Chapter 11: Problem 20
Graph each parabola. Give the vertex, axis of symmetry, domain, and range. $$ f(x)=\frac{2}{3} x^{2}-4 $$
Short Answer
Expert verified
Vertex: (0, -4). Axis of symmetry: x = 0. Domain: (-∞, ∞). Range: [-4, ∞).
Step by step solution
01
- Identify the parabola form
The given function is in the form of a standard quadratic function: \[ f(x) = \frac{2}{3} x^{2} - 4 \]Identify the coefficients: \[ a = \frac{2}{3}, b = 0, c = -4 \]
02
- Find the vertex
The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a} \]Since \( b = 0 \), the x-coordinate of the vertex is: \[ x = 0 \]To find the y-coordinate, substitute \( x = 0 \) back into the equation:\[ f(0) = \frac{2}{3}(0)^2 - 4 = -4 \]So, the vertex is: \( (0, -4) \)
03
- Determine the axis of symmetry
The axis of symmetry of a parabola is given by the line \( x = h \), where \( (h, k) \) is the vertex. Therefore, the axis of symmetry is: \[ x = 0 \]
04
- State the domain
The domain of any quadratic function \( f(x) = ax^2 + bx + c \) is all real numbers since a parabola extends infinitely in both directions along the x-axis: \[ \text{Domain: } (-\infty, \infty) \]
05
- Determine the range
Since the coefficient \( a = \frac{2}{3} \) is positive, the parabola opens upwards. Therefore, the range is all y-values greater than or equal to the y-coordinate of the vertex. Thus, the range is: \[ \text{Range: } [-4, \infty) \]
06
- Graph the parabola
To graph the parabola, plot the vertex \( (0, -4) \). Since the parabola opens upwards, draw a U-shaped curve that extends infinitely upwards. Use additional points if needed to assist in drawing an accurate curve.
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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, as it represents the minimum or maximum value of the quadratic function. In the equation \( f(x) = \frac{2}{3}x^2 - 4 \), the polynomial is of the form \( ax^2 + bx + c \). For our function:
\[ f(0) = \frac{2}{3}(0)^2 - 4 = -4 \]
So, the vertex of our parabola is at \((0, -4)\). This point is where the parabola either reaches its lowest or highest value. Since \( a > 0 \), it's the lowest point and the parabola opens upwards.
- \( a = \frac{2}{3} \)
- \( b = 0 \)
- \( c = -4 \)
\[ f(0) = \frac{2}{3}(0)^2 - 4 = -4 \]
So, the vertex of our parabola is at \((0, -4)\). This point is where the parabola either reaches its lowest or highest value. Since \( a > 0 \), it's the lowest point and the parabola opens upwards.
axis of symmetry
The axis of symmetry of a parabola is a vertical line that passes through the vertex. It divides the parabola into two mirror-image halves. For a quadratic function like \( f(x) = ax^2 + bx + c \), the axis of symmetry can be found using the same formula used to find the x-coordinate of the vertex, \( x = -\frac{b}{2a} \).
Since the x-coordinate of the vertex is \( x = 0 \), the axis of symmetry for our function \( f(x) = \frac{2}{3} x^2 - 4 \) is the vertical line:
\[ x = 0 \]
This line goes through the vertex \( (0, -4) \) and helps in graphing the parabola symmetrically.
Since the x-coordinate of the vertex is \( x = 0 \), the axis of symmetry for our function \( f(x) = \frac{2}{3} x^2 - 4 \) is the vertical line:
\[ x = 0 \]
This line goes through the vertex \( (0, -4) \) and helps in graphing the parabola symmetrically.
domain and range of a function
Understanding the domain and range of functions is essential for graphing.
Domain: For any quadratic function such as \( f(x) = ax^2 + bx + c \), the domain is all real numbers, because parabolas extend indefinitely along the x-axis. Hence, the domain of our function \( f(x) = \frac{2}{3} x^2 - 4 \) is:
\[ \text{Domain: } (-\infty, \infty) \]
Range: The range depends on the direction that the parabola opens. If \( a > 0 \), the parabola opens upwards, and the range starts from the y-coordinate of the vertex and goes to positive infinity. Since our \( a = \frac{2}{3} \) is positive, the parabola opens upwards, and the vertex is at \( (0, -4) \). Therefore, the range of our function is:
\[ \text{Range: } [-4, \, \infty) \]
This means the y-values start from -4 and go upwards to infinity.
Domain: For any quadratic function such as \( f(x) = ax^2 + bx + c \), the domain is all real numbers, because parabolas extend indefinitely along the x-axis. Hence, the domain of our function \( f(x) = \frac{2}{3} x^2 - 4 \) is:
\[ \text{Domain: } (-\infty, \infty) \]
Range: The range depends on the direction that the parabola opens. If \( a > 0 \), the parabola opens upwards, and the range starts from the y-coordinate of the vertex and goes to positive infinity. Since our \( a = \frac{2}{3} \) is positive, the parabola opens upwards, and the vertex is at \( (0, -4) \). Therefore, the range of our function is:
\[ \text{Range: } [-4, \, \infty) \]
This means the y-values start from -4 and go upwards to infinity.
quadratic function
A quadratic function is a second-degree polynomial and its graph is a parabola. The general form is \( f(x) = ax^2 + bx + c \), where:
- \( a \) determines the width and direction of the parabola (upwards if \( a > 0 \), downwards if \( a < 0 \))
- \( b \) affects the position of the vertex along the x-axis
- \( c \) is the y-intercept, where the graph crosses the y-axis
- \( a = \frac{2}{3} \), so the parabola opens upwards and is relatively wide
- \( b = 0 \), placing the vertex along the y-axis
- \( c = -4 \), meaning the graph intersects the y-axis at -4