1. [tex]f[/tex] has a horizontal asymptote at [tex]y=-4[/tex]
This means that
[tex]\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0[/tex]
(for at least one of these limits)
2. [tex]f[/tex] has a vertical asymptote at [tex]x=3[/tex]
This means that [tex]f[/tex] has a non-removable discontinuity at [tex]x=3[/tex]. Since [tex]f[/tex] is some rational function, there must be a factor of [tex]x-3[/tex] in its denominator.
3. [tex]f[/tex] has an [tex]x[/tex]-intercept at (1, 0)
This means [tex]f(1)=0[/tex].
(a) With
[tex]f(x)=\dfrac{ax+b}{x+c}[/tex]
the second point above suggests [tex]c=-3[/tex]. The first point tells us that
[tex]\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0[/tex]
In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if [tex]a=-4[/tex] so that the linear terms vanish.
The third point tells us that
[tex]f(1)=\dfrac{a+b}{1-3}=0\implies a=-b\implies b=4[/tex]
So
[tex]f(x)=\dfrac{-4x+4}{x-3}[/tex]
(b) Since
[tex]f(x)=\dfrac{rx+s}{2x+t}=\dfrac12\dfrac{rx+s}{x+\frac t2}[/tex]
we find that [tex]\dfrac t2=-3\implies t=-6[/tex], and [tex]r=a=-4[/tex] and [tex]s=b=4[/tex].
