1. **Problem Statement:** We are given a parabola that opens upward with vertex near $(-1, -9)$ and need to find: (a) Whether it opens upward or downward. (b) The equation of the axis of symmetry. (c) The coordinates of the vertex. (d) The x- and y-intercepts. 2. **Key Formulas and Rules:** - A parabola in vertex form is given by $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. - The axis of symmetry is the vertical line $$x = h$$. - The vertex is the point $(h, k)$. - The x-intercepts are found by solving $$y=0$$. - The y-intercept is found by evaluating $$y$$ at $$x=0$$. 3. **Given Information:** - Vertex: $(-1, -9)$ so $h = -1$, $k = -9$. - Parabola opens upward, so $a > 0$. 4. **Find the equation of the parabola:** Since the vertex is $(-1, -9)$ and it opens upward, assume $a=1$ for simplicity: $$y = (x + 1)^2 - 9$$ 5. **(a) Direction of opening:** Since $a=1 > 0$, the parabola opens **upward**. 6. **(b) Axis of symmetry:** The axis of symmetry is the vertical line through the vertex: $$x = -1$$ 7. **(c) Vertex coordinates:** The vertex is: $$(-1, -9)$$ 8. **(d) Find intercepts:** - **x-intercepts:** Set $y=0$: $$0 = (x + 1)^2 - 9$$ $$ (x + 1)^2 = 9$$ $$x + 1 = \\pm 3$$ So, $$x = -1 + 3 = 2$$ $$x = -1 - 3 = -4$$ - **y-intercept:** Set $x=0$: $$y = (0 + 1)^2 - 9 = 1 - 9 = -8$$ **Summary:** - (a) Upward - (b) Axis of symmetry: $x = -1$ - (c) Vertex: $(-1, -9)$ - (d) x-intercepts: $2, -4$ - (d) y-intercept: $-8$