The general vertex form for a quadratic is
#color(white)("XXX")color(brown)y=color(green)m(color(orange)x-color(red)a)^2+color(blue)b#
with vertex at #(color(red)a,color(blue)b)#
and a "spread factor" of #color(green)m#
Given the vertex: #(color(red)2,color(blue)3)#
this becomes
#color(white)("XXX")color(brown)y=color(green)m(color(orange)x-color(red)2)^2+color(blue)3#
We are given that one solution point is #(color(orange)x,color(brown)y)=(color(orange)4,color(brown)11)#
So we have
#color(white)("XXX")color(brown)11=color(green)m(color(orange)4-color(red)2)^2+color(blue)3#
Simplifying:
#color(white)("XXX")8=color(green)m(2)^2#
#color(white)("XXX")8 =4color(green)m#
#color(white)("XXX")color(green)m=2#
We can substitute this back into our vertex equation, to get
#color(white)("XXX")color(brown)y=color(green)2(color(orange)x-color(red)2)^2+color(blue)3#
If your instructor prefers this in "standard form" we can expand the right side to get:
#color(white)("XXX")y=2x^2-8x+11#
For verification purposes, here is the graph of #y=2(x-2)^2+3#
