Domain of a rational function

Factor the denominator: $x^2 - 4 = (x-2)(x+2)$. This expression is zero when $x = 2$ or $x = -2$, so $h$ is undefined at these two points.

Since the numerator $x-1$ does not share these zeros, both points are excluded from the domain. Thus the domain is $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$.

The denominator $x^2-4$ is zero at $x=2$ and $x=-2$, and the numerator $x-1$ is nonzero at both points, so $x=2$ and $x=-2$ are vertical asymptotes.

Since the degree of the denominator (2) exceeds the degree of the numerator (1), $\lim_{x\to\pm\infty} h(x) = 0$, so $y=0$ is the horizontal asymptote.

Compute $h(-x) = \dfrac{-x-1}{x^2-4}$, using that $(-x)^2-4 = x^2-4$.

Since $h(-x) = \dfrac{-x-1}{x^2-4}$ differs from $h(x) = \dfrac{x-1}{x^2-4}$ and $-h(x) = \dfrac{-(x-1)}{x^2-4} = \dfrac{1-x}{x^2-4}$, we see that $h(-x) \neq h(x)$ and $h(-x) \neq -h(x)$. Therefore $h$ is neither even nor odd.

By the quotient rule, $h'(x) = \dfrac{(x^2-4)(1) - (x-1)(2x)}{(x^2-4)^2} = \dfrac{x^2-4-2x^2+2x}{(x^2-4)^2} = \dfrac{-x^2+2x-4}{(x^2-4)^2}$.

The denominator $(x^2-4)^2$ is always positive on the domain, so the sign of $h'(x)$ matches the sign of $-x^2+2x-4 = -(x^2-2x+4)$. The discriminant of $x^2-2x+4$ is $4-16=-12<0$, so $x^2-2x+4>0$ for all $x$, making $-(x^2-2x+4)<0$ everywhere.

Since $h'(x) < 0$ for every $x$ in the domain, $h$ is decreasing on each interval of its domain: $(-\infty,-2)$, $(-2,2)$, and $(2,\infty)$.

No. Part d established that $h'(x) < 0$ everywhere on the domain. On the interval $(-\infty, -2)$, the function is strictly decreasing. On the interval $(-2, 2)$, the function is also strictly decreasing. Since $h'(x)$ is negative throughout each of these intervals and does not change sign, the function never turns around to increase near $x = -2$. The numerator of $h'(x)$ is always negative (as shown by the discriminant argument), so the function remains monotonically decreasing on both sides of this asymptote.