1. **State the problem:** We are given the function $f(x) = 3x^2 - 2x + \frac{1}{2}$ and want to understand its properties. 2. **Formula and rules:** This is a quadratic function of the form $f(x) = ax^2 + bx + c$ where $a=3$, $b=-2$, and $c=\frac{1}{2}$. Quadratic functions graph as parabolas. 3. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a} = -\frac{-2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$. 4. **Calculate the vertex $y$-coordinate:** $$f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) + \frac{1}{2} = 3\times \frac{1}{9} - \frac{2}{3} + \frac{1}{2} = \frac{1}{3} - \frac{2}{3} + \frac{1}{2}$$ 5. **Simplify the vertex $y$-coordinate:** $$\frac{1}{3} - \frac{2}{3} + \frac{1}{2} = -\frac{1}{3} + \frac{1}{2} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}$$ 6. **Find the $y$-intercept:** Set $x=0$, then $f(0) = \frac{1}{2}$. 7. **Find the $x$-intercepts:** Solve $3x^2 - 2x + \frac{1}{2} = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times \frac{1}{2}}}{2 \times 3} = \frac{2 \pm \sqrt{4 - 6}}{6} = \frac{2 \pm \sqrt{-2}}{6}$$ Since the discriminant is negative, there are no real $x$-intercepts. **Final answer:** The parabola opens upward (since $a=3>0$), has vertex at $\left(\frac{1}{3}, \frac{1}{6}\right)$, $y$-intercept at $\frac{1}{2}$, and no real $x$-intercepts.