1. The problem is to graph the function $f(x) = x^2$ with the domain restriction $-2 < x < 1$. 2. The function $f(x) = x^2$ is a parabola opening upwards with vertex at the origin $(0,0)$. 3. The domain restriction means we only consider $x$ values strictly between $-2$ and $1$. 4. To graph, calculate key points within the domain: - At $x = -2$, $f(-2) = (-2)^2 = 4$ (not included because domain is strict inequality). - At $x = -1$, $f(-1) = 1$. - At $x = 0$, $f(0) = 0$. - At $x = 1$, $f(1) = 1$ (not included because domain is strict inequality). 5. Plot points $(-1,1)$, $(0,0)$, and points close to $-2$ and $1$ but not including them. 6. Draw the parabola segment connecting these points smoothly, excluding the endpoints at $x=-2$ and $x=1$. 7. The graph is the portion of the parabola between $x=-2$ and $x=1$, open at both ends. Final answer: The graph is the parabola $y = x^2$ restricted to $-2 < x < 1$ with open endpoints at $x=-2$ and $x=1$.