Find the vertex of the following function:

$f\left( x \right) \quad =\quad { 9x }^{ 2 }\quad +\quad 3x\quad +\quad 1$

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To find the $x$ value of the vertex, we must use the formula $x = -\frac {b}{2a}$ . We get $x = -\frac {3}{18} = -\frac {1}{6}$ .

To get the $y$ value of the vertex we must plug in the $x$ value of the vertex in the equation. We get $f\left( x \right) =9( -\frac {1}{6} )^{ 2 } + 3(-\frac {1}{6}) +1 = 9(\frac {1}{36}) + 3(-\frac {1}{6}) + 1 = \frac {9}{36} - \frac {3}{6} + 1 = \frac {1}{4} - \frac {1}{2} + 1 = \frac {2}{8} - \frac {4}{8} + \frac {8}{8} = \frac {6}{8} = \frac {3}{4}$ .

Therefore, the vertex of this function is $(-\frac {1}{6} , \frac {3}{4})$ .