Answer in Calculus for CCW #236685
f(x)= 3(x-1)(x+2) / (x-4) (x+2)
f) state horizontal asymptotes
g) sketch of graph
“x-4not=0, x+2not=0”
“Domain: (-infin, -2)cap (-2, 4)cap (4, infin)”
(f)
“limlimits_{xto-infin}f(x)=limlimits_{xto-infin}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”
“=limlimits_{xto-infin}dfrac{3(x-1)}{x-4}=dfrac{3}{1}=3”
“limlimits_{xtoinfin}f(x)=limlimits_{xtoinfin}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”
“=limlimits_{xtoinfin}dfrac{3(x-1)}{x-4}=dfrac{3}{1}=3”
Horizontal asymptote: “y=3”
(g)
“limlimits_{xto-2^-}f(x)=limlimits_{xto-2^-}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”
“=limlimits_{xto-2^-}dfrac{3(x-1)}{x-4}=dfrac{3(-2-1)}{-2-4}=dfrac{3}{2}”
“limlimits_{xto-2^+}f(x)=limlimits_{xto-2^+}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”
“=limlimits_{xto-2^+}dfrac{3(x-1)}{x-4}=dfrac{3(-2-1)}{-2-4}=dfrac{3}{2}”
“limlimits_{xto-2^-}f(x)=dfrac{1}{2}=limlimits_{xto-2^+}f(x)=>limlimits_{xto-2}f(x)=dfrac{1}{2}”
The function “f(x)” has a removable discontinuity at “x=-2.”
“limlimits_{xto4^-}f(x)=limlimits_{xto4^-}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”
“=limlimits_{xto4^-}dfrac{3(x-1)}{x-4}=-infin”
“limlimits_{xto4^+}f(x)=limlimits_{xto4^+}dfrac{3(x-1)(x+2)}{(x-4)(x+2)}”“=limlimits_{xto4^+}dfrac{3(x-1)}{x-4}=infin”
Vertical asymptote: “x=4”
“x=0, f(0)=dfrac{3(0-1)(0+2)}{(0-4)(0+2)}=dfrac{3}{4}”
Point “(0, dfrac{3}{4})” is “y” -intercept.
“y=0, 0=dfrac{3(x-1)(x+2)}{(x-4)(x+2)}=>x=1”
Point “(1, 0)” is “x” -intercept.
