Given the function $f(x) = 2 - x^2$, let's analyze some key aspects of it:

1. Domain:

The domain of $f(x)$ includes all real numbers since there are no restrictions on the values of $x$ (like division by zero or square roots of negative numbers). Therefore, the domain is: $\text{Domain} = (-\infty, \infty)$

2. Range:

To find the range, observe that the function is a downward-opening parabola (since the coefficient of $x^2$ is negative). The maximum value occurs at the vertex of the parabola.

• The vertex is found at $x = 0$ because the general form $ax^2 + bx + c$ simplifies to $-x^2 + 2$.
• Evaluating $f(0)$: $f(0) = 2 - 0^2 = 2$
• Since the parabola opens downward, the range is: $\text{Range} = (-\infty, 2]$

3. Intercepts:

• x-intercepts: Set $f(x) = 0$. $2 - x^2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2}$ So, the x-intercepts are $(\sqrt{2}, 0)$ and $(-\sqrt{2}, 0)$.

• y-intercept: Set $x = 0$. $f(0) = 2$ So, the y-intercept is $(0, 2)$.

4. Vertex:

As noted, the vertex is at $(0, 2)$.

5. Graph Behavior:

• The graph is symmetric about the y-axis because it's an even function (replacing $x$ with $-x$ gives the same result).
• As $x \to \infty$ or $x \to -\infty$, $f(x) \to -\infty$.

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Related Questions:

1. How do you find the maximum or minimum of a function using calculus?
2. What are the properties of even and odd functions?
3. How can you determine concavity and inflection points of a function?
4. What is the significance of intercepts in graph analysis?
5. How do transformations affect the graph of a parabola?

Tip: When analyzing quadratic functions, the sign of the coefficient of $x^2$ determines if the parabola opens upwards (positive) or downwards (negative).