Finding vertical and horizontal asymptotes in curve sketching (1 Viewer)

[whoisurdaddy]

Need help in this area. Thanks in advance.

 

[lyounamu]

Need help in this area. Thanks in advance.

Post up a question that you don't know.

I will provide a step-by-step solution, if I can.

 

[vds700]

Need help in this area. Thanks in advance.

heres an example i dreamt up....

Graph y = (x + 1)/(x + 2)

x cannot = -2, as that would make the denominator 0. So x = -2 is a vertical asymptote.

To find the horizontal asymptote, we can rearrange the equation a little.

y = (x + 2 - 1)/(x + 2)
=(x + 2)/(x + 2) - 1(x + 2)
=1 - 1/(x + 2). As x approaches +/- infinity, 1/(x + 2) will appproach 0 and thu8s the graph will approach y = 1. So y = 1 is a horizontal asymptote.

Hope this helps

 

[bored of sc]

Need help in this area. Thanks in advance.

Basically write the denominator part only as an equation equal to zero. Solve for x or y and that is the equation of the asymptote.

 

[whoisurdaddy]

Very helpful. Thanks to everyone.

 

[noobonastick]

hi, i need help with oblique asymptotes =[. Say for x+ 1/x, the oblique asymptote is y = x. How does this work? thanks.

 

[tacogym27101990]

hi, i need help with oblique asymptotes =[. Say for x+ 1/x, the oblique asymptote is y = x. How does this work? thanks.

because in the case of y=x+1/x, as x>>infinity 1/x becomes negligible
so y>>x