SOLUTION: Hi, I have to find the vertical and horizontal asymptotes, as well as the y-intercept for the following: {{{y=(1 + 2x)/(x+2)}}} If I'm not mistaken, vertical asymptote = {

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Question 369255: Hi, I have to find the vertical and horizontal asymptotes, as well as the y-intercept for the following:

If I'm not mistaken,
vertical asymptote =
horizontal asymptotes are and
I calculated the y-intercept by plugging 0 in for x and got 1/2, but when I graph the equation it doesn't look like 1/2. I appreciate any help you can provide :)

Found 3 solutions by stanbon, lefty4ever26, mananth:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
find the vertical and horizontal asymptotes, as well as the y-intercept for the following:
y=(1 + 2x)/(x+2)
If I'm not mistaken,
vertical asymptote =
horizontal asymptote: y = 2x/x = 2
--------------------------------------------
I calculated the y-intercept by plugging 0 in for x and got 1/2, but when I graph the equation it doesn't look like 1/2.
---
x-int: Solve [(1+2x)/(x+2)] = 0
True when 1+2x = 0
2x = -1
x = -1/2
=========================

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Cheers,
Stan H.
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Answer by lefty4ever26(59)   (Show Source):
You can put this solution on YOUR website!
Vertical asymptote is when the denominator equals 0. so x+2=0
x=-2
Horizontal asymptote in this problem would be the leading coefficients of the x values or so the Horizontal asymptote is at y=2.
You are right about the y-intercept at 1/2. The x-intercept is where y=0 so set the equation equal to 0 and solve for x.

The graph below shows the intercepts well:

The graph below shows the asymptotes well:

Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
(1 + 2x)/(x+2)
x+2 = 0
x=-2
plug y =0
1+2x=0
x=-1/2
..