Question 745660
I thought I just answered this one.  Apparently my post stalled?

Standard form {{{y=a(x-h)^2+k}}} has center (h,k).  Yours has center (3,-5), so you get {{{y=a(x-3)^2-5}}}.  To find 'a', use the other given point (5,-4).
{{{y=a(x-3)^2-5}}} will become {{{-4=a(5-3)^2-5}}}, and just solve for a.


Since you have the value now for 'a', you convert {{{y=a(x-3)^2-5}}}, but with the VALUE for a, and do the multiplications and simplification to get general form.  Once done, just read the values of a, b, and c, and do what you need with them.