1. **State the problem:** Find the axis of symmetry, vertex, y-intercept, and x-intercepts of the parabola given by the equation $y = 3x^2 + 12x + 9$. 2. **Axis of symmetry formula:** For a quadratic $y = ax^2 + bx + c$, the axis of symmetry is given by $$x = -\frac{b}{2a}$$ where $a = 3$ and $b = 12$. 3. **Calculate axis of symmetry:** $$x = -\frac{12}{2 \times 3} = -\frac{12}{6} = -2$$ 4. **Find the vertex:** The vertex lies on the axis of symmetry. Substitute $x = -2$ into the equation: $$y = 3(-2)^2 + 12(-2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3$$ So, the vertex is at $(-2, -3)$. 5. **Find the y-intercept:** Set $x=0$: $$y = 3(0)^2 + 12(0) + 9 = 9$$ So, the y-intercept is at $(0, 9)$. 6. **Find the x-intercepts:** Set $y=0$ and solve for $x$: $$0 = 3x^2 + 12x + 9$$ Divide both sides by 3: $$0 = \cancel{3}x^2 + \cancel{3}4x + \cancel{3}3 \Rightarrow 0 = x^2 + 4x + 3$$ Factor the quadratic: $$0 = (x + 3)(x + 1)$$ Set each factor to zero: $$x + 3 = 0 \Rightarrow x = -3$$ $$x + 1 = 0 \Rightarrow x = -1$$ 7. **Write x-intercepts as coordinates:** Left x-intercept: $(-3, 0)$ Right x-intercept: $(-1, 0)$ **Final answers:** - Axis of symmetry: $x = -2$ - Vertex: $(-2, -3)$ - Y-intercept: $(0, 9)$ - Left x-intercept: $(-3, 0)$ - Right x-intercept: $(-1, 0)$