1. **State the problem:** Given a downward-opening parabola with vertex approximately at $(-3,4)$, axis of symmetry $x=-3$, and zeros at $(-5,0)$ and $(-1,0)$, identify the axis of symmetry, vertex, domain, range, zeros, and the correct equation from the options. 2. **Axis of symmetry formula:** For a parabola $y = ax^2 + bx + c$, the axis of symmetry is given by $$x = -\frac{b}{2a}$$ 3. **Vertex:** The vertex is at $(h,k)$ where $h$ is the axis of symmetry and $k$ is the value of $y$ at $x=h$. 4. **Domain:** The domain of any quadratic function is all real numbers: $$\text{Domain} = (-\infty, \infty)$$ 5. **Range:** Since the parabola opens downward (as given), the range is all $y$ values less than or equal to the vertex's $y$-coordinate: $$\text{Range} = (-\infty, 4]$$ 6. **Zeros:** The zeros are the $x$-intercepts where $y=0$, given approximately as $-5$ and $-1$. 7. **Identify the equation:** Check each option's axis of symmetry and vertex: - Option A: $y = x^2 + 6x - 5$ - $a=1$, $b=6$ - Axis: $x = -\frac{6}{2(1)} = -3$ - Vertex $y$-value: $y(-3) = (-3)^2 + 6(-3) - 5 = 9 - 18 - 5 = -14$ - Vertex is $(-3, -14)$, which does not match. - Option B: $y = x^2 - 6x - 5$ - $a=1$, $b=-6$ - Axis: $x = -\frac{-6}{2(1)} = 3$ - Vertex $y$-value: $y(3) = 9 - 18 - 5 = -14$ - Vertex is $(3, -14)$, does not match. - Option C: $y = -x^2 + 6x - 5$ - $a=-1$, $b=6$ - Axis: $x = -\frac{6}{2(-1)} = -\frac{6}{-2} = 3$ - Vertex $y$-value: $y(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4$ - Vertex is $(3,4)$, but given vertex is $(-3,4)$, so no. - Option D: $y = -x^2 - 6x - 5$ - $a=-1$, $b=-6$ - Axis: $x = -\frac{-6}{2(-1)} = -\frac{-6}{-2} = -3$ - Vertex $y$-value: $y(-3) = -(-3)^2 - 6(-3) - 5 = -9 + 18 - 5 = 4$ - Vertex is $(-3,4)$, matches perfectly. 8. **Final answers:** - Axis of Symmetry: $x = -3$ - Vertex: $(-3, 4)$ - Domain: $(-\infty, \infty)$ - Range: $(-\infty, 4]$ - Zeros: $-5$ and $-1$ - Equation: $y = -x^2 - 6x - 5$ (Option D)