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To graph the function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex], let's follow these steps:
### 1. Identify the General Shape
The function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex] is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex]. Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( \frac{1}{2} \)[/tex]) is positive, the parabola opens upwards.
### 2. Find the Vertex
The vertex form of a quadratic function is [tex]\( f(x) = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. For the standard form [tex]\( ax^2 + bx + c \)[/tex], the vertex [tex]\((h, k)\)[/tex] can be calculated using:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) \][/tex]
In this case, [tex]\( a = \frac{1}{2} \)[/tex] and [tex]\( b = 2 \)[/tex], so:
[tex]\[ h = -\frac{2}{2 \cdot \frac{1}{2}} = -\frac{2}{1} = -2 \][/tex]
Next, substitute [tex]\( x = -2 \)[/tex] into the function to find [tex]\( k \)[/tex]:
[tex]\[ k = f(-2) = \frac{1}{2}(-2)^2 + 2(-2) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(4) - 4 - 6 \][/tex]
[tex]\[ = 2 - 4 - 6 \][/tex]
[tex]\[ = -8 \][/tex]
So, the vertex of the parabola is at [tex]\((-2, -8)\)[/tex].
### 3. Identify the Axis of Symmetry
The axis of symmetry for a parabola in the form [tex]\( ax^2 + bx + c \)[/tex] is the vertical line passing through the vertex. For this function, the axis of symmetry is:
[tex]\[ x = -2 \][/tex]
### 4. Find the y-Intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{2}(0)^2 + 2(0) - 6 \][/tex]
[tex]\[ = -6 \][/tex]
So, the y-intercept is at (0, -6).
### 5. Find Additional Points
Let's calculate [tex]\( f(x) \)[/tex] for a couple more values of [tex]\( x \)[/tex].
For [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{1}{2}(-4)^2 + 2(-4) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(16) - 8 - 6 \][/tex]
[tex]\[ = 8 - 8 - 6 \][/tex]
[tex]\[ = -6 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{1}{2}(2)^2 + 2(2) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(4) + 4 - 6 \][/tex]
[tex]\[ = 2 + 4 - 6 \][/tex]
[tex]\[ = 0 \][/tex]
### 6. Plotting the Points
Now we have the following points to plot:
- Vertex at [tex]\((-2, -8)\)[/tex]
- Axis of symmetry at [tex]\(x = -2\)[/tex]
- y-intercept at [tex]\((0, -6)\)[/tex]
- Additional point at [tex]\((-4, -6)\)[/tex], which helps us see symmetry on the other side of the axis.
- Additional point at [tex]\((2, 0)\)[/tex]
### 7. Sketch the Graph
1. Plot the vertex [tex]\((-2, -8)\)[/tex].
2. Draw the axis of symmetry [tex]\(x = -2\)[/tex].
3. Plot the y-intercept [tex]\((0, -6)\)[/tex].
4. Plot the point [tex]\((-4, -6)\)[/tex].
5. Plot the point [tex]\((2, 0)\)[/tex].
6. Draw a smooth parabolic curve through these points and extending upwards, respecting the symmetry about the axis.
This gives us the graph of [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex]. The shape should be a standard upward-opening parabola with its lowest point (vertex) at [tex]\((-2, -8)\)[/tex].
### 1. Identify the General Shape
The function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex] is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex]. Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( \frac{1}{2} \)[/tex]) is positive, the parabola opens upwards.
### 2. Find the Vertex
The vertex form of a quadratic function is [tex]\( f(x) = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. For the standard form [tex]\( ax^2 + bx + c \)[/tex], the vertex [tex]\((h, k)\)[/tex] can be calculated using:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) \][/tex]
In this case, [tex]\( a = \frac{1}{2} \)[/tex] and [tex]\( b = 2 \)[/tex], so:
[tex]\[ h = -\frac{2}{2 \cdot \frac{1}{2}} = -\frac{2}{1} = -2 \][/tex]
Next, substitute [tex]\( x = -2 \)[/tex] into the function to find [tex]\( k \)[/tex]:
[tex]\[ k = f(-2) = \frac{1}{2}(-2)^2 + 2(-2) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(4) - 4 - 6 \][/tex]
[tex]\[ = 2 - 4 - 6 \][/tex]
[tex]\[ = -8 \][/tex]
So, the vertex of the parabola is at [tex]\((-2, -8)\)[/tex].
### 3. Identify the Axis of Symmetry
The axis of symmetry for a parabola in the form [tex]\( ax^2 + bx + c \)[/tex] is the vertical line passing through the vertex. For this function, the axis of symmetry is:
[tex]\[ x = -2 \][/tex]
### 4. Find the y-Intercept
The y-intercept occurs where [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{2}(0)^2 + 2(0) - 6 \][/tex]
[tex]\[ = -6 \][/tex]
So, the y-intercept is at (0, -6).
### 5. Find Additional Points
Let's calculate [tex]\( f(x) \)[/tex] for a couple more values of [tex]\( x \)[/tex].
For [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = \frac{1}{2}(-4)^2 + 2(-4) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(16) - 8 - 6 \][/tex]
[tex]\[ = 8 - 8 - 6 \][/tex]
[tex]\[ = -6 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{1}{2}(2)^2 + 2(2) - 6 \][/tex]
[tex]\[ = \frac{1}{2}(4) + 4 - 6 \][/tex]
[tex]\[ = 2 + 4 - 6 \][/tex]
[tex]\[ = 0 \][/tex]
### 6. Plotting the Points
Now we have the following points to plot:
- Vertex at [tex]\((-2, -8)\)[/tex]
- Axis of symmetry at [tex]\(x = -2\)[/tex]
- y-intercept at [tex]\((0, -6)\)[/tex]
- Additional point at [tex]\((-4, -6)\)[/tex], which helps us see symmetry on the other side of the axis.
- Additional point at [tex]\((2, 0)\)[/tex]
### 7. Sketch the Graph
1. Plot the vertex [tex]\((-2, -8)\)[/tex].
2. Draw the axis of symmetry [tex]\(x = -2\)[/tex].
3. Plot the y-intercept [tex]\((0, -6)\)[/tex].
4. Plot the point [tex]\((-4, -6)\)[/tex].
5. Plot the point [tex]\((2, 0)\)[/tex].
6. Draw a smooth parabolic curve through these points and extending upwards, respecting the symmetry about the axis.
This gives us the graph of [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex]. The shape should be a standard upward-opening parabola with its lowest point (vertex) at [tex]\((-2, -8)\)[/tex].
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