Question

State the domain and range of y=2x+8/(3x^2-9)?
I know the domain is all real numbers except the positive and negative square roots of 3, but I don't know how to find the exact range?
I'm allowed to graph, but the range should be put into the thousandths place.

Answer 1

when x -> -infinity
then y is about 2*x (the fractional part gets very small)
so from x = near -sqr(3_ to -infinity the range is all negative numbers.
for x approaching -sqr(3) (say -sqr(3) - dx) where dx is tiny
we get
y = - 6 + 8 / 3(3+dx^2 -2*sqr(3)dx) - 9
the bottom is 9 + 3 dx^2 - 2*sqr(3) dx - 9 = 3 dx^2 - 2*sqr(3) dx =
(3*dx - 2 sqr(3)) * dx
as x gets very close to -sqr(3) dx becomes very tiny (think very near zero)
the 8/tiny number becomes very big. In other words, y -> + infinity

so the range is all real numbers