[tex]\text{Answer:}}\quad \boxed{\begin{array}{l||c|c|c}\underline{\qquad \qquad }&\underline{Zero}&\underline{\stackrel {Vertical}{Asymptote}}&\underline{\stackrel{Removable}{Discontinuity}}\\&&&\\x=-3&&&X\\&\quad&\quad&\quad\\x=1&&X\end{array}}[/tex]
Step-by-step explanation:
[tex]g(x)=\dfrac{x+3}{x^2+2x-2}\qquad =\dfrac{x+3}{(x+3)(x-1)}[/tex]
Set the denominator equal to zero and solve for x. These are the asymptotes: x = -3 and x = 1
Notice that (x+3) cancels out from the numerator and denominator
This becomes a Removable Discontinuity, instead of an asymptote.
⇒ Therefore, x = 1 is the vertical asymptote
x = -3 is a removable discontinuity
