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.

Start by factoring the numerator and denominator of the function.

$\displaystyle j(x)=\frac{x^2+7x+6}{x^2-1}=\frac{(x+6)(x+1)}{(x+1)(x-1)}=\frac{x+6}{x-1}$

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.

$\displaystyle \frac{-1+6}{-1-1}=-\frac{5}{2}$

$\displaystyle \left(-1, -\frac{5}{2}\right)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle k(x)=\frac{2x^2-x-1}{x^2-6x+5}=\frac{(2x+1)(x-1)}{(x-1)(x-5)}=\frac{2x+1}{x-5}$

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.

$\displaystyle \frac{2(1)+1}{1-5}=-\frac{3}{4}$

$\displaystyle \left(1, -\frac{3}{4}\right)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle l(x)=\frac{2x^2+5x-12}{x^2+3x-4}=\frac{(2x-3)(x+4)}{(x+4)(x-1)}=\frac{2x-3}{x-1}$

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.

$\displaystyle \frac{2(-4)-3}{-4-1}=\frac{11}{5}$

$\displaystyle \left(-4, \frac{11}{5}\right)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

$\displaystyle n(x)=\frac{x^3+2x^2-3x}{x^2+8x+15}=\frac{x(x^2+2x-3)}{(x+3)(x+5)}=\frac{x(x-1)(x+3)}{(x+3)(x+5)}=\frac{x(x-1)}{x+5}$

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.

$\displaystyle \frac{-3(-3-1)}{-3+5}=6$

$\displaystyle (-3, 6)$ is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

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

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

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.

$\displaystyle \frac{-1}{(-1-1)(-1+3)}=\frac{1}{4}$

$\displaystyle \left(-1, \frac{1}{4}\right)$ is the point of discontinuity.

Before we simplify, set the denominator equal to zero to determine where $\displaystyle x$ is invalid.  The value of the denominator cannot equal to zero.

$\displaystyle 4x-2 \neq0$

$\displaystyle 4x \neq2$

$\displaystyle x\neq \frac{2}{4} \neq \frac{1}{2}$

The value at $\displaystyle x=\frac{1}{2}$ is invalid in the domain.

Pull out a greatest common factor for the numerator and the denominator and simplify.

$\displaystyle y=\frac{8x-4}{4x-2} =\frac{4(2x-1)}{2(2x-1)} = 2$

Since the terms $\displaystyle 2x-1$ can be cancelled, there will not be any vertical asymptotes.  Even though the rational function simplifies to $\displaystyle y=2$, there will be instead a hole at $\displaystyle x=\frac{1}{2}$ on the graph.  

The answer is:  $\displaystyle \textup{Hole at }x=\frac{1}{2}$