Graph the ellipse.

(x+3)² + 4(y-4)²=16    

First....let's get this in stamdard form.....divide everything by 16....and we have

(x  +3)^2  /   16    +  (y - 4)^2 / 4    = 1

This is an ellipse centered at (-3, 4).....the major axis lies parallel to the x axis  and is 8 units long....the minor axis is parallel to the y axis and is 4 units long

Here's the graph :  https://www.desmos.com/calculator/iecr3eenhy

I appreciate that. I am trying to find a step by step. My old and addled brain is not seeing how this turns into (-3,4). It's been too long since I've had to take this kind of math. While Desmos is good at showing the end result, it sucks for show how it got there.

Remember, vickeryw......  if we have the standard form of the ellipse

( x - h)^2 / a^2   + ( y - k)^2 / b^2  = 1 

Then the center lies at ( h ,k )   ....look at your problem  in standard form 

(x  + 3)^2  /   16    +  (y - 4)^2 / 4    = 1       and we can write this as

( x - (-3) )^2  / 16   +  ( y - 4)^2 / 4    = 1

So......  h = -3 and k = 4     and the center  is at  (-3, 4)

Does that help ???

I can see how 4(y-4)²/16 can become (4) now, but how does (x+3)²/16 become -3?

Your original problem  can be wriiten as   [ x - (-3)]^2  + 4 [y - 4] ^2  = 16

In this form....we actually have  the form   ( x - h)^2  + 4 ( y - k)^2   = 16  

Note that we can write   (x + 3)^2   as  ( x - (-3) )^2   .....so    [x - h] ^2  means that h   = -3

And the center is at   ( h ,k)   =  (-3, 4)

Don't get confused ...we're dividing both sides by 16   to get  the right side to equal 1........this has nothing to do with finding the center !!!!

That actually does help CPhill. I'm sorry I didn't respond sooner. I had to come to work and am just now getting a chance to see this. I really do appreciate the help.