Answer to Question #236685 in Calculus for CCW

Question #236685

f(x)= 3(x-1)(x+2) / (x-4) (x+2)

f) state horizontal asymptotes

g) sketch of graph


1
Expert's answer
2021-09-16T00:46:49-0400
"f(x)=\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"

"x-4\\not=0, x+2\\not=0"


"Domain: (-\\infin, -2)\\cap (-2, 4)\\cap (4, \\infin)"

(f)


"\\lim\\limits_{x\\to-\\infin}f(x)=\\lim\\limits_{x\\to-\\infin}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"

"=\\lim\\limits_{x\\to-\\infin}\\dfrac{3(x-1)}{x-4}=\\dfrac{3}{1}=3"


"\\lim\\limits_{x\\to\\infin}f(x)=\\lim\\limits_{x\\to\\infin}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"

"=\\lim\\limits_{x\\to\\infin}\\dfrac{3(x-1)}{x-4}=\\dfrac{3}{1}=3"

Horizontal asymptote: "y=3"


(g)


"\\lim\\limits_{x\\to-2^-}f(x)=\\lim\\limits_{x\\to-2^-}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"

"=\\lim\\limits_{x\\to-2^-}\\dfrac{3(x-1)}{x-4}=\\dfrac{3(-2-1)}{-2-4}=\\dfrac{3}{2}"



"\\lim\\limits_{x\\to-2^+}f(x)=\\lim\\limits_{x\\to-2^+}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"


"=\\lim\\limits_{x\\to-2^+}\\dfrac{3(x-1)}{x-4}=\\dfrac{3(-2-1)}{-2-4}=\\dfrac{3}{2}"


"\\lim\\limits_{x\\to-2^-}f(x)=\\dfrac{1}{2}=\\lim\\limits_{x\\to-2^+}f(x)=>\\lim\\limits_{x\\to-2}f(x)=\\dfrac{1}{2}"

The function "f(x)" has a removable discontinuity at "x=-2."



"\\lim\\limits_{x\\to4^-}f(x)=\\lim\\limits_{x\\to4^-}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}"

"=\\lim\\limits_{x\\to4^-}\\dfrac{3(x-1)}{x-4}=-\\infin"





"\\lim\\limits_{x\\to4^+}f(x)=\\lim\\limits_{x\\to4^+}\\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}""=\\lim\\limits_{x\\to4^+}\\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.


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