1. **Problem:** Sketch the graph of the function $f(x) = x^2 - 1$. 2. **Formula and explanation:** This is a quadratic function of the form $f(x) = ax^2 + bx + c$ where $a=1$, $b=0$, and $c=-1$. 3. **Key features:** - The graph is a parabola opening upwards because $a=1 > 0$. - The vertex is at the point $(h, k)$ where $h = -\frac{b}{2a} = 0$ and $k = f(h) = f(0) = 0^2 - 1 = -1$. - The vertex is at $(0, -1)$. - The axis of symmetry is the vertical line $x=0$. - The y-intercept is $f(0) = -1$. - To find x-intercepts, solve $x^2 - 1 = 0$: $$ \begin{aligned} x^2 - 1 &= 0 \\ x^2 &= 1 \\ x &= \pm 1 \end{aligned} $$ 4. **Plot points:** - At $x=0$, $f(0) = -1$ (vertex and y-intercept). - At $x=1$, $f(1) = 1^2 - 1 = 0$ (x-intercept). - At $x=-1$, $f(-1) = (-1)^2 - 1 = 0$ (x-intercept). - At $x=2$, $f(2) = 4 - 1 = 3$. - At $x=-2$, $f(-2) = 4 - 1 = 3$. 5. **Graph description:** The parabola opens upward with vertex at $(0,-1)$, crossing the x-axis at $x=1$ and $x=-1$, and the y-axis at $y=-1$. This completes the sketch and analysis of the graph of $f(x) = x^2 - 1$.