F(x) = (x+2)/(x+3) where x is not equal to 0. Parentheses are not in the original problem, I included them to show what is over what.
I'm glad you put those parentheses there. Most students would
have written " x+2/x+3 " which really means , an
altogether different expression.
But you didn't state what you were to find. I assume it is
the domain and range.
The denominator x+3 must not equal 0, so
x+3 ≠ 0
x ≠ -3, so we have a vertical asymptote at x = -3
The degree of the numerator and denominator are both 1, so
the horizontal asymptote is
y =
y =
y = 1 , so we have a horizontal asymptote at y = 1
You also state "x is not equal to 0", so we must leave out the
point (0,) and put an open circle there. The two asymptotes
are in green:
So the domain is (-∞,-3)U(-3,0)U(0,∞) and
the range is (-∞,)U(,1)U(1,∞)
Edwin