rational functions--->R(x) = 1 / (x-1) ^2

[uhmmlin]

equation---> R(x) = 1 / (x-1) ^2

how do i find the...

-horizontal asymptote--->would it be "1" since "x-1=0" is "1"?
-domain---->would it be (-infinity, 1)hole(1, infinity) assuming the horiz. asymp.= 1?
-veritcal asymptote----->i know the rules for finding the asymptote but how do i find the degree for the numerator?

thanks for your help.

 

[tkhunny]

equation---> R(x) = 1 / (x-1) ^2

-horizontal asymptote--->would it be "1" since "x-1=0" is "1"?

You're a bit confused on this one. Your calculation suggests you are looking for a VERTICAL asymptote.

-domain---->would it be (-infinity, 1)hole(1, infinity) assuming the horiz. asymp.= 1?

You are correct except for your use of the word "horizontal". You mean "vertical".

-veritcal asymptote----->i know the rules for finding the asymptote but how do i find the degree for the numerator?

Still confused. The degree of numerator and denominator go to HORIZONTAL asymptotes. Think HORIZON - you look off into the distance and see the line from side to side across your gaze. Vertical is up and down.

x^2 -- Degree 2
x^1 = x -- Degree 1
x^0 = 1 (for x not zero)) -- Degree 0

Your R(x) has degree of denominator = 2 and degree of numerator = 0. This makes an HORIZONTAL asymptote at R(x) = 0, the x-axis.

One cannot do the math if one doesn't know which way is up.

 

[galactus]

If the degree of the numerator is less than the degree of the denominator, then the x-axis is the horizontal asymptote.

Your numerator has power 0: \(\displaystyle \frac{1x^{0}}{(x-1)^{2}}\)


The domain is all except that which makes the denominator 0.

As for the vertical asymptote, the line x=a is the asymptote if the graph of the function approaches infinity or neg. infinity as x approaches a from the right or the left.

Where is your vertical asymptote?. \(\displaystyle \L\\\lim_{x\rightarrow{a}}\frac{1}{(x-1)^{2}}={\infty}\). What is 'a'?.