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Sagot :
Sure! Let's break down the given function [tex]\( y = 2(x-1)^2 - 8 \)[/tex].
### 1. Co-ordinates of the Vertex
The given function is in the vertex form of a parabola, which is [tex]\( y = a(x - h)^2 + k \)[/tex]. In this form:
- [tex]\( (h, k) \)[/tex] represents the vertex of the parabola.
- [tex]\( a \)[/tex] determines the direction and the width of the parabola.
Here, by comparing [tex]\( y = 2(x - 1)^2 - 8 \)[/tex] with [tex]\( y = a(x - h)^2 + k \)[/tex], we can see that:
- [tex]\( h = 1 \)[/tex]
- [tex]\( k = -8 \)[/tex]
Thus, the vertex of the parabola is at [tex]\( (1, -8) \)[/tex].
### 2. Direction of Opening
To determine the direction in which the parabola opens, we look at the coefficient [tex]\( a \)[/tex] in front of the squared term [tex]\( (x - h)^2 \)[/tex].
For the given function [tex]\( y = 2(x - 1)^2 - 8 \)[/tex]:
- The coefficient [tex]\( a = 2 \)[/tex].
Since [tex]\( a \)[/tex] is positive, the parabola opens upward.
### 3. Equation of Axis of Symmetry
The axis of symmetry for a parabola in the form [tex]\( y = a(x - h)^2 + k \)[/tex] is given by the vertical line [tex]\( x = h \)[/tex].
For the given function [tex]\( h = 1 \)[/tex]. Therefore, the equation of the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### 4. Domain
The domain of a quadratic function is all possible values of [tex]\( x \)[/tex].
For any parabola, regardless of its orientation or position, the domain is all real numbers:
[tex]\[ \text{Domain} = \text{all real numbers} \][/tex]
### 5. Range
The range of the function is determined by the direction in which the parabola opens and the y-coordinate of the vertex.
Since the parabola opens upward ([tex]\( a \)[/tex] is positive) and the vertex is at [tex]\( (1, -8) \)[/tex]:
- The minimum value of [tex]\( y \)[/tex] is [tex]\(-8\)[/tex] (the y-coordinate of the vertex).
- As the parabola opens upward, [tex]\( y \)[/tex] can take any value greater than or equal to -8.
Thus, the range is:
[tex]\[ \text{Range} = \{ y \mid y \geq -8 \} \][/tex]
### Summary
- Co-ordinates of the vertex: [tex]\( (1, -8) \)[/tex]
- Direction of opening: Upward
- Equation of axis of symmetry: [tex]\( x = 1 \)[/tex]
- Domain: All real numbers
- Range: [tex]\( y \geq -8 \)[/tex]
This concludes the detailed solution for the given function [tex]\( y = 2(x-1)^2 - 8 \)[/tex].
### 1. Co-ordinates of the Vertex
The given function is in the vertex form of a parabola, which is [tex]\( y = a(x - h)^2 + k \)[/tex]. In this form:
- [tex]\( (h, k) \)[/tex] represents the vertex of the parabola.
- [tex]\( a \)[/tex] determines the direction and the width of the parabola.
Here, by comparing [tex]\( y = 2(x - 1)^2 - 8 \)[/tex] with [tex]\( y = a(x - h)^2 + k \)[/tex], we can see that:
- [tex]\( h = 1 \)[/tex]
- [tex]\( k = -8 \)[/tex]
Thus, the vertex of the parabola is at [tex]\( (1, -8) \)[/tex].
### 2. Direction of Opening
To determine the direction in which the parabola opens, we look at the coefficient [tex]\( a \)[/tex] in front of the squared term [tex]\( (x - h)^2 \)[/tex].
For the given function [tex]\( y = 2(x - 1)^2 - 8 \)[/tex]:
- The coefficient [tex]\( a = 2 \)[/tex].
Since [tex]\( a \)[/tex] is positive, the parabola opens upward.
### 3. Equation of Axis of Symmetry
The axis of symmetry for a parabola in the form [tex]\( y = a(x - h)^2 + k \)[/tex] is given by the vertical line [tex]\( x = h \)[/tex].
For the given function [tex]\( h = 1 \)[/tex]. Therefore, the equation of the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### 4. Domain
The domain of a quadratic function is all possible values of [tex]\( x \)[/tex].
For any parabola, regardless of its orientation or position, the domain is all real numbers:
[tex]\[ \text{Domain} = \text{all real numbers} \][/tex]
### 5. Range
The range of the function is determined by the direction in which the parabola opens and the y-coordinate of the vertex.
Since the parabola opens upward ([tex]\( a \)[/tex] is positive) and the vertex is at [tex]\( (1, -8) \)[/tex]:
- The minimum value of [tex]\( y \)[/tex] is [tex]\(-8\)[/tex] (the y-coordinate of the vertex).
- As the parabola opens upward, [tex]\( y \)[/tex] can take any value greater than or equal to -8.
Thus, the range is:
[tex]\[ \text{Range} = \{ y \mid y \geq -8 \} \][/tex]
### Summary
- Co-ordinates of the vertex: [tex]\( (1, -8) \)[/tex]
- Direction of opening: Upward
- Equation of axis of symmetry: [tex]\( x = 1 \)[/tex]
- Domain: All real numbers
- Range: [tex]\( y \geq -8 \)[/tex]
This concludes the detailed solution for the given function [tex]\( y = 2(x-1)^2 - 8 \)[/tex].
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