i) The given function is
[tex]f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}[/tex]
The domain is
[tex](x-5)(x+4)^2\ne 0[/tex]
[tex](x-5)\ne0,(x+4)^2\ne 0[/tex]
[tex]x\ne5,x\ne -4[/tex]
ii) For vertical asymptotes, we simplify the function to get;
[tex]f(x)=\frac{(2x+1)}{(x+4)^2}[/tex]
The vertical asymptote occurs at
[tex](x+4)^2=0[/tex]
[tex]x=-4[/tex]
iii) The roots are the x-intercepts of the reduced fraction.
Equate the numerator of the reduced fraction to zero.
[tex]2x+1=0[/tex]
[tex]2x=-1[/tex]
[tex]x=-\frac{1}{2}[/tex]
iv) To find the y-intercept, we substitute [tex]x=0[/tex] into the reduced fraction.
[tex]f(0)=\frac{(2(0)+1)}{(0+4)^2}[/tex]
[tex]f(0)=\frac{(1)}{(4)^2}[/tex]
[tex]f(0)=\frac{1}{16}[/tex]
v) The horizontal asymptote is given by;
[tex]lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0[/tex]
The horizontal asymptote is [tex]y=0[/tex].
vi) The function has a hole at [tex]x-5=0[/tex].
Thus at [tex]x=5[/tex].
This is the factor common to both the numerator and the denominator.
vii) The function is a proper rational function.
Proper rational functions do not have oblique asymptotes.
