1. The problem is to analyze the quadratic function $f(x) = 3x^2 + 5x + 2$. 2. Recall the general form of a quadratic function: $$f(x) = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. Important rules: - If $a > 0$, the graph is a smooth U shape opening upwards with a minimum point. - If $a < 0$, the graph is a smooth ∩ shape opening downwards with a maximum point. - The axis of symmetry is given by $$x = -\frac{b}{2a}$$ 4. For $f(x) = 3x^2 + 5x + 2$, identify $a=3$, $b=5$, and $c=2$. 5. Since $a=3 > 0$, the graph opens upwards and has a minimum point. 6. Calculate the axis of symmetry: $$x = -\frac{b}{2a} = -\frac{5}{2(3)} = -\frac{5}{6}$$ 7. To find the minimum point, substitute $x = -\frac{5}{6}$ back into $f(x)$: $$f\left(-\frac{5}{6}\right) = 3\left(-\frac{5}{6}\right)^2 + 5\left(-\frac{5}{6}\right) + 2$$ $$= 3\times \frac{25}{36} - \frac{25}{6} + 2$$ $$= \frac{75}{36} - \frac{150}{36} + \frac{72}{36} = \frac{75 - 150 + 72}{36} = \frac{-3}{36} = -\frac{1}{12}$$ 8. Therefore, the minimum point is at $$\left(-\frac{5}{6}, -\frac{1}{12}\right)$$. Final answer: The quadratic function $f(x) = 3x^2 + 5x + 2$ has a minimum point at $$\left(-\frac{5}{6}, -\frac{1}{12}\right)$$ and its axis of symmetry is $$x = -\frac{5}{6}$$.