Domain and vertical asymptote of r(x)

The denominator $x-2$ equals zero when $x=2$, so $x=2$ must be excluded from the domain. The domain is therefore all real numbers except $x=2$, written $(-\infty,2)\cup(2,\infty)$.

At $x=2$, the numerator is $2^2+1=5\neq 0$, so the zero of the denominator is not canceled by a matching zero of the numerator. This confirms that $x=2$ gives a true vertical asymptote (not a removable hole), so the vertical asymptote is $x=2$.

Dividing $x^2+1$ by $x-2$: $x^2 \div x = x$, and $x(x-2) = x^2-2x$, leaving a remainder of $2x+1$. Next, $2x \div x = 2$, and $2(x-2) = 2x-4$, leaving a remainder of $5$.

So $r(x) = x + 2 + \dfrac{5}{x-2}$. As $x\to\pm\infty$, the term $\dfrac{5}{x-2}\to 0$, so $r(x)$ approaches the line $y = x+2$, which is the slant asymptote.

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

Setting the numerator equal to zero, $x^2-4x-1=0$, gives $x = \dfrac{4\pm\sqrt{16+4}}{2} = 2\pm\sqrt{5}$. These are the critical points (both in the domain, since neither equals 2).

Testing sign of $r'(x)$ on the resulting intervals (noting the denominator $(x-2)^2$ is always positive): $r$ is increasing on $(-\infty, 2-\sqrt5)$ and $(2+\sqrt5,\infty)$, and decreasing on $(2-\sqrt5,2)$ and $(2,2+\sqrt5)$.

The graph hugs a slanted straight line far from the vertical asymptote. This is the slant asymptote found in part b, the line y equals x plus two, since the remainder term vanishes as x grows large in either direction.