We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
43
3
avatar

 

So I know that the verticle asymptote is at x = -2, the horizontal asymptote at y = 1, and the point discontinuity is at (0,3).

 

So far, I have \({x - 3 \over (x + 2)(x - 3)}\). I'm not sure about what to do next?

 

Thank you! :)

 May 22, 2019
 #1
avatar+101086 
+3

If the horizontal asymptote = 1  then  the degree of the numerator = the degree of the denominator and the coefficients on the lead terms will be the same....

 

If we have a vertical asymptote at x = -2  then  ( x + 2)  must be a linear factor in the denominator

 

Note that   x = - 6 is a root....therefore  (x + 6)  must be a factor in the numerator

 

So.....the polynomial  that we are looking for is

 

x ( x + 6)

_______

x (x + 2)

 

 

When the x"s   are cancelled,  we end up with  (x + 6) /(x + 2) ....and when x = 0, y = 3

But x cannot = 0  because the original function is undefined at that point....so....the "hole" occurs at (0, 3)

 

Here's the graph  with the asymptotes and pertinent points : https://www.desmos.com/calculator/wge4wepw0i

 

 

cool cool cool

 May 22, 2019
edited by CPhill  May 22, 2019
 #2
avatar
+1

Just wanted to clarify, will the horizontal asymptote always equal the degree of the numerator and denominator? :)

Guest May 22, 2019
 #3
avatar+101086 
+2

Sorry.....I meant to say that if the horizontal asymptote  is y = 1.....the degree of the polynomials in the numerator and denominator is the same and their lead coefficients are also the same....

 

cool cool cool

CPhill  May 22, 2019

15 Online Users

avatar
avatar