Chapter 5: Problem 12
Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=2 x^{2} $$
Short Answer
Expert verified
Y-intercept: (0,0); Axis of symmetry: x=0; Vertex: (0,0).
Step by step solution
01
Find the y-intercept
To find the y-intercept of the function, evaluate the function at \(x = 0\). This gives us \(f(0) = 2(0)^2 = 0\). Thus, the y-intercept is \((0, 0)\).
02
Find the Axis of Symmetry and X-Coordinate of the Vertex
For a quadratic function in the form \(f(x) = ax^2 + bx + c\), the axis of symmetry is given by the formula \(x = -\frac{b}{2a}\). For \(f(x) = 2x^2\), \(a = 2\) and \(b = 0\). Thus, the axis of symmetry is \(x = -\frac{0}{2 \times 2} = 0\). The x-coordinate of the vertex is the same, \(x = 0\).
03
Find the Vertex
Since the x-coordinate of the vertex is \(0\) and we have already calculated \(f(0) = 0\), the vertex of the function is \((0, 0)\).
04
Create a Table of Values
To create a table of values, select a few points around the vertex \(x = 0\), for example, \(x = -2, -1, 0, 1, 2\). Calculate \(f(x)\) for each value of \(x\): - \(f(-2) = 2(-2)^2 = 8\) - \(f(-1) = 2(-1)^2 = 2\) - \(f(0) = 0\) - \(f(1) = 2(1)^2 = 2\) - \(f(2) = 2(2)^2 = 8\). This gives the table: \[\begin{array}{c|c}x & f(x) \\hline-2 & 8 \-1 & 2 \0 & 0 \1 & 2 \2 & 8 \\end{array}\]
05
Graph the Function
Use the information to plot the graph. Start by plotting the y-intercept and vertex \((0, 0)\). Then plot the points from the table: \((-2, 8), (-1, 2), (1, 2), (2, 8)\). Draw a symmetric curve through these points, which open upwards as the coefficient of \(x^2\) is positive. The line \(x=0\) is the axis of symmetry.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Y-Intercept
The y-intercept of a quadratic function is the point where the graph crosses the y-axis. It occurs when the value of \(x\) is zero. For any quadratic equation in the form \(f(x) = ax^2 + bx + c\), the y-intercept is simply \(f(0) = c\). This is because when you substitute \(x = 0\) into the equation, the terms involving \(x\) disappear, leaving only the constant term.
In our example, the function is \(f(x) = 2x^2\). Thus, \(f(0) = 2(0)^2 = 0\), and the y-intercept is at the origin, \((0, 0)\). This means that the graph intersects the y-axis at this point. The y-intercept provides a starting point for graphing the parabola.
In our example, the function is \(f(x) = 2x^2\). Thus, \(f(0) = 2(0)^2 = 0\), and the y-intercept is at the origin, \((0, 0)\). This means that the graph intersects the y-axis at this point. The y-intercept provides a starting point for graphing the parabola.
Axis of Symmetry
The axis of symmetry of a parabola is a vertical line that divides the graph into two mirror-image halves. It passes through the vertex of the parabola. For any quadratic function in standard form \(f(x) = ax^2 + bx + c\), the axis of symmetry can be found using the formula \(x = -\frac{b}{2a}\).
In the example \(f(x) = 2x^2\), the coefficients are \(a = 2\) and \(b = 0\). Plugging these into the formula gives \(x = -\frac{0}{2 \times 2} = 0\). Thus, the axis of symmetry is the line \(x = 0\), which is actually the y-axis in this case.
In the example \(f(x) = 2x^2\), the coefficients are \(a = 2\) and \(b = 0\). Plugging these into the formula gives \(x = -\frac{0}{2 \times 2} = 0\). Thus, the axis of symmetry is the line \(x = 0\), which is actually the y-axis in this case.
- Helps determine the location of the vertex.
- Ensures that the parabola is symmetric on either side of this line.
Vertex of Parabola
The vertex of a parabola is its highest or lowest point, depending on whether it opens upwards or downwards. It gives the graph its distinct U-shape. For a function in the form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex is found using the axis of symmetry, as \(x = -\frac{b}{2a}\).
Once the x-coordinate is known, substitute it back into the function to find the y-coordinate. For \(f(x) = 2x^2\), we have already determined that \(x = 0\). Substituting \(x = 0\) back into the function, \(f(0) = 0\). Therefore, the vertex of this parabola is \((0, 0)\).
Once the x-coordinate is known, substitute it back into the function to find the y-coordinate. For \(f(x) = 2x^2\), we have already determined that \(x = 0\). Substituting \(x = 0\) back into the function, \(f(0) = 0\). Therefore, the vertex of this parabola is \((0, 0)\).
- Indicates the minimum or maximum point of the parabola.
- Essential for creating the table of values for graphing.
Graphing Parabolas
Graphing a parabola involves several steps that tie together the concepts of y-intercept, vertex, and axis of symmetry. Once these are found, you can plot the parabola on a graph. First, place the y-intercept and vertex on the graph. Then, choose a few points on either side of the vertex along the x-axis to help determine the shape.
For \(f(x) = 2x^2\), we can create a table by substituting different values of \(x\) and finding \(f(x)\). For example:
For \(f(x) = 2x^2\), we can create a table by substituting different values of \(x\) and finding \(f(x)\). For example:
- \(x = -2\), \(f(x) = 8\)
- \(x = -1\), \(f(x) = 2\)
- \(x = 0\), \(f(0) = 0\)
- \(x = 1\), \(f(1) = 2\)
- \(x = 2\), \(f(2) = 8\)