1. The problem is to analyze the function $f(x) = x^2 - 4$ using calculus. 2. We start by finding the critical points where the derivative is zero or undefined. The derivative of $f(x)$ is given by the power rule: $$f'(x) = 2x$$ 3. Set the derivative equal to zero to find critical points: $$2x = 0 \implies x = 0$$ 4. To determine the nature of the critical point, compute the second derivative: $$f''(x) = 2$$ Since $f''(0) = 2 > 0$, the function has a local minimum at $x=0$. 5. Evaluate the function at the critical point: $$f(0) = 0^2 - 4 = -4$$ 6. The function $f(x) = x^2 - 4$ is a parabola opening upwards with vertex at $(0, -4)$, which is the minimum point. 7. The $x$-intercepts are found by solving $f(x) = 0$: $$x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2$$ 8. The $y$-intercept is $f(0) = -4$. Summary: The function has a minimum at $(0, -4)$, $x$-intercepts at $x = -2$ and $x = 2$, and opens upwards.