To determine the axis of symmetry and the vertex of the function [tex]\( f(x) = 3(x-2)^2 + 4 \)[/tex], we will analyze the given function which is a quadratic function in vertex form. The vertex form of a quadratic function is given by:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
where:
- [tex]\( a \)[/tex] is the coefficient of the quadratic term,
- [tex]\( h \)[/tex] is the x-coordinate of the vertex,
- [tex]\( k \)[/tex] is the y-coordinate of the vertex.
From the given function [tex]\( f(x) = 3(x-2)^2 + 4 \)[/tex], we can see that:
- [tex]\( a = 3 \)[/tex],
- [tex]\( h = 2 \)[/tex],
- [tex]\( k = 4 \)[/tex].
### Axis of Symmetry
The axis of symmetry of a parabola in vertex form is a vertical line that passes through the vertex and its equation is given by [tex]\( x = h \)[/tex].
In this case, [tex]\( h = 2 \)[/tex], so the axis of symmetry is:
[tex]\[ x = 2 \][/tex]
### Vertex
The vertex of the parabola is the point [tex]\((h, k)\)[/tex]. For this function, the vertex is:
[tex]\[ (h, k) = (2, 4) \][/tex]
So, the vertex is at:
[tex]\[ (2, 4) \][/tex]
Thus, the solutions for the axis of symmetry and the vertex are:
[tex]\[ x = 2 \][/tex]
Vertex: [tex]\( (2, 4) \)[/tex]
