# Check my precalc

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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 +26x-7)/4x-1

y=2x+7 and x=1/4

#3. (x^4-16)/(x^2-2x)

x= 0 and 2
y= x^2 +2x+4
thanks!

• Check my precalc -

#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"

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