Chapter 8: Problem 30
Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples 1 through 4 . $$ f(x)=2 x^{2}-x-3 $$
Short Answer
Expert verified
Vertex: \( (\frac{1}{4}, -\frac{25}{8}) \); Opens upward; x-intercepts: \( (-1,0) \), \( (\frac{3}{2},0) \); y-intercept: \( (0,-3) \).
Step by step solution
01
Identify the Coefficients
The quadratic function is given by \( f(x) = ax^2 + bx + c \). In this case, \( a = 2 \), \( b = -1 \), and \( c = -3 \). These coefficients will help us find various properties of the function.
02
Determine the Direction of Opening
The sign of \( a \) determines whether the parabola opens upwards or downwards. Since \( a = 2 \) and it is positive, the parabola opens upwards.
03
Find the Vertex Using the Vertex Formula
The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substitute \( b = -1 \) and \( a = 2 \): \[ x = -\frac{-1}{2 \times 2} = \frac{1}{4}. \] Now, find \( y \) by substituting \( x = \frac{1}{4} \) into the function:\[ f\left(\frac{1}{4}\right) = 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} - 3. \]Calculate this to find: \[ f\left(\frac{1}{4}\right) = 2\left(\frac{1}{16}\right) - \frac{1}{4} - 3 = \frac{1}{8} - \frac{2}{8} - \frac{24}{8} = -\frac{25}{8}. \]The vertex is \( \left( \frac{1}{4}, -\frac{25}{8} \right) \).
04
Find the y-intercept
The y-intercept is where \( x = 0 \). Substitute \( x = 0 \) into the function \( f(x) = 2x^2 - x - 3 \): \[ f(0) = 2(0)^2 - 0 - 3 = -3. \]So, the y-intercept is \( (0, -3) \).
05
Find the x-intercepts (if any)
The x-intercepts occur where \( f(x) = 0 \). Set the equation \( 2x^2 - x - 3 = 0 \) and solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = -1 \), and \( c = -3 \):\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}. \]Therefore, the x-intercepts are \( x = \frac{6}{4} = \frac{3}{2} \) and \( x = \frac{-4}{4} = -1 \). The points are \( (\frac{3}{2}, 0) \) and \( (-1, 0) \).
06
Sketch the Graph
Plot the vertex \( \left( \frac{1}{4}, -\frac{25}{8} \right) \), the y-intercept \( (0, -3) \), and the x-intercepts \( (\frac{3}{2}, 0) \) and \( (-1, 0) \). Since the parabola opens upward, draw a smooth curve through these points. The graph is symmetric around the line \( x = \frac{1}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex of a Parabola
The vertex of a parabola is a crucial point that indicates where the graph reaches its maximum or minimum value. In the standard quadratic function format, \( f(x) = ax^2 + bx + c \), the vertex can be found using the vertex formula, \( x = -\frac{b}{2a} \). This formula gives us the x-coordinate of the vertex.
To find the y-coordinate, substitute this x-value back into the function. For the equation \( f(x) = 2x^2 - x - 3 \), the x-coordinate is calculated as \( x = \frac{1}{4} \).
Substituting back into the equation, the y-coordinate turns out to be \( y = -\frac{25}{8} \). Thus, the vertex is \( \left( \frac{1}{4}, -\frac{25}{8} \right) \). The vertex is vital as it helps in determining the parabola's symmetry and direction.
To find the y-coordinate, substitute this x-value back into the function. For the equation \( f(x) = 2x^2 - x - 3 \), the x-coordinate is calculated as \( x = \frac{1}{4} \).
Substituting back into the equation, the y-coordinate turns out to be \( y = -\frac{25}{8} \). Thus, the vertex is \( \left( \frac{1}{4}, -\frac{25}{8} \right) \). The vertex is vital as it helps in determining the parabola's symmetry and direction.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots, or x-intercepts, of a quadratic equation. These are the points where the graph crosses the x-axis. The formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula works for any quadratic equation of the form \( ax^2 + bx + c = 0 \).
By substituting the coefficients from our example \( a = 2 \), \( b = -1 \), and \( c = -3 \), into the formula, we solve to find the x-intercepts as \( \frac{3}{2} \) and \( -1 \).
By substituting the coefficients from our example \( a = 2 \), \( b = -1 \), and \( c = -3 \), into the formula, we solve to find the x-intercepts as \( \frac{3}{2} \) and \( -1 \).
- The "\( \pm \)" signifies that there are usually two solutions or roots.
- In our case, these roots confirm the graph's intersections with the x-axis at points \( \left( \frac{3}{2}, 0 \right) \) and \( \left( -1, 0 \right) \).
Graphing Parabolas
Graphing a parabola involves plotting points such as the vertex, y-intercept, and x-intercepts, and then drawing a smooth curve through these points.
The vertex \( \left( \frac{1}{4}, -\frac{25}{8} \right) \) is the starting point since it represents the peak or trough of the parabola. The y-intercept \( (0, -3) \) is where the parabola crosses the y-axis.
Utilizing the x-intercepts found using the quadratic formula, the graph crosses the x-axis at \( \left( \frac{3}{2}, 0 \right) \) and \( \left( -1, 0 \right) \).
The vertex \( \left( \frac{1}{4}, -\frac{25}{8} \right) \) is the starting point since it represents the peak or trough of the parabola. The y-intercept \( (0, -3) \) is where the parabola crosses the y-axis.
Utilizing the x-intercepts found using the quadratic formula, the graph crosses the x-axis at \( \left( \frac{3}{2}, 0 \right) \) and \( \left( -1, 0 \right) \).
- The symmetry of the parabola is depicted on the line \( x = \frac{1}{4} \), called the axis of symmetry.
- All these points provide a clear path to sketch an accurate graph.
x-intercepts and y-intercepts
In a parabola, intercepts are where the graph crosses the axes.
The y-intercept happens where the graph meets the y-axis, typically found by setting \( x = 0 \) in the function. For \( f(x) = 2x^2 - x - 3 \), the y-intercept is at \( (0, -3) \).
x-intercepts occur where \( f(x) = 0 \). Solved by the quadratic formula, they inform us where the parabola crosses the x-axis; in this case, at the points \( x = \frac{3}{2} \) and \( x = -1 \).
The y-intercept happens where the graph meets the y-axis, typically found by setting \( x = 0 \) in the function. For \( f(x) = 2x^2 - x - 3 \), the y-intercept is at \( (0, -3) \).
- Y-intercepts denote initial value or starting position of a function.
x-intercepts occur where \( f(x) = 0 \). Solved by the quadratic formula, they inform us where the parabola crosses the x-axis; in this case, at the points \( x = \frac{3}{2} \) and \( x = -1 \).
- X-intercepts signify the solution to the equation \( ax^2 + bx + c = 0 \).
Direction of Parabola Opening
The direction in which a parabola opens is determined by the coefficient \( a \) in a quadratic function \( f(x) = ax^2 + bx + c \). If \( a \) is positive, the parabola opens upwards like a smile, and if \( a \) is negative, it opens downwards like a frown.
In our example, \( a = 2 \), which is positive, so the parabola opens upwards. This indicates that the vertex represents the lowest point (a minimum) on the graph.
In our example, \( a = 2 \), which is positive, so the parabola opens upwards. This indicates that the vertex represents the lowest point (a minimum) on the graph.
- Understanding the direction assists in predicting the graph's overall shape and aiding in the sketching process.