Chapter 11: Problem 53
Sketch the graph of the function. $$y=\frac{1}{2} x^{2}$$
Short Answer
Expert verified
The graph of the function \(y=\frac{1}{2}x^{2}\) is an upward-opening parabola centered at the origin with points (-2,2), (-1,0.5), (0,0), (1,0.5) and (2,2).
Step by step solution
01
Identify Vertex
The vertex of the function \(y=ax^{2}+bx+c\) is given by the point \((-b/(2a), c-b^{2}/(4a))\). For the given equation, it can be seen that \(a=\frac{1}{2}\), \(b=0\) and \(c=0\). Therefore, the vertex is \((-0/(2*\frac{1}{2}), 0-0^{2}/(4*\frac{1}{2}))\) which simplifies to \((0,0)\).
02
Plot Some Points
Start with the vertex, (0,0), and choose some points to the left and right of the vertex to plot. For instance, let's choose x=-2, -1, 1, 2. When x=-2, y=\(\frac{1}{2}*(-2)^{2}=2\); When x=-1, y=\(\frac{1}{2}*(-1)^{2}=0.5\); When x=1, y=\(\frac{1}{2}*(1)^{2}=0.5\); and when x=2, y=\(\frac{1}{2}*(2)^{2}=2\). Plot these points along with the vertex.
03
Sketch the Parabola
Now we have sufficient points to sketch the parabola. The parabola opens upwards, and it is symmetric with respect to the y-axis. Therefore, plot the points and draw smooth curves on both sides of the y-axis passing through these points. This will result in an upward-opening parabola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola Sketching
Parabola sketching involves several key steps, beginning with understanding the general shape and symmetry of a parabola. In the exercise, the equation given is a simple quadratic function of the form \( y = ax^2 \), where \( a \) determines the direction in which the parabola opens. If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
For our equation, \( y = \frac{1}{2}x^2 \), since \( a = \frac{1}{2} \) and that's positive, the parabola opens upwards. This is essential to know before you begin plotting. The next step is to identify the vertex, which is the peak or the lowest point of the parabola, depending on the direction it opens. Then, choose a set of x-values and calculate their corresponding y-values to plot points on a graph. It's particularly helpful to choose symmetric x-values around the vertex, just as in the exercise, where x-values of -2, -1, 1, and 2 were selected.
Once the points are plotted on the graph, draw a smooth curve through them, making sure the parabola is symmetric across the axis of symmetry, which for our function is the y-axis. Remember that sketching a parabola is not about creating a perfect curve, but rather understand and represent the function's general shape and symmetrical properties.
For our equation, \( y = \frac{1}{2}x^2 \), since \( a = \frac{1}{2} \) and that's positive, the parabola opens upwards. This is essential to know before you begin plotting. The next step is to identify the vertex, which is the peak or the lowest point of the parabola, depending on the direction it opens. Then, choose a set of x-values and calculate their corresponding y-values to plot points on a graph. It's particularly helpful to choose symmetric x-values around the vertex, just as in the exercise, where x-values of -2, -1, 1, and 2 were selected.
Once the points are plotted on the graph, draw a smooth curve through them, making sure the parabola is symmetric across the axis of symmetry, which for our function is the y-axis. Remember that sketching a parabola is not about creating a perfect curve, but rather understand and represent the function's general shape and symmetrical properties.
Vertex of a Parabola
The vertex of a parabola is a crucial point that serves as the 'turning point' of the graph. Finding the vertex of a parabola given by \( y = ax^2 + bx + c \) is straightforward once you understand the formula \( (-\frac{b}{2a}, \frac{4ac - b^2}{4a}) \). The vertex is at the axis of symmetry of the parabola, and it represents the minimum or maximum point of the graph.
In our case, the function is \( y = \frac{1}{2} x^2 \), simplifying the process since \( b = 0 \) and \( c = 0 \). Here, the x-coordinate of the vertex is \( -\frac{0}{2*\frac{1}{2}} \), which equals zero and the y-coordinate also calculates to zero, placing our vertex at \( (0, 0) \).
The vertex is an integral part of sketching the parabola as it guides the direction in which the curve extends and is pivotal for symmetry. Whether tackling homework or plotting graphs in more advanced applications, knowing how to find a parabola’s vertex is an invaluable skill for making accurate and informative sketches.
In our case, the function is \( y = \frac{1}{2} x^2 \), simplifying the process since \( b = 0 \) and \( c = 0 \). Here, the x-coordinate of the vertex is \( -\frac{0}{2*\frac{1}{2}} \), which equals zero and the y-coordinate also calculates to zero, placing our vertex at \( (0, 0) \).
The vertex is an integral part of sketching the parabola as it guides the direction in which the curve extends and is pivotal for symmetry. Whether tackling homework or plotting graphs in more advanced applications, knowing how to find a parabola’s vertex is an invaluable skill for making accurate and informative sketches.
Plotting Quadratic Functions
Plotting quadratic functions is a methodical process which involves taking the standard form of a quadratic equation, \( y = ax^2 + bx + c \), and translating it into a visual representation on a graph, known as a parabola. This process starts with identifying the vertex, as discussed previously. Following this, you select a range of x-values, and calculate the corresponding y-values to get points that will lie on the graph.
In the exercise provided, the points (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2) were calculated and hence plotted. When plotting these points, it is useful to remember that quadratic functions are symmetrical, so points equidistant from the vertex on the x-axis will have the same y-values.
Once sufficient points are plotted, the pattern or shape begins to emerge, which is the recognizable 'U' shape of the parabola. By connecting these points with a smooth, continuous curve, we complete the plot of the quadratic function. It's important to ensure that the graph is a mirror image on either side of the vertex to maintain the correct symmetry of the parabola. With practice, plotting quadratic functions becomes more intuitive, making it easier to create accurate and informative graphs.
In the exercise provided, the points (-2, 2), (-1, 0.5), (0, 0), (1, 0.5), and (2, 2) were calculated and hence plotted. When plotting these points, it is useful to remember that quadratic functions are symmetrical, so points equidistant from the vertex on the x-axis will have the same y-values.
Once sufficient points are plotted, the pattern or shape begins to emerge, which is the recognizable 'U' shape of the parabola. By connecting these points with a smooth, continuous curve, we complete the plot of the quadratic function. It's important to ensure that the graph is a mirror image on either side of the vertex to maintain the correct symmetry of the parabola. With practice, plotting quadratic functions becomes more intuitive, making it easier to create accurate and informative graphs.