Question

For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote.

f(x) = 1/x -0 2

Answer 1

The function provided is given below,

f(x) = 1/(x – 2)

 

To find the vertical asymptote follow as given below,

It is seen that the denominator of the given fraction, that is, x - 2 will be zero at x = 2.

 

So there is a vertical asymptote at x = 2.

 

Table showing the values of f(x) when the input values are smaller than 2 and become negative is given below,

x	1.5	1	0	-1	-2
f(x) = 1/(x – 2)	-2	-1	-0.5	-0.33	-0.25

 

It can be seen from the table that as x → 2-, f(x) → -∞.

 

Table showing the values of f(x) when the input values are greater than 2 and become larger is given below,

x	2.5	3	3.5	4	5
f(x) = 1/(x – 2)	2	1	0.67	0.5	0.33

 

It can be seen from the table that as x → 2+, f(x) →-∞.

 

To find the horizontal asymptote follow as given below,

Relate the given function f(x) = 1/(x – 2) with f(x) = p(x)/q(x), q(x) ≠ 0.

So it is seen that degree of p = 0 and degree of q = 1.

 

Since p < q, there is a horizontal asymptote found at y = 0.

So, as x → ∞ or x → -∞ then f(x) → 0.

So, as x → ∞ or x → -∞ then f(x) → 0.