i) The given function is
![f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}](/tpl/images/0480/2125/4f6b7.png)
The domain is
![(x-5)(x+4)^2\ne 0](/tpl/images/0480/2125/ae230.png)
![(x-5)\ne0,(x+4)^2\ne 0](/tpl/images/0480/2125/05961.png)
![x\ne5,x\ne -4](/tpl/images/0480/2125/05bf4.png)
ii) For vertical asymptotes, we simplify the function to get;
![f(x)=\frac{(2x+1)}{(x+4)^2}](/tpl/images/0480/2125/e36ca.png)
The vertical asymptote occurs at
![(x+4)^2=0](/tpl/images/0480/2125/3ddef.png)
![x=-4](/tpl/images/0480/2125/14c42.png)
iii) The roots are the x-intercepts of the reduced fraction.
Equate the numerator of the reduced fraction to zero.
![2x+1=0](/tpl/images/0480/2125/afdb7.png)
![2x=-1](/tpl/images/0480/2125/ed1a1.png)
![x=-\frac{1}{2}](/tpl/images/0480/2125/2e63a.png)
iv) To find the y-intercept, we substitute
into the reduced fraction.
![f(0)=\frac{(2(0)+1)}{(0+4)^2}](/tpl/images/0480/2125/154ba.png)
![f(0)=\frac{(1)}{(4)^2}](/tpl/images/0480/2125/d8428.png)
![f(0)=\frac{1}{16}](/tpl/images/0480/2125/92313.png)
v) The horizontal asymptote is given by;
![lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0](/tpl/images/0480/2125/3bf9e.png)
The horizontal asymptote is
.
vi) The function has a hole at
.
Thus at
.
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.
![domain: v.a: roots: y-int: h.a: holes: o.a: also, draw on the graph attached.](/tpl/images/0480/2125/067fc.jpg)