Question

What are the vertices of the hyperbola given by the equation (y-4)^2/121 - (x+9)^2/141 = 1?

Answer 1

first you need the center
so you know that one ?

Answer 2

No, I really don't understand this section which is why I was asking for help.

Answer 3

center of
\[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\]is \((h,k)\)

Answer 4

in your case it is \((-9,4)\)

Answer 5

because the \(y\) part comes first (before the minus sign) this hyperbola looks like this (roughly)

Answer 6

to find the vertices, go \(a\) units up up and down from the center. that is why you needed the center first
in your case \(a^2=121\) and so \(a=11\)

Answer 7

since the center is \((-9,4)\) if you go up 11 units you are at \((-9,15)\) and if you go down 11 units you are at \((-9,-7)\)

Answer 8

hope that is clear, and also that you see it is not that hard once you know what you are doing

Answer 9

I'm still extremely lost, as you can tell math isn't my strong point. I appreciate the help however.