Discontinuities

Vertical Asymptote: $x=7$ since $x-7=0$

Hole: None

Vertical Asymptote: $x=\frac{3}{2}$ since $3-2x=0$ , $x=\frac{3}{2}$

Hole: None

Vertical Asymptote: $x=9$ , $x=-1$ since $x=9$ and $x=-1$

Hole: None

Vertical Asymptote: $x=\frac{1}{2}$ , $x=3$ since $2x^2-7x+3=0$ , $(2x-1)(x-3)=0$ , $2x-1=0$ and $x-3=0$ , $x=\frac{1}{2}$ and $x=3$

Hole: None

Vertical Asymptote: $x=-5$ since $(x+5)^2=0$ , $x+5=0$ , $x=-5$

Hole: None

Vertical Asymptote: None since $x^2+25=0$ , $x^2=-25$ , a number squared will never be negative

Hole: None

Vertical Asymptote: None since $x^2+2=0$ , $x^2=-2$ and any number squared will never be a negative number

Hole: None

Vertical Asymptote: $x=3$ since $\left|x-3\right|=0$ , $x-3=0$ , $x=3$

Hole: None

Vertical asymptotes: $x=-4$ and $x=4$ since $\left|x\right|-4=0$ , $\left|x\right|=4$ , $x=-4$ and $x=4$

Hole: None

Vertical Asymptote: $x=-3$

Hole: (3, $\frac{5}{8}$ ) since $\frac{3(x^2-x-6)}{4(x^2-9)}=\frac{3(x-3)(x+2)}{4(x+3)(x-3)}=\frac{3(x+2)}{4(x+3)}$ , (x-3) was cancelled, so the hole is at x=3. To find the y-coordinate, plug 3 into the reduced equation: $\frac{3(3+2)}{4(3+3)}=\frac{3\times 5}{4\times 6}=\frac{15}{24}=\frac{5}{8}$

$\frac{-2(x^2-4)}{3(x^2+4x+4)}=\frac{-2(x+2)(x-2)}{3(x+2)^2}=\frac{-2(x-2)}{3(x+2)}$

Vertical Asymptote: $x=-2$

Hole: None since the vertical asymptote takes care of the hole.

Vertical Asymptote: None

Hole: (-2,-4) since $\frac{x^2-4}{x+2}=\frac{(x+2)(x-2)}{x+2}=x-2$ , (x+2) was cancelled, so the hole is at x = -2. To find the y-coordinate, plug -2 into the reduced equation: $-2-2=-4$

Vertical Asymptotes: None

Holes: (3,3), (0,0) since $\frac{x^2(x-3)}{x^2-3x}=\frac{x^2(x-3)}{x(x-3)}=x$ , x and (x-3) were cancelled, so the holes are at x=0 and x=3. To find the y-coordinate, plug 0 and 3 into the reduced equation: 0, 3

Vertical Asymptote: None

Hole: (1,3) since $\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1$ , (x-1) was cancelled, so the hole is at x=1. To find the y-coordinate, plug 1 into the reduced equation: $1^2+1+1=3$

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

Vertical asymptote: $x=1$ since $x-1=0$

Hole: (-1, $\frac{7}{2}$ ) Since (x+1) was cancelled, the hole is at x= -1. To find the y-coordinate, plug -1 into the reduced equation: $\frac{2(-1-5)}{-1-1}=\frac{7}{2}$