What is the vertex form of #y=x^2 - 19x +14 #?

1 Answer

The vertex form is #(x-k)^2=4p(y-k)#
#(x-19/2)^2=y--305/4#
with vertex at #(h, k)=(19/2, (-305)/4)#

Explanation:

Start from the given equation #y=x^2-19x+14#

Divide 19 by 2 then square the result to obtain #361/4#. Add and subtract #361/4# to the right side of the equation right after #-19x#

#y=x^2-19x+14#

#y=x^2-19x+361/4-361/4+14#

the first three terms form a PERFECT SQUARE TRINOMIAL

#y=(x^2-19x+361/4)-361/4+14#

#y=(x-19/2)^2-361/4+14#

#y=(x-19/2)^2-361/4+56/4#

#y=(x-19/2)^2-305/4#

#y--305/4=(x-19/2)^2#

#(x-19/2)^2=y--305/4#

God bless....I hope the explanation is useful.