transformation of graphs / slopes of asymptotes

[kathy's]

1) Describe the transformation neccessary to obtain the graph of y^2/16 - x^2/4 = 1 from the graph of y^2 - x^2 = 1

My answer: Because y^2/16 - x^2/4 = 1 is the same as (y/4)^2 - (x/2)^2 = 1. the graph of y^2 - x^2 = 1 is strecthed horizontally by a factor of 2 and vertically by a factor of 4.

2) What are the slopes of the asymptotes of y^2/16 - x^2/4 = 1?

My answer: The slope of the asymptotes of y^2 - x^2 = 1, using "rise over run" are + ( plus or minus) 1/1. Therefore the slopes of the new asymptotes are m = + ( plus or minus) 1X vertical strecth/ 1X horizontail strecth

m = + (plus or minus) 4/2
m = + (plus or minus) 2

are theses right? thanks!

 

[tkhunny]

I think you have answered correctly, but I am just a little disappointed that the word "hyperbola" didn't appear in your answers. I realize it wasn't asked for, as presented. It just seems like it would be more complete and thorough to provide that identifiaction explicitly.

Good work.

 

[kathy's]

Thank you

Thanks!

P.S
when I re-wrote my answer I put the word "hyperbola" into my answer :wink: !