Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
