Given:
[tex]y=\frac{2x^2-5x+2}{x^2-4}[/tex]
The vertical asymptotes, x= -2
The horizontal asymptotes y=2.
Aim:
We need to graph the function and find the end behavior.
Explanation:
The graph of the function:
x-intercept is (0.5,0) and y-intercept is (0, -0.5)
End behavior:
Take the limit of the function:
[tex]\lim _{x\to\infty}y=\lim _{x\to\infty}\frac{2x^2-5x+2}{x^2-4}[/tex]
[tex]\lim _{x\to\infty}y=\lim _{x\to\infty}\frac{2-\frac{5}{x}+\frac{2}{x^2}}{1^{}-\frac{4}{x^2}}[/tex]
[tex]\lim _{x\to\infty}y=\frac{2-0+0}{1^{}-0}=2[/tex]
We get
[tex]\lim _{x\to\infty}y=2\text{ }[/tex]
Taking the limit to negative infinity
[tex]\lim _{x\to-\infty}y=\lim _{x\to-\infty}\frac{2x^2-5x+2}{x^2-4}[/tex]
[tex]\lim _{x\to-\infty}y=\frac{2-0+0}{1^{}-0}=2[/tex]
We get
[tex]\lim _{x\to-\infty}y=2\text{ }[/tex]
Taking the limit to -2.
[tex]\lim _{x\to-2^+}y=\lim _{x\to-2^+}\frac{2x^2-5x+2}{x^2-4}[/tex]
[tex]\lim _{x\to-2^+}y=\frac{2(-2)^2-5(-2)+2}{(-2)^2_{}-4}=\frac{20}{0}=\infty[/tex]
We get
[tex]\lim _{x\to-2^+}y=\infty[/tex]
[tex]\lim _{x\to-2^-}y=\lim _{x\to-2^-}\frac{2x^2-5x+2}{x^2-4}[/tex]
[tex]\lim _{x\to-2^-}y=\frac{2(-2)^2-5(-2)+2}{(-2)^2_{}-4}=\frac{20}{0}=\infty[/tex]
We get
[tex]\lim _{x\to-2^-}y=\infty[/tex]
Final answer:
All limits:
[tex]\lim _{x\to\infty}y=2\text{ }[/tex]
[tex]\lim _{x\to-\infty}y=2\text{ }[/tex]
[tex]\lim _{x\to-2^+}y=\infty[/tex]
[tex]\lim _{x\to-2^-}y=\infty[/tex]
