Question

Which one of these functions does not have a vertical asymptote? Please explain why!!!!!!!!
y=5x/x+x^2
y=x/1-x^2
y=5x/1-2x^2
y=5x-1/3+x^2

Answer 1

for vertical asymptotes set the denominator = 0

Answer 2

So then it would be
x+x^2=0
1-x^2=0
1-2x^2=0
x^2+3=0
Now I am having a brain fart and have no idea how to solve any of these.....

Answer 3

\(x+x^2=x(x+1)=0\\x=0,-1\\...\\1-x^2=0\\1=x^2\\\sqrt1=x\\x=1
\\...\\1-2x^2=0\\1=2x^2\\x^2=\dfrac{1}{2}\)

Answer 4

we're trying to find the denominator with no real roots

Answer 5

Oh yeah wow! So wait this is going to sound dumb, but is a fraction a real number?

Answer 6

that's not dumb :)
assuming the fraction is rational, yeah http://www.regentsprep.org/regents/math/algebra/AOP1/Lrat.htm

Answer 7

the square root of a negative number isn't real btw

Answer 8

yeah thats an imaginary/complex number, i know that

Answer 9

So since after multiplying I get
x=sqrt-x
x=sqrt of 1
x= sqrt of 1/2
and x=sqrt of -3
Would the answer be the last one since it is an irrational number?

Answer 10

since the square root of -3 isnt a real number??

Answer 11

yeah :D
http://www.wolframalpha.com/input/?i=y%3D%285x-1%29%2F%283%2Bx%5E2%29 this says the domain is all \(\mathbb{R}\) too

Answer 12

sweet thanks