A hole when x = 2

A vertical asymptote at x = 2

there is no hole or vertical asymptote

All real numbers

All real numbers except 4

All real numbers except 2

All real numbers except 2 and -2

the degree of the numerator and denominator are equal

the degree of the numerator is less than the degree of the denominator

the degree of the numerator is greater than the degree of the denominator

the numerator equals zero

y= 5

y= -6

y= 1

y= 5/-6

V.A. at x = 2

No Holes

H.A. at y = -1/3

y-intercepts at (0, 1/2)

V.A. at x = -3 and x = 1

Hole at (1, 0)

H.A. at y = 2

x-intercepts at (3,0) and (1, 0)

$$\frac{\left(2x-5\right)}{2x+8}$$  

$$\frac{\left(x-5\right)}{x+4}$$  

$$\frac{\left(2x-5\right)}{x+4}$$  

$$\frac{\left(x^2-6\right)}{x+4}$$  

$$-\frac{\left(3x+5\right)}{x^2+1}$$  

$$\frac{\left(3x+5\right)}{x^2+1}$$  

$$-\frac{\left(3x+3\right)}{x^2+1}$$  

$$\frac{\left(-3x-5\right)}{2x^2+2}$$  

$$\frac{29}{21x}$$  

$$\frac{7}{10x}$$  

$$\frac{1}{21x}$$  

$$\frac{19}{7x}$$  

$$-\frac{\left(x+6\right)}{\left(x+3\right)\left(x-3\right)}$$  

$$\frac{2\left(x+6\right)}{\left(x+3\right)\left(x-3\right)}$$  

$$-\frac{5x-6}{\left(x+3\right)\left(x-3\right)}$$  

$$\frac{\left(5x-6\right)}{\left(x+3\right)\left(x-3\right)}$$  

$$\frac{\left(x^3+3x-4\right)}{2\left(x^2-1\right)}$$  

$$\frac{\left(x^3+4x+12\right)}{2\left(x+1\right)^2}$$  

$$\frac{\left(3x-4\right)}{2\left(x-1\right)^2}$$  

$$\frac{\left(x^2+3x-4\right)}{2\left(x-1\right)^{ }}$$  

(3x² + 6x) / (2x² + 4)

(3x² + 6x) / (2x + 4)

(3x + 6) / (2x + 4)

(3x + 6) / (2x² + 4x)

$$\left\{x\left|x\ne8,\ x\ne6\right|\right\}$$

$$\left\{x\left|x\ne8,\ x\ne-6\right|\right\}$$

$$\left\{x\left|x\ne-8,\ x\ne6\right|\right\}$$

$$all\ real\ numbers$$