Factorize the numerator for the function: 

$\displaystyle y=\frac{x^2-1}{x-1}$

$\displaystyle y=\frac{(x+1)(x-1)}{x-1} = x+1$

The removable discontinuity is $\displaystyle x-1$ since this is a term that can be eliminated from the function.  There are no vertical asymptotes.

Set the removable discontinutity to zero and solve for the location of the hole.

 $\displaystyle x-1=0$

$\displaystyle x=1$

The hole is located at:  $\displaystyle x=1$

Rewrite the function $\displaystyle y=\frac{x^2-6x+9}{x^2-9}$ in its factored form.

$\displaystyle y=\frac{x^2-6x+9}{x^2-9}=\frac{(x-3)(x-3)}{(x+3)(x-3)}$

Since the $\displaystyle x-3$ term can be cancelled, there is a removable discontinuity, or a hole, at $\displaystyle x=3$.

The remaining denominator of $\displaystyle x+3$ indicates a vertical asymptote at $\displaystyle x=-3$.

By looking at the denominator of $\displaystyle y= \frac{6x-9}{2x-3}$, there will be a discontinuity.

Since the denominator cannot be zero, set the denominator not equal to zero and solve the value of $\displaystyle x$.

$\displaystyle 2x-3\neq 0$

$\displaystyle 2x\neq 3$

$\displaystyle x\neq\frac{3}{2}$

There is a discontinuity at $\displaystyle x=\frac{3}{2}$.

To determine what type of discontinuity, check if there is a common factor in the numerator and denominator of $\displaystyle y= \frac{6x-9}{2x-3}$.

Since the common factor is existent, reduce the function.

$\displaystyle y= \frac{6x-9}{2x-3}=\frac{(3)(2x-3)}{2x-3}=3$

Since the $\displaystyle 2x-3$ term can be cancelled, there is a removable discontinuity, or a hole, at $\displaystyle x=\frac{3}{2}$.

Start by factoring the numerator and denominator of the function.

$\displaystyle m(x)=\frac{x^2+4x-5}{x^2+7x+10}=\frac{(x+5)(x-1)}{(x+5)(x+2)}=\frac{x-1}{x+2}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=-5$ is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the $\displaystyle y$ value, plug in $\displaystyle -5$ into the final simplified equation.

$\displaystyle \frac{-5-1}{-5+2}=2$

$\displaystyle (-5, 2)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}=\frac{x(x^2-4x-5)}{(x-5)(x-4)}=\frac{x(x-5)(x+1)}{(x-5)(x-4)}=\frac{x(x+1)}{x-4}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=5$ is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the $\displaystyle y$ value, plug in $\displaystyle 5$ into the final simplified equation.

$\displaystyle \frac{5(5+1)}{5-4}=30$

$\displaystyle (5, 30)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle r(x)=\frac{16-4x^2}{x^2-4}=\frac{-4(x^2-4)}{(x+2)(x-2)}=\frac{-4(x-2)(x+2)}{(x-2)(x+2)}=-4$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=\pm2$ is a zero for both the numerator and denominator, there is a point of discontinuity there. Since the final function is $\displaystyle r(x)=-4$, $\displaystyle (2, -4)$ and $\displaystyle (-2, -4)$ are points of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle s(x)=\frac{180-5x^2}{x^2-36}=\frac{-5(x^2-36)}{(x-6)(x+6)}=\frac{-5(x-6)(x+6)}{(x-6)(x+6)}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=\pm 6$ is a zero for both the numerator and denominator, there is a point of discontinuity there. Since the final function is $\displaystyle s(x)=-5$, $\displaystyle (6, -5)$ and $\displaystyle (-6, -5)$ are points of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle f(x)=\frac{x^2+x-12}{x^2+2x-8}=\frac{(x+4)(x-3)}{(x+4)(x-2)}=\frac{(x-3)}{(x-2)}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=-4$ is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the $\displaystyle y$ value, plug in $\displaystyle -4$ into the final simplified equation.

$\displaystyle \frac{-4-3}{-4-2}=\frac{7}{6}$

$\displaystyle \left(-4, \frac{7}{6}\right)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle g(x)=\frac{x^2-1}{x^2+8x+7}=\frac{(x-1)(x+1)}{(x+1)(x+7)}=\frac{x-1}{x+7}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=-1$ is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the $\displaystyle y$ value, plug in $\displaystyle -1$ into the final simplified equation.

$\displaystyle \frac{-1-1}{-1+7}=-\frac{1}{3}$

$\displaystyle \left(-1, -\frac{1}{3}\right)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle h(x)=\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3}$

A point of discontinuity occurs when a number $\displaystyle a$ is both a zero of the numerator and denominator.

Since $\displaystyle x=3$ is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the $\displaystyle y$ value, plug in $\displaystyle 3$ into the final simplified equation.

$\displaystyle \frac{3-2}{3+3}=\frac{1}{6}$

$\displaystyle \left(3, \frac{1}{6}\right)$ is the point of discontinuity.