Answer :
Certainly! Let's analyze and graph the function [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex] step by step, and then determine its domain and range.
### 1. Understanding the Function
The function [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex] is a quadratic function. This type of function is typically represented by [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = \frac{1}{6} \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 0 \)[/tex].
### 2. Shape of the Graph
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( \frac{1}{6} \)[/tex]) is positive, the parabola opens upwards.
### 3. Vertex of the Parabola
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], [tex]\( b = 0 \)[/tex] and [tex]\( a = \frac{1}{6} \)[/tex]:
[tex]\[ x = -\frac{0}{2 \cdot \frac{1}{6}} = 0 \][/tex]
So, the vertex is at the point [tex]\((0, 0)\)[/tex], which is the minimum point of the parabola.
### 4. Plotting the Function
To graph [tex]\( f(x) \)[/tex], we can plot a few points around the vertex to get the shape of the parabola:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{1}{6}(-2)^2 = \frac{4}{6} = \frac{2}{3} \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{1}{6}(-1)^2 = \frac{1}{6} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{6}(0)^2 = 0 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{6}(1)^2 = \frac{1}{6} \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{1}{6}(2)^2 = \frac{4}{6} = \frac{2}{3} \][/tex]
### 5. Graphing
We plot the points [tex]\((0,0)\)[/tex], [tex]\((1, \frac{1}{6})\)[/tex], [tex]\((2, \frac{2}{3})\)[/tex], [tex]\((-1, \frac{1}{6})\)[/tex], and [tex]\((-2, \frac{2}{3})\)[/tex]. Since the function is quadratic and symmetric about the y-axis, we connect these points with a smooth upward curve to form the parabola.
### 6. Identifying the Domain and Range
Domain:
The domain of a function is the set of all possible input values (x-values) that the function can accept. Since there is no restriction on the x-values for the quadratic function [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], it can accept any real number:
[tex]\[ \text{Domain: All real numbers} \quad \text{or} \quad (-\infty, \infty) \][/tex]
Range:
The range of a function is the set of all possible output values (y-values). For [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], since the parabola opens upwards and the vertex is at the minimum point [tex]\((0, 0)\)[/tex], the function will only take non-negative values:
[tex]\[ \text{Range: } [0, \infty) \][/tex]
### Summary
- The graph of [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex] is a parabola opening upwards with its vertex at (0,0).
- The domain of the function is all real numbers.
- The range of the function is all non-negative real numbers, [0, ∞).
Thus,
[tex]\[ \text{Domain: All real numbers} \quad \text{or} \quad (-\infty, \infty) \][/tex]
[tex]\[ \text{Range: } [0, \infty) \][/tex]
### 1. Understanding the Function
The function [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex] is a quadratic function. This type of function is typically represented by [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = \frac{1}{6} \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 0 \)[/tex].
### 2. Shape of the Graph
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( \frac{1}{6} \)[/tex]) is positive, the parabola opens upwards.
### 3. Vertex of the Parabola
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], [tex]\( b = 0 \)[/tex] and [tex]\( a = \frac{1}{6} \)[/tex]:
[tex]\[ x = -\frac{0}{2 \cdot \frac{1}{6}} = 0 \][/tex]
So, the vertex is at the point [tex]\((0, 0)\)[/tex], which is the minimum point of the parabola.
### 4. Plotting the Function
To graph [tex]\( f(x) \)[/tex], we can plot a few points around the vertex to get the shape of the parabola:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{1}{6}(-2)^2 = \frac{4}{6} = \frac{2}{3} \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{1}{6}(-1)^2 = \frac{1}{6} \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{6}(0)^2 = 0 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{6}(1)^2 = \frac{1}{6} \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{1}{6}(2)^2 = \frac{4}{6} = \frac{2}{3} \][/tex]
### 5. Graphing
We plot the points [tex]\((0,0)\)[/tex], [tex]\((1, \frac{1}{6})\)[/tex], [tex]\((2, \frac{2}{3})\)[/tex], [tex]\((-1, \frac{1}{6})\)[/tex], and [tex]\((-2, \frac{2}{3})\)[/tex]. Since the function is quadratic and symmetric about the y-axis, we connect these points with a smooth upward curve to form the parabola.
### 6. Identifying the Domain and Range
Domain:
The domain of a function is the set of all possible input values (x-values) that the function can accept. Since there is no restriction on the x-values for the quadratic function [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], it can accept any real number:
[tex]\[ \text{Domain: All real numbers} \quad \text{or} \quad (-\infty, \infty) \][/tex]
Range:
The range of a function is the set of all possible output values (y-values). For [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex], since the parabola opens upwards and the vertex is at the minimum point [tex]\((0, 0)\)[/tex], the function will only take non-negative values:
[tex]\[ \text{Range: } [0, \infty) \][/tex]
### Summary
- The graph of [tex]\( f(x) = \frac{1}{6} x^2 \)[/tex] is a parabola opening upwards with its vertex at (0,0).
- The domain of the function is all real numbers.
- The range of the function is all non-negative real numbers, [0, ∞).
Thus,
[tex]\[ \text{Domain: All real numbers} \quad \text{or} \quad (-\infty, \infty) \][/tex]
[tex]\[ \text{Range: } [0, \infty) \][/tex]
