find the focus and directrix of each parabola and graph the parabola. **Equation of the Parabola:**\[ (x + 1)^2 = -8(y + 1) \]**Vertex:**The vertex of the parabola is \((-1, -1)\).**Focus:**The focus of the parabola is \((-1, -3)\).**Directrix:**The directrix of the parabola is the line \(y = 1\).**Two Points:**Two points on the parabola are \((-3, -2)\) and \((1, -2)\).**Graph Description:**The graph displays a coordinate plane with a grid. The x-axis and y-axis intersect at the origin \((0,0)\), with markings for both positive and negative values. The parabola opens downward, with its vertex at \((-1, -1)\). The points \((-3, -2)\) and \((1, -2)\) lie on the parabola, and the axis of symmetry is vertical, passing through the vertex. The image shows a Cartesian coordinate system graph with both x-axis and y-axis ranging from -10 to 10. The grid is made up of squares, each representing one unit. The x-axis is the horizontal line and the y-axis is the vertical line, intersecting at the origin (0, 0). Numbers are marked at intervals of one along both axes. This type of graph is often used to plot points, lines, and curves in mathematics and can be used in various subjects such as algebra and geometry. The thick lines at integers help in easily identifying and plotting coordinates.

Consider the given equation. x+12=-8y+1 The general form of the parabola is 4py-k=x-h2. Where, (h,k) is vertex and 4p is focal length. SO, compere the equation and find the vertex . h,k=-1,-1Now, find the focal length of the equation. 4p=-84p=8p=2 Thus, the focal is -2. And the focus is defined as, Focus=h,k+p=-1,-1-2=-1,-3The given equation is symmetric around the y - axis. So the directrix is a line parallel to the x - axis, a distance - p from the center (- 1,- 1) y - coordinate. y=-1-p=-1+2y=1 And, the graph is,