We have the expression:
[tex]y=\frac{x+2}{x^2+5x+6}[/tex]
We will start factorizing the denominator:
[tex]\begin{gathered} x=\frac{-5\pm\sqrt[]{5^2-4\cdot1\cdot6}}{2} \\ x=\frac{-5\pm\sqrt[]{25-24}}{2} \\ x=\frac{-5\pm\sqrt[]{1}}{2} \\ x=\frac{-5\pm1}{2} \\ x_1=\frac{-5-1}{2}=-\frac{6}{2}=-3 \\ x_2=\frac{-5+1}{2}=-\frac{4}{2}=-2 \\ \Rightarrow x^2+5x+6=(x+3)(x+2) \end{gathered}[/tex]
Then, we can simplify the expression as:
[tex]y=\frac{x+2}{x^2+5x+6}=\frac{x+2}{(x+3)(x+2)}=\frac{1}{x+3}[/tex]
We will have a vertical assymptote at the discontinuity x=-3:
[tex]\begin{gathered} \lim _{x\to-3+}y=\infty \\ \lim _{x\to-3-}y=-\infty \end{gathered}[/tex]
As holes, we can consider x=-2, because in the original equation it would make a discontinuity (it would make zero the quadratic denominator and the numerator), but not an assymptote. The limit exists but the function is not defined for x=-2.
Then, as x=-3 and x=-2 are discontinuities, the domain is D: {x in R | x≠-3, x≠-2}.
We can calculate if there is an horizontal assymptote calculating the limit of y when x approaches infinity:
[tex]\lim _{x\to\infty}\frac{1}{x+3}=0[/tex]
Then, as 0 is a finite value, we have an HA at y=0.
The range is all the values that y can take given the domain.
The only value that y can not take is the HA y=0, so the range is R: {y in R | y≠0}
[tex]R\colon\mleft\lbrace y\in R\mright|y\neq0\}[/tex]
Answer:
VA: x=-3
Holes: x=-2
HA: y=0
Domain: {x | x≠-3, x≠-3}
Range: {y | y≠0}
