what are the key aspects of the graph of $f(x)=x^{2}-b^{2}$, where $b$ is a real number?

Explanation:

Step1: Identify the function type

The function $f(x)=x^{2}-b^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$ (here $a = 1$, $b = 0$, $c=-b^{2}$).

Step2: Find the vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. Since $b = 0$ and $a = 1$, the x - coordinate of the vertex is $x = 0$. Substitute $x = 0$ into the function: $f(0)=0^{2}-b^{2}=-b^{2}$. So the vertex is $(0,-b^{2})$.

Step3: Find the y - intercept

The y - intercept is found by setting $x = 0$. We already found that when $x = 0$, $y=-b^{2}$, so the y - intercept is $(0,-b^{2})$.

Step4: Find the x - intercepts

Set $y = 0$, so $x^{2}-b^{2}=0$. Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we have $(x + b)(x - b)=0$. Then $x=-b$ or $x = b$. The x - intercepts are $(-b,0)$ and $(b,0)$.

Step5: Determine the shape

Since $a = 1>0$, the parabola opens upward.

Answer:

The vertex is $(0,-b^{2})$, the y - intercept is $(0,-b^{2})$, the x - intercepts are $(-b,0)$ and $(b,0)$, and the parabola opens upward.