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.
Explanation:
The vertical asymtote is found at values of
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.
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]}