as you know, the hyperbola
x^2/a^2 - y^2/b^2 = 1
has asymptotes of
y = ± b/a x
You have a=2, b=1, so the asymptotes have slope ±1/2.
Also, your hyperbola is shifted, so its center is not at (0,0). Thus, your asymptotes are
y-2 = 1/2 (x+1)
y-2 = -1/2 (x+1)
or,
y = x/2 + 5/2
y = -x/2 + 3/2
google and wikipedia are your friends. A simple search will turn up many useful articles.
Also, wolframalpha is very useful for solving equations and plotting graphs. For this one, see
http://www.wolframalpha.com/input/?i=plot+%28x%2B1%29^2%2F4+-+%28y-2%29^2%2F1+%3D1+%2C+y+%3D+x%2F2+%2B+5%2F2%2C+y+%3D+-x%2F2+%2B+3%2F2
What are the asymptotes of the hyperbola (x+1)^2/4 - (y-2)^2/1 =1
I'm homeschooled, so the concept is hard to grasp.
Show steps please.
2 answers
Thank you very much, I understand it a little bit better now...