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Magazine Chapter 8: Problem 22
Graph each quadratic function, and state its domain and range. $$y=\frac{1}{3} x^{2}-6$$
Short Answer
Expert verified
Domain: \( (-\infty, +\infty) \), Range: \( [-6, +\infty) \)
Step by step solution
01
Identify the function's general form
The function is given as \( y = \frac{1}{3} x^2 - 6 \). This is a quadratic function, which generally has the form \( y = ax^2 + bx + c \). Here, \( a = \frac{1}{3} \), \( b = 0 \), and \( c = -6 \).
02
Determine the vertex of the parabola
For a quadratic function \( y = ax^2 + bx + c \), the vertex form is given by \( y = a(x - h)^2 + k \). The x-coordinate of the vertex (\(h\)) is found using the formula \( h = -\frac{b}{2a} \). Substituting \( a = \frac{1}{3} \) and \( b = 0 \), we get \( h = 0 \). The y-coordinate (\(k\)) is found by substituting \( h \) back into the function: \( k = \frac{1}{3} \times 0^2 - 6 = -6 \). So, the vertex is at \( (0, -6) \).
03
Determine the shape and direction of the parabola
The coefficient \( a = \frac{1}{3} \) is positive, so the parabola opens upwards. The graph is a U-shaped curve with its minimum point at the vertex \( (0, -6) \).
04
Plot points on either side of the vertex
Choose values of \( x \) around the vertex to plot points. For \( x = -3 \), \( y = \frac{1}{3}(-3)^2 - 6 = -3. \) For \( x = 3 \), \( y = \frac{1}{3}(3)^2 - 6 = -3 \). Plot these points and more to get the curve.
05
State the domain
The domain of a quadratic function is all real numbers because quadratic functions are defined for every real number \( x \). Thus, the domain is \( (-\infty, +\infty) \).
06
State the range
Since the parabola opens upwards and the minimum value (at the vertex) is \( -6 \), the range is all real numbers greater than or equal to \( -6 \). Thus, the range is \( [-6, +\infty) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable. It is generally written in the form: \[ y = ax^2 + bx + c \] where:
- \(a\), \(b\), and \(c\) are constants
- \(a eq 0\) (if \(a\) equals zero, the equation becomes linear, not quadratic)
- \(a = \frac{1}{3}\)
- \(b = 0\)
- \(c = -6\)
Parabola
The graph of a quadratic equation is a parabola. Parabolas have a characteristic U-shape and are symmetric about a vertical axis called the axis of symmetry. The direction of the parabola (upwards or downwards) is determined by the sign of the coefficient \(a\):
- If \(a > 0\), the parabola opens upwards.
- If \(a < 0\), the parabola opens downwards.
Vertex Form
The vertex form of a quadratic function is a useful format for graphing and understanding the properties of the parabola. It is given by the formula: \[ y = a(x - h)^2 + k \] Here, \((h, k)\) is the vertex of the parabola.To convert the standard form to the vertex form, we use the following steps:
- Calculate \( h \) using the formula \( h = -\frac{b}{2a} \).
- Calculate \( k \) by substituting \( h \) back into the equation to get \( y = a(h)^2 + bx + c \).
- \( h = 0 \) because \( b = 0 \).
- \( k = -6 \) after substituting \( x = 0 \) in the function.
Domain and Range
The domain and range of a function provide information about the input (x-values) and output (y-values) respectively. The domain of all quadratic functions is always all real numbers because any value of \(x\) will produce a valid \(y\) value. Hence, for our function \( y = \frac{1}{3} x^2 - 6 \), the domain is \((-\infty, +\infty)\).The range depends on the vertex and the direction in which the parabola opens. Since our parabola opens upwards and the vertex is at \( (0, -6) \) (the minimum point), the range is all real numbers greater than or equal to -6. Thus, the range is: \([-6, +\infty)\).
