1. The problem is to sketch the graph of the function $y = x^2 - 1$. 2. This is a quadratic function in the form $y = ax^2 + bx + c$ where $a = 1$, $b = 0$, and $c = -1$. 3. The graph of $y = x^2 - 1$ is a parabola opening upwards because $a > 0$. 4. The vertex of the parabola is at the point $(h, k)$ where $h = -\frac{b}{2a} = 0$ and $k = c - \frac{b^2}{4a} = -1$. 5. So, the vertex is at $(0, -1)$. 6. The y-intercept is found by evaluating $y$ at $x=0$: $y = 0^2 - 1 = -1$. 7. The x-intercepts are found by solving $x^2 - 1 = 0$ which gives $x^2 = 1$ so $x = \pm 1$. 8. Plotting these points and the shape of the parabola, the graph opens upwards with vertex at $(0, -1)$ and crosses the x-axis at $x = -1$ and $x = 1$. Final answer: The graph of $y = x^2 - 1$ is a parabola with vertex at $(0, -1)$, y-intercept at $(0, -1)$, and x-intercepts at $(-1, 0)$ and $(1, 0)$.