\( f(x)=-(x+1)^{2}+2 \) Make sure that you graph the vertex, the AOS, the 2 extra points and refelect them across the AOS.

1. **Identify the vertex form of the parabola** The given function is \[ f(x)=-(x+1)^2+2. \] It is already in vertex form, \( f(x)=a(x-h)^2+k \), where \[ a=-1,\quad h=-1,\quad k=2. \] 2. **Determine the vertex** The vertex is at \[ (h, k)=(-1, 2). \] 3. **Find the Axis of Symmetry (AOS)** The axis of symmetry is the vertical line passing through the vertex, so it is \[ x=-1. \] 4. **Find 2 extra points on one side and reflect them across the AOS** We'll start by choosing two \(x\)-values on one side of the AOS. For convenience, let’s choose values to the right of \(x=-1\). - **Extra Point 1: Using \(x=0\)** Evaluate \( f(0) \): \[ f(0)=-(0+1)^2+2= -1+2=1. \] This gives the point \[ (0, 1). \] *Reflect across the AOS:* The reflection of a point \((x, y)\) across the line \(x=-1\) is \[ (x', y) \quad \text{where} \quad x' = -1 - (x - (-1)) = -1 - (x+1). \] For \((0, 1)\), \[ x' = -1 - (0+1) = -2. \] So the reflected point is \[ (-2, 1). \] - **Extra Point 2: Using \(x=1\)** Evaluate \( f(1) \): \[ f(1)=-(1+1)^2+2= -4+2=-2. \] This gives the point \[ (1, -2). \] *Reflect across the AOS:* For \((1, -2)\), \[ x' = -1 - (1+1) = -1 - 2 = -3. \] So the reflected point is \[ (-3, -2). \] 5. **Summary of the key points for the graph** - **Vertex:** \(\,(-1, 2)\) - **Axis of Symmetry (AOS):** \(\,x=-1\) - **Extra Points and their Reflections:** - Point: \((0, 1)\) and its reflection: \((-2, 1)\) - Point: \((1, -2)\) and its reflection: \((-3, -2)\) Plot these points on the coordinate plane and draw a smooth parabolic curve opening downward (since \(a=-1\)). Mark the vertex, draw the vertical line \(x=-1\) to indicate the axis of symmetry, and ensure the extra points are plotted along with their reflections across the AOS.