# How do you find the vertical, horizontal and slant asymptotes of: # g(x)= (x+2 )/( 2x^2)#?

##### 2 Answers

#### Answer:

The vertical asymptote is

No slant asymptote

The horizontal asymptote is

#### Explanation:

The domain of

As you cannot divide by

The vertical asymptote is

The degree of the numerator is

Now, we compute the limits of

The horizontal asymptote is

graph{(x+2)/(2x^2) [-7.9, 7.9, -3.95, 3.95]}

#### Answer:

vertical asymptote at x = 0

horizontal asymptote at y = 0

#### Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to x and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve :

#2x^2=0rArrx^2=0rArrx=0#

#rArrx=0" is the asymptote"# Horizontal asymptotes occur as

#lim_(xto+-oo),g(x)toc" ( a constant)"# divide terms on numerator/denominator by the highest power of x, that is

#x^2#

#g(x)=(x/x^2+2/x^2)/((2x^2)/x^2)=(1/x+2/x^2)/2# as

#xto+-oo,g(x)to(0+0)/2#

#rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.

graph{(x+2)/(2x^2) [-10, 10, -5, 5]}