Question

y = x^2 / (x^2 +4)
Is the vertical asymptote = 0
horizontal asymtote = 1
domain = XER

and one question asks for "intervals where y is above x-axis" , all intervals are above ..how do i say it?

Answer 1

There is no verticle asymptote, the denominator never equals 0 on the real number system.

Answer 2

So the domain is all real.

Answer 3

XER represents all real right?

Answer 4

so there cant be V.A of 0?

Answer 5

y = x at x = 0, so your interval for y>x-axis should be\[x \epsilon (- \infty, 0) U (0, \infty)\].

Answer 6

Yes, \[x \epsilon ℝ\] represents x is an element of all real numbers.

Answer 7

ohh ok thanks!

Answer 8

yw :)