Answer
(a) $r(x),u(x)$
(b) $s(x)$
(c) $s(x),w(x)$
(d) $w(x)$
(e) $x=-1,2$ (vertical) and $y=0$ (horizontal)
(f) see graph, symptotes $x=\pm5$, x-intercepts $-4.372, 1.372$, y-intercept $0.24$
(g) $P(x)=x^2-2x-5$
Work Step by Step
(a) as $x\to \pm \infty, r(x)\to 0, u(x)\to 1$ they both have horizontal asymptotes.
(b) $s(x)$ will has a slant asymptote because the quotient will be a line.
(c) $s(x)$ has no vertical asymptote because its divisor is always larger than 4, $w(x)$ has no vertical asymptote because the dividend contains a factor which is the same as the divisor.
(d) $w(x)$ has a "hole" because both the dividend and the divisor has a common factor of $x+3$
(e) The asymptotes of function $r(x)$ are $x=-1,x=2$ (vertical) and $y=0$ (horizontal)
(f) See graph, asymptotes are $x=\pm5$, x-intercepts $-3,2$, y-intercept $0.24$
(g) Using long division as shown in the figure (g), we can find the quotient of $t(x)$ as $x^2-2x-5$, let $P(x)=x^2-2x-5$ and plot both $P(x), t(x)$ and it can be seen that they have the same end behavior.