QPaperGen

Complete the square:
\[
f(x) = 9x^2 + 12x + 2 = 9\left(x^2 + \frac{4}{3}x\right) + 2
= 9\left(x + \frac{2}{3}\right)^2 - 4 + 2 = 9\left(x + \frac{2}{3}\right)^2 - 2
\]
\[
\text{Minimum value occurs at } x = -\frac{2}{3}
\]
\[
f\left(-\frac{2}{3}\right) = -2
\]
No maximum value exists since the parabola opens upwards.