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To find which graph correctly represents the quadratic function [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex], we will follow a step-by-step approach to understand its key features and how it would be plotted on the graph. Here’s how you can determine the correct graph:
### 1. Determine the Vertex
A quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] has a vertex at [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = -4 \)[/tex]
Calculate the x-coordinate of the vertex:
[tex]\[ x = -\frac{-4}{2 \cdot 4} = \frac{4}{8} = \frac{1}{2} \][/tex]
Substitute [tex]\( x = \frac{1}{2} \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 4 \cdot \frac{1}{4} - 2 + 1 \][/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 1 - 2 + 1 = 0 \][/tex]
So, the vertex of the graph is at [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex].
### 2. Determine the Axis of Symmetry
The axis of symmetry for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the line [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex], the axis of symmetry is:
[tex]\[ x = \frac{1}{2} \][/tex]
### 3. Determine the Y-Intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 4(0)^2 - 4(0) + 1 = 1 \][/tex]
So, the y-intercept is at [tex]\( (0, 1) \)[/tex].
### 4. Determine the Shape (Direction of Opening)
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 4 \)[/tex]) is positive, the parabola opens upwards.
### 5. Identify Key Points
We can find additional points by substituting other values of [tex]\( x \)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 4(1)^2 - 4(1) + 1 = 4 - 4 + 1 = 1 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 4(-1)^2 - 4(-1) + 1 = 4 + 4 + 1 = 9 \][/tex]
### Summary of Key Features:
- Vertex: [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex]
- Axis of Symmetry: [tex]\( x = \frac{1}{2} \)[/tex]
- Y-Intercept: [tex]\( (0, 1) \)[/tex]
- Opens upwards
- Additional points: [tex]\( (1, 1) \)[/tex] and [tex]\( (-1, 9) \)[/tex]
### Sketching the Graph
1. Draw a vertical line at [tex]\( x = \frac{1}{2} \)[/tex] for the axis of symmetry.
2. Plot the vertex [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex].
3. Plot the y-intercept at [tex]\( (0, 1) \)[/tex].
4. Plot additional points such as [tex]\( (1, 1) \)[/tex] and [tex]\( (-1, 9) \)[/tex].
5. Draw a symmetrical parabola opening upwards and passing through the plotted points.
By comparing with available graphs, the correct graph will be the one matching the characteristics listed above. The graph of [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex] should be a parabolic curve opening upwards with a vertex at [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex] and passing through the points [tex]\( (0, 1) \)[/tex], [tex]\( (1, 1) \)[/tex], and [tex]\( (-1, 9) \)[/tex].
### 1. Determine the Vertex
A quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] has a vertex at [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = -4 \)[/tex]
Calculate the x-coordinate of the vertex:
[tex]\[ x = -\frac{-4}{2 \cdot 4} = \frac{4}{8} = \frac{1}{2} \][/tex]
Substitute [tex]\( x = \frac{1}{2} \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 1 \][/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 4 \cdot \frac{1}{4} - 2 + 1 \][/tex]
[tex]\[ f\left(\frac{1}{2}\right) = 1 - 2 + 1 = 0 \][/tex]
So, the vertex of the graph is at [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex].
### 2. Determine the Axis of Symmetry
The axis of symmetry for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the line [tex]\( x = -\frac{b}{2a} \)[/tex].
For [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex], the axis of symmetry is:
[tex]\[ x = \frac{1}{2} \][/tex]
### 3. Determine the Y-Intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 4(0)^2 - 4(0) + 1 = 1 \][/tex]
So, the y-intercept is at [tex]\( (0, 1) \)[/tex].
### 4. Determine the Shape (Direction of Opening)
Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a = 4 \)[/tex]) is positive, the parabola opens upwards.
### 5. Identify Key Points
We can find additional points by substituting other values of [tex]\( x \)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 4(1)^2 - 4(1) + 1 = 4 - 4 + 1 = 1 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 4(-1)^2 - 4(-1) + 1 = 4 + 4 + 1 = 9 \][/tex]
### Summary of Key Features:
- Vertex: [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex]
- Axis of Symmetry: [tex]\( x = \frac{1}{2} \)[/tex]
- Y-Intercept: [tex]\( (0, 1) \)[/tex]
- Opens upwards
- Additional points: [tex]\( (1, 1) \)[/tex] and [tex]\( (-1, 9) \)[/tex]
### Sketching the Graph
1. Draw a vertical line at [tex]\( x = \frac{1}{2} \)[/tex] for the axis of symmetry.
2. Plot the vertex [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex].
3. Plot the y-intercept at [tex]\( (0, 1) \)[/tex].
4. Plot additional points such as [tex]\( (1, 1) \)[/tex] and [tex]\( (-1, 9) \)[/tex].
5. Draw a symmetrical parabola opening upwards and passing through the plotted points.
By comparing with available graphs, the correct graph will be the one matching the characteristics listed above. The graph of [tex]\( f(x) = 4x^2 - 4x + 1 \)[/tex] should be a parabolic curve opening upwards with a vertex at [tex]\( \left(\frac{1}{2}, 0\right) \)[/tex] and passing through the points [tex]\( (0, 1) \)[/tex], [tex]\( (1, 1) \)[/tex], and [tex]\( (-1, 9) \)[/tex].
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