Answer-
The vertex will be at [tex](-\dfrac{1}{3},\dfrac{2}{3})[/tex]
Solution-
The parabolic equation given is,
[tex]y=3x^2+2x+1[/tex]
Here,
a = 3, b = 2, c = 1
The equation of the axis of symmetry is,
[tex]x=-\dfrac{b}{2a}[/tex]
Putting the values,
[tex]\Rightarrow x=-\dfrac{2}{2\times 3}\\\\\Rightarrow x=-\dfrac{1}{3}[/tex]
As we know, the axis of symmetry and the parabola meet only at one point and that is the vertex.
So, the vertex will be at [tex](-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]
Hence,
[tex]f(-\frac{b}{2a})=f(-\frac{1}{3})=3(-\frac{1}{3})^2+2(-\frac{1}{3})+1\\\\=3(\frac{1}{9})-2(\frac{1}{3})+1\\\\=\dfrac{2}{3}[/tex]
