What is the vertex form of #2y=5x^2-3x+11 #?

...I can never remember it, so I always have to look it up.

The vertex form of a quadratic equation is:

#f(x) = a(x - h)^2 + k#

So, for your original equation #2y = 5x^2 - 3x + 11#, you have to do some algebraic manipulation.

Firstly, you need the #x^2# term to have a multiple of 1, not 5.
So divide both sides by 5:

#2/5y = x^2 - 3/5x + 11/5#

...now you have to perform the infamous "complete the square" maneuver. Here's how I go about it:

Say that your #-3/5# coefficient is #2a#. Then #a = -3/5 * 1/2 = -3/10#

And #a^2# would be #9/100#.

So, if we add and subract this from the quadratic equation, we'd have:

#2/5y = x^2 - 3/5x + 9/100 - 9/100 + 11/5#

...and now the 1st 3 terms of the right side are a perfect square in form #(x - a)^2 = x^2 - 2ax + a^2#

...so you can write:

#2/5y = (x - 3/10)^2 + (11/5 - 9/100)#

#2/5y = (x - 3/10)^2 + (220 - 9)/100#

#2/5y = (x - 3/10)^2 + 211/100#

So now, all you gotta do is multiply through by #5/2#, giving:

#y = 5/2(x-3/10)^2 + 5/2*211/100#

#y = 5/2(x-3/10)^2 + 211/40#

which is vertex form, #y = a(x-h)^2 + k#

where #a = 5/2#, #h = 3/10#, and #k = 211/40#