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To determine which graph represents the function [tex]\( f(x) = \frac{1}{2} x^2 + 2x - 6 \)[/tex], let's analyze the key features of the function step-by-step.
### Step 1: Determine the Roots of the Quadratic Function
To find the roots of the quadratic function [tex]\( \frac{1}{2} x^2 + 2x - 6 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = \frac{1}{2} \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -6 \)[/tex], we calculate:
- Discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4 \cdot \frac{1}{2} \cdot (-6) = 4 + 12 = 16 \][/tex]
- Roots:
[tex]\[ x_1 = \frac{-b + \sqrt{16}}{2a} = \frac{-2 + 4}{1} = 2 \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{16}}{2a} = \frac{-2 - 4}{1} = -6 \][/tex]
The roots of the function are [tex]\( x_1 = 2 \)[/tex] and [tex]\( x_2 = -6 \)[/tex].
### Step 2: Determine the Vertex of the Parabola
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is at [tex]\( x = -\frac{b}{2a} \)[/tex]:
- Vertex [tex]\( x \)[/tex]-coordinate:
[tex]\[ x = -\frac{b}{2a} = -\frac{2}{2 \cdot \frac{1}{2}} = -2 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the vertex, substitute [tex]\( x = -2 \)[/tex] back into the function:
[tex]\[ f(-2) = \frac{1}{2} (-2)^2 + 2(-2) - 6 = \frac{1}{2} \cdot 4 - 4 - 6 = 2 - 4 - 6 = -8 \][/tex]
So, the vertex of the parabola is at [tex]\( (-2, -8) \)[/tex].
### Step 3: Match the Features with the Given Options
The parabolas given in the options have the following features:
1. Option 1:
- Passes through [tex]\( (-6, 0) \)[/tex]
- Vertex at [tex]\( (-2, -8) \)[/tex]
- Passes through [tex]\( (2, 0) \)[/tex]
2. Option 2:
- Passes through [tex]\( (-2, 0) \)[/tex]
- Vertex at [tex]\( (2, -8) \)[/tex]
- Passes through [tex]\( (6, 0) \)[/tex]
3. Option 3:
- Passes through [tex]\( (-4, 6) \)[/tex]
- Vertex at [tex]\( (4, -10) \)[/tex]
- Passes through [tex]\( (8, -6) \)[/tex]
4. Option 4:
- Passes through [tex]\( (-8, -6) \)[/tex]
- Vertex at [tex]\( (-4, -10) \)[/tex]
- Passes through [tex]\( (4, 6) \)[/tex]
From our calculations:
- The roots are [tex]\( x = 2 \)[/tex] and [tex]\( x = -6 \)[/tex], which corresponds to passing through [tex]\( (2, 0) \)[/tex] and [tex]\( (-6, 0) \)[/tex].
- The vertex is at [tex]\( (-2, -8) \)[/tex].
### Conclusion
The correct graph must pass through [tex]\( (-6, 0) \)[/tex] and [tex]\( (2, 0) \)[/tex], and have a vertex at [tex]\( (-2, -8) \)[/tex].
Thus, the correct graph of the function [tex]\( f(x) = \frac{1}{2} x^2 + 2x - 6 \)[/tex] is described in Option 1:
> On a coordinate plane, a parabola opens up. It goes through [tex]\( (-6, 0) \)[/tex], has a vertex at [tex]\( (-2, -8) \)[/tex], and goes through [tex]\( (2, 0) \)[/tex].
### Step 1: Determine the Roots of the Quadratic Function
To find the roots of the quadratic function [tex]\( \frac{1}{2} x^2 + 2x - 6 = 0 \)[/tex], we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = \frac{1}{2} \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -6 \)[/tex], we calculate:
- Discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4 \cdot \frac{1}{2} \cdot (-6) = 4 + 12 = 16 \][/tex]
- Roots:
[tex]\[ x_1 = \frac{-b + \sqrt{16}}{2a} = \frac{-2 + 4}{1} = 2 \][/tex]
[tex]\[ x_2 = \frac{-b - \sqrt{16}}{2a} = \frac{-2 - 4}{1} = -6 \][/tex]
The roots of the function are [tex]\( x_1 = 2 \)[/tex] and [tex]\( x_2 = -6 \)[/tex].
### Step 2: Determine the Vertex of the Parabola
The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is at [tex]\( x = -\frac{b}{2a} \)[/tex]:
- Vertex [tex]\( x \)[/tex]-coordinate:
[tex]\[ x = -\frac{b}{2a} = -\frac{2}{2 \cdot \frac{1}{2}} = -2 \][/tex]
To find the [tex]\( y \)[/tex]-coordinate of the vertex, substitute [tex]\( x = -2 \)[/tex] back into the function:
[tex]\[ f(-2) = \frac{1}{2} (-2)^2 + 2(-2) - 6 = \frac{1}{2} \cdot 4 - 4 - 6 = 2 - 4 - 6 = -8 \][/tex]
So, the vertex of the parabola is at [tex]\( (-2, -8) \)[/tex].
### Step 3: Match the Features with the Given Options
The parabolas given in the options have the following features:
1. Option 1:
- Passes through [tex]\( (-6, 0) \)[/tex]
- Vertex at [tex]\( (-2, -8) \)[/tex]
- Passes through [tex]\( (2, 0) \)[/tex]
2. Option 2:
- Passes through [tex]\( (-2, 0) \)[/tex]
- Vertex at [tex]\( (2, -8) \)[/tex]
- Passes through [tex]\( (6, 0) \)[/tex]
3. Option 3:
- Passes through [tex]\( (-4, 6) \)[/tex]
- Vertex at [tex]\( (4, -10) \)[/tex]
- Passes through [tex]\( (8, -6) \)[/tex]
4. Option 4:
- Passes through [tex]\( (-8, -6) \)[/tex]
- Vertex at [tex]\( (-4, -10) \)[/tex]
- Passes through [tex]\( (4, 6) \)[/tex]
From our calculations:
- The roots are [tex]\( x = 2 \)[/tex] and [tex]\( x = -6 \)[/tex], which corresponds to passing through [tex]\( (2, 0) \)[/tex] and [tex]\( (-6, 0) \)[/tex].
- The vertex is at [tex]\( (-2, -8) \)[/tex].
### Conclusion
The correct graph must pass through [tex]\( (-6, 0) \)[/tex] and [tex]\( (2, 0) \)[/tex], and have a vertex at [tex]\( (-2, -8) \)[/tex].
Thus, the correct graph of the function [tex]\( f(x) = \frac{1}{2} x^2 + 2x - 6 \)[/tex] is described in Option 1:
> On a coordinate plane, a parabola opens up. It goes through [tex]\( (-6, 0) \)[/tex], has a vertex at [tex]\( (-2, -8) \)[/tex], and goes through [tex]\( (2, 0) \)[/tex].
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