Parabola Graph 1524A4

1. **State the problem:** We need to graph the parabola given by the equation $$y = (x + 4)^2 + 1$$ and plot five points: the vertex, two points to the left, and two points to the right of the vertex. 2. **Identify the vertex:** The parabola is in vertex form $$y = (x - h)^2 + k$$ where the vertex is at $$(h, k)$$. Here, $$y = (x + 4)^2 + 1 = (x - (-4))^2 + 1$$, so the vertex is at $$(-4, 1)$$. 3. **Plot points around the vertex:** Choose two points to the left of $$x = -4$$: $$x = -6$$ and $$x = -5$$. Calculate their $$y$$ values: $$y(-6) = (-6 + 4)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5$$ $$y(-5) = (-5 + 4)^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$$ Choose two points to the right of $$x = -4$$: $$x = -3$$ and $$x = -2$$. Calculate their $$y$$ values: $$y(-3) = (-3 + 4)^2 + 1 = (1)^2 + 1 = 1 + 1 = 2$$ $$y(-2) = (-2 + 4)^2 + 1 = (2)^2 + 1 = 4 + 1 = 5$$ 4. **Summary of points:** - Vertex: $$(-4, 1)$$ - Left points: $$(-6, 5)$$, $$(-5, 2)$$ - Right points: $$(-3, 2)$$, $$(-2, 5)$$ 5. **Graph description:** The parabola opens upward with vertex at $$(-4, 1)$$ and is symmetric about the vertical line $$x = -4$$. These points confirm the shape and symmetry of the parabola. Final answer: The parabola $$y = (x + 4)^2 + 1$$ with vertex $$(-4, 1)$$ and points $$(-6, 5)$$, $$(-5, 2)$$, $$(-3, 2)$$, $$(-2, 5)$$ plotted.