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

2 Answers
Jan 1, 2017

Answer:

The vertical asymptote is #x=0#
No slant asymptote
The horizontal asymptote is #y=0#

Explanation:

The domain of #g(x)# is #D_g(x)=RR-{0}#

As you cannot divide by #0#, #x!=0#

The vertical asymptote is #x=0#

The degree of the numerator is #<# than the degree of the numerator, there is no slant asymptote.

Now, we compute the limits of #g(x)# as #x->+-oo#

#lim_(x->-oo)g(x)=lim_(x->-oo)x/(2x^2)=lim_(x->-oo)1/(2x)=0^(-)#

#lim_(x->+oo)g(x)=lim_(x->+oo)x/(2x^2)=lim_(x->+oo)1/(2x)=0^(+)#

The horizontal asymptote is #y=0#

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

Jan 1, 2017

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]}