[tex]y=\frac{2x^2-5x+2}{x^2-4}[/tex]
Limits:
To + infinite and - infinite limit you can evaluate the limit as follow: As you can see the limit is is the horizontal asymptote.
[tex]\begin{gathered} \lim _{x\to\pm\infty}(\frac{2x^2-5x+2}{x^2-4}) \\ \\ \text{Transform the equation by factoring x}^2\text{ in numerator and denominator:} \\ \lim _{x\to\pm\infty}(\frac{x^2(2-\frac{5}{x}+\frac{2}{x^2})}{x^2(1-\frac{4}{x^2})}) \\ \\ \lim _{x\to\pm\infty}(\frac{2-\frac{5}{x}+\frac{2}{x^2}}{1-\frac{4}{x^2}}) \\ \\ \text{Evaluate the limit knowinf that:} \\ \lim _{n\to\pm\infty}(\frac{1}{n^p})=0 \\ \\ \\ \lim _{x\to\pm\infty}(\frac{2-\frac{5}{x}+\frac{2}{x^2}}{1-\frac{4}{x^2}})=\frac{2-0+0}{1-0}=\frac{2}{1}=2 \end{gathered}[/tex][tex]\begin{gathered} \lim _{x\to-2^+}(\frac{2x^2-5x+2}{x^2-4})=-\infty \\ \\ \lim _{x\to-2^-}(\frac{2x^2-5x+2}{x^2-4})=+\infty \\ \\ \end{gathered}[/tex]
For the limits above you use the vertical asymptote (x=-2), the function tends to -infinite to the right of the asymptote and tends to + infinite to the left of the symptote
Graph: Function in red
Vertical asymptote in blue
Horizontal asymptote in green
x- Intercept (0.5,0)
y-intercept (0,0.5)
