To analyze the function [tex]\( f(x) = 3(x-1)^2 + 2 \)[/tex], let’s identify its key aspects:
### Step-by-Step Solution:
1. Vertex:
The given function [tex]\( f(x) = 3(x-1)^2 + 2 \)[/tex] is in the vertex form of a quadratic function, [tex]\( a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] represents the vertex of the parabola.
By comparing [tex]\( f(x) = 3(x-1)^2 + 2 \)[/tex] with the general vertex form [tex]\( a(x-h)^2 + k \)[/tex]:
- [tex]\( h = 1 \)[/tex]
- [tex]\( k = 2 \)[/tex]
Therefore, the vertex of the function is [tex]\((1, 2)\)[/tex].
2. Domain:
For any quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], the domain is all real numbers. This implies that the function is defined for all values of [tex]\( x \)[/tex].
Hence, the domain is "all real numbers."
3. Range:
To determine the range, we need to consider the direction in which the parabola opens:
- Since the coefficient of the squared term (which is 3) is positive, the parabola opens upwards.
- The minimum value of the function occurs at the vertex.
For a parabola that opens upwards, the range includes all [tex]\( y \)[/tex] values greater than or equal to the [tex]\( y \)[/tex]-coordinate of the vertex.
Therefore, since the vertex is [tex]\((1, 2)\)[/tex]:
- The [tex]\( y \)[/tex]-coordinate of the vertex is 2.
- Consequently, the range is [tex]\( y \geq 2 \)[/tex].
### Summary:
- Vertex: [tex]\((1, 2)\)[/tex]
- Domain: All real numbers
- Range: [tex]\( y \geq 2 \)[/tex]
### Conclusion:
The information matches the option:
"The vertex is [tex]\((1, 2)\)[/tex], the domain is all real numbers, and the range is [tex]\( y \geq 2 \)[/tex]."
