[tex]f(x) = \frac{(2x+1)(x-5)}{(x-5)(x+4)^{2} }[/tex]Domain:V.A:Roots:Y-int:H.A:Holes:O.A:Also, draw on the graph attached.

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.