1) \), which coincidentally in this function coincides with the vertex.
Knowing the y-intercept aids in sketching the graph, providing a pivotal point for alignment.
Graphing
Graphing quadratic functions involves plotting a curve based on key points and properties such as the direction, vertex, and y-intercept. These elements create a clear roadmap for sketching.
Here's a simple step-by-step guide:
Here's a simple step-by-step guide:
- Identify if the parabola opens upwards or downwards by checking the sign of \( a \).
- Calculate the vertex and plot it on the graph.
- Find the y-intercept by evaluating \( f(0) \), and plot it.
- Choose additional easy-to-calculate x-values to get extra points on the graph, ensuring symmetry around the vertex.
- Connect all these points smoothly with a curve to form the parabola.
This helps you visualize the function's behavior precisely.
Other exercises in this chapter
Problem 25
Show that \((f \circ g)(x)=x\) and \((g \circ f)\) \((x)=x\) for each pair of functions. \(f(x)=-\frac{1}{4} x-\frac{1}{2}\) and \(g(x)=-4 x-2\)
View solution Problem 25
Graph each of the functions. $$f(x)=|x-1|+2$$
View solution Problem 25
Specify the domain for each of the functions. $$4f(t)=\frac{3 t}{t^{2}-4}$$
View solution Problem 26
If \(y\) is directly proportional to \(x\), and \(y=42\) when \(x=28\), find the value of \(y\) when \(x=38\).
View solution