Check my precalc
posted by Rebekah .
Find the verticle, horizontal, and oblique asymtopes if any.
#1 (x^2 +6x+5)/2x^2 +7x+5
y= (1/2) and x= 5/2 and 1
#2 (8x^2 +26x7)/4x1
y=2x+7 and x=1/4
#3. (x^416)/(x^22x)
x= 0 and 2
y= x^2 +2x+4
thanks!

#1, I think you meant:
(x^2 + 6x+5)/(2x^2 + 7x + 5)
then
y = (x+1)(x+5)/((x+1)(2x+5))
= (x+5)/(2x+5)
horizontal: y = 1/2 , you had that
vertical x = 5/2
a "hole" at x = 1 , not a vertical asymptote
#2, again, you need brackets for the denominator, the way it stands , only the 4x is divided.
the rest is correct.
#3. The asymptote is actually the parabola
y = x^2 + 2x + 4 , so it is a "curved asymptote"