1. The problem asks which points lie on the graph of $$y = f(x) + 4$$, where $$f(x)$$ is a parabola with vertex at $$(-4,1)$$. 2. The graph $$y = f(x) + 4$$ is the original parabola shifted up by 4 units. 3. This means for any $$x$$, $$y = f(x) + 4$$, so if a point $$(x,y)$$ is on the graph, then $$y - 4 = f(x)$$. 4. We know the vertex of $$f(x)$$ is at $$(-4,1)$$, so $$f(-4) = 1$$. 5. Check each point $$(x,y)$$ by calculating $$y - 4$$ and comparing to $$f(x)$$: - For $$(-6,5)$$: $$5 - 4 = 1$$. We need to check if $$f(-6) = 1$$. - For $$(-4,5)$$: $$5 - 4 = 1$$. Since $$f(-4) = 1$$, this point is on the graph. - For $$(-8,1)$$: $$1 - 4 = -3$$. Check if $$f(-8) = -3$$. - For $$(-2,1)$$: $$1 - 4 = -3$$. Check if $$f(-2) = -3$$. - For $$(-4,-3)$$: $$-3 - 4 = -7$$. Check if $$f(-4) = -7$$ (no, since $$f(-4) = 1$$). - For $$(-2,9)$$: $$9 - 4 = 5$$. Check if $$f(-2) = 5$$. 6. Using the vertex form of the parabola $$f(x) = a(x+4)^2 + 1$$ and the point $$(-2,5)$$ on $$f(x)$$ (given in the problem), find $$a$$: $$5 = a(-2+4)^2 + 1$$ $$5 = a(2)^2 + 1$$ $$5 = 4a + 1$$ $$4a = 4$$ $$a = 1$$ 7. So, $$f(x) = (x+4)^2 + 1$$. 8. Calculate $$f(x)$$ for each $$x$$: - $$f(-6) = (-6+4)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5$$ - $$f(-8) = (-8+4)^2 + 1 = (-4)^2 + 1 = 16 + 1 = 17$$ - $$f(-2) = (-2+4)^2 + 1 = (2)^2 + 1 = 4 + 1 = 5$$ 9. Now compare $$y - 4$$ to $$f(x)$$: - $$(-6,5)$$: $$5 - 4 = 1$$ but $$f(-6) = 5$$, so no. - $$(-4,5)$$: $$5 - 4 = 1$$ and $$f(-4) = 1$$, yes. - $$(-8,1)$$: $$1 - 4 = -3$$ but $$f(-8) = 17$$, no. - $$(-2,1)$$: $$1 - 4 = -3$$ but $$f(-2) = 5$$, no. - $$(-4,-3)$$: $$-3 - 4 = -7$$ but $$f(-4) = 1$$, no. - $$(-2,9)$$: $$9 - 4 = 5$$ and $$f(-2) = 5$$, yes. 10. Therefore, the points on the graph of $$y = f(x) + 4$$ are $$(-4,5)$$ and $$(-2,9)$$. **Final answer:** $$(-4,5)$$ and $$(-2,9)$$ are on the graph of $$y = f(x) + 4$$.