Let's analyze the given quadratic function [tex]\( y = 2(x-1)^2 - 8 \)[/tex].
### Co-ordinates of vertex
The standard form of a quadratic function is [tex]\( y = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
For the given function:
- Coefficient [tex]\( a = 2 \)[/tex]
- [tex]\( h = 1 \)[/tex]
- [tex]\( k = -8 \)[/tex]
Therefore, the vertex is:
[tex]\[ (h, k) = (1, -8) \][/tex]
### Direction of opening
The direction in which the parabola opens is determined by the coefficient [tex]\( a \)[/tex]:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards.
Since [tex]\( a = 2 \)[/tex] is positive, the parabola opens upwards.
### Equation of axis of symmetry
The axis of symmetry in a quadratic function [tex]\( y = a(x-h)^2 + k \)[/tex] is given by the line [tex]\( x = h \)[/tex].
For our function:
[tex]\[ x = h = 1 \][/tex]
So, the axis of symmetry is:
[tex]\[ x = 1 \][/tex]
### Domain
The domain of any quadratic function is all real numbers, as it can take any [tex]\( x \)[/tex]-value.
Thus, the domain is:
[tex]\[ \text{All real numbers} \][/tex]
### Range
The range of the quadratic function depends on the direction in which the parabola opens:
- If the parabola opens upwards, the range starts from the [tex]\( y \)[/tex]-value of the vertex and goes to [tex]\( \infty \)[/tex].
- If the parabola opens downwards, the range starts from [tex]\( -\infty \)[/tex] and goes up to the [tex]\( y \)[/tex]-value of the vertex.
Since our parabola opens upwards and the vertex occurs at [tex]\( y = -8 \)[/tex], the range is:
[tex]\[ [-8, \infty) \][/tex]
Summarizing all the findings:
| | |
|--------------------------|----------------------------|
| Co-ordinates of vertex | [tex]\( (1, -8) \)[/tex] |
| Direction of opening | Upwards |
| Equation of axis of symmetry | [tex]\( x = 1 \)[/tex] |
| Domain | All real numbers |
| Range | [tex]\([-8, \infty)\)[/tex] |
