1. **State the problem:** We need to graph the function $$f(x) = (x-1)^2 - 1$$. 2. **Recall the formula:** This is a quadratic function in vertex form $$f(x) = a(x-h)^2 + k$$ where the vertex is at $$(h, k)$$. 3. **Identify the vertex:** Here, $a=1$, $h=1$, and $k=-1$, so the vertex is at $$(1, -1)$$. 4. **Plot the vertex:** This is the lowest point since $a>0$ (parabola opens upwards). 5. **Find the y-intercept:** Set $x=0$, then $$f(0) = (0-1)^2 - 1 = 1 - 1 = 0$$, so the y-intercept is at $$(0,0)$$. 6. **Find the x-intercepts:** Set $f(x)=0$: $$0 = (x-1)^2 - 1$$ $$ (x-1)^2 = 1$$ $$x-1 = \pm 1$$ $$x = 1 \pm 1$$ So, $x=0$ and $x=2$ are the roots. 7. **Sketch the graph:** Plot points at $(0,0)$, $(1,-1)$, and $(2,0)$ and draw a smooth parabola opening upwards. **Final answer:** The graph of $$f(x) = (x-1)^2 - 1$$ is a parabola with vertex at $(1,-1)$, passing through $(0,0)$ and $(2,0)$.