Answer with explanation:
The given curve has two vertical Asymptote,
First, x=2
And,Second , x= -2
So,the equation of the curve can be written as
1.
[tex]f(x)=\frac{k}{(x-2)(x+2)}\\\\ f(x)=\frac{k}{x^2-4}\\\\ Now,\lim_{x \to \infty} \frac{k}{x^2-4}= \frac{k}{(\infty)^2-4}\\\\\lim_{x \to \infty} f(x)=\frac{k}{\infty}=0[/tex]
,2.
[tex]\lim_{x \to -\infty} f(x)= \lim_{x \to -\infty}\frac{k}{x^2-4}\\\\=\frac{k}{(-\infty)^2-4} \\\\ =\frac{k}{\infty} \\\\=0[/tex]
3.
[tex]\lim_{x \to 0} f(x)= \lim_{x \to 0}\frac{k}{x^2-4}\\\\=\frac{k}{(0)^2-4} \\\\ =\frac{k}{-4}[/tex]
Option C: As the x-values go to negative infinity, the function's values are equal to zero