I am having trouble figuring out this problem, I already have theanswer, but it

Question

I am having trouble figuring out this problem, I already have theanswer, but it does me little good not knowing how to get it.I need to solve this hyperbola equation and determine the foci,vertices, and the direction it opens.

x^2-16y^2+6x+32y-23=0

according to my study sheet the center is at (-3,1), verticesat (-7,1)&(1,1), foci at(-3-17,1)&(-3-17,1).

this is what I am coming up with
(x^2+3)/16 - (y^2+1), and when I plug it in I get an entirelydifferent answer. Could someone please show me where I am messingthis up?

Explanation / Answer

x^2-16y^2+6x+32y-23=0 well first you have to group them & solve it: (x^2 +6x __) - (16y^2 + 32y __ ) = 23 {to figure out whatgoes in the blank, divide the middle term by 2, then square it} (x^2 +6x +9) = (x+3)^2 Make sure you add 9 to the other side to, so23+9= 32 for the y part, simplify it. so far, youhave:(x+3)^2 - 16(y^2 - 2y +1) = 32-16 (the -16comes from the -16*1) (x+3)^2 - 16(y^2 - 2y +1) = 16 Factor our the y part...Divide both sides by 16, so the finalshould be: [(x+3)^2 over 16] - [(y-1)^2] = 1 then you graph it, and since x is first in the equation, your graphwould look like that: ) (

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