HELP ASAP, PLEASE RIGHT NOW. CONIC SECTIONS, GOING TO SCHOOL IN A FEW MINUTES.

1. For 9y^2-4x^2-18y+24x-63=0 (y-1)^2/4-(x-3)^2/9=1 What is the domain and range?
2. For y^2=8x, the vertex is (0,0, and focus is (2,0) and the directrix is x=-2 right? PLEASE ANSWER, I WOULD HAVE ASKED LAST NIGHT BUT OPEN STUDY WAS CLOSED! THANK YOU!

Question

HELP ASAP, PLEASE RIGHT NOW. CONIC SECTIONS, GOING TO SCHOOL IN A FEW MINUTES.

1. For 9y^2-4x^2-18y+24x-63=0 (y-1)^2/4-(x-3)^2/9=1 What is the domain and range?
2. For y^2=8x, the vertex is (0,0, and focus is (2,0) and the directrix is x=-2 right? PLEASE ANSWER, I WOULD HAVE ASKED LAST NIGHT BUT OPEN STUDY WAS CLOSED! THANK YOU!

e.mccormick · Answer

2) That is correct.

e.mccormick · Answer

1) That is a hyperbola, which is not a function, so I am not sure how you would get a domain and range...  However, I can tell you it has no y values between 3 and -3.  There it does not exist.  The x values from $\pm \infty$ are valid, but they produce multiple y values....

e.mccormick · Answer

Oops.. wait a sec, 3 and -1 on the y.  Still not awake yet.

anonymous · Answer

Thank you so much for helping. I'm not sure how to find the domain and range either but I think it has something to do with the x-values or vertices.

e.mccormick · Answer

It has vertices at (3,3) and (3,-1) with nothing between them being valid.  So probaly they mean a domain of $(-\infty,\infty$) and a range of $(-\infty,-1)(3,\infty)$

anonymous · Answer

Thank you so much! You have no idea how much you've saved me! :D

e.mccormick · Answer

Actually, inclusive.  so -1]U[3

e.mccormick · Answer

Because the -1 and 3 are on the curve.

e.mccormick · Answer

$(-\infty,-1]\cup[3,\infty)$  Yah, that would be the proper notation.