1. **State the problem:** Find the vertex, direction of opening, intercepts, and sketch the graph of the quadratic function $$f(x) = 9x^2 + 18x + 14$$. 2. **Formula for vertex:** The vertex of a parabola given by $$f(x) = ax^2 + bx + c$$ is at $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$. 3. **Calculate the vertex:** Here, $$a=9$$ and $$b=18$$. $$x_{vertex} = -\frac{18}{2 \times 9} = -\frac{18}{18} = -1$$ Evaluate $$f(-1)$$: $$f(-1) = 9(-1)^2 + 18(-1) + 14 = 9 - 18 + 14 = 5$$ So, the vertex is $$(-1, 5)$$. 4. **Direction of opening:** Since $$a=9 > 0$$, the parabola opens **upward**. 5. **Find intercepts:** - **y-intercept:** Set $$x=0$$: $$f(0) = 14$$ So, y-intercept is $$(0, 14)$$. - **x-intercepts:** Solve $$9x^2 + 18x + 14 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-18 \pm \sqrt{18^2 - 4 \times 9 \times 14}}{2 \times 9}$$ Calculate discriminant: $$18^2 - 4 \times 9 \times 14 = 324 - 504 = -180$$ Since the discriminant is negative, there are **no real x-intercepts**. 6. **Summary:** - Vertex: $$(-1, 5)$$ - Opens upward - y-intercept: $$(0, 14)$$ - No real x-intercepts 7. **Sketch:** The parabola has vertex at $$(-1,5)$$, opens upward, crosses y-axis at $14$, and does not cross x-axis.