Answer to Question #236685 in Calculus for CCW

Question

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

f) state horizontal asymptotes

g) sketch of graph

Answer 1

$$f(x)=\dfrac{3(x-1)(x+2)}{(x-4)(x+2)}$$

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

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

(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.