How do you identify all vertical asymptotes for #f(x)=(3x^2+x-5)/(x^2+1)#?

2 Answers
Jan 10, 2017

Your function shouldn't have any vertical asymptote.


The vertical asymtote is found at values of #x# that makes the denominator equal to zero (and so creating a discontinuity) but the denominator of your function will never be zero regardless of the real value of #x# you may choose.

You can also see this graphically:
graph{(3x^2+x-5)/(x^2+1) [-9.12, 10.88, -13.76, -3.76]}

Jan 10, 2017

None. See the graph and explanation.


By actual division,

f = 3+(x-8)/(x^2+1)

y = quotient = 3 and

the factors of the denominator (x+i)(x-i) of the remainder = 0 give

the asymptotes.

The graph is asymptote-inclusive.

So, the only real asymptote is the horizontal asymptote y = 3.
graph{(y(x^2+1)-3x^2-x+5)(y-3)=0 [-20, 20, -10, 10]}