Question

Please I need help it’s urgent

Answer 1

`\text{Answer:}}\quad \boxed{\begin{array}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}}`

Explanation:

`g(x)=(x+3)/(x^2+2x-2)\qquad =(x+3)/((x+3)(x-1))`

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