What is the vertex form of #y=3x^2 - 4x +17?

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Answer 1 · 2016-05-12T07:13:25

The Vertex Form

#color(red)((x-2/3)^2=1/3(y-47/3))#

Start from the given

#y=3x^2-4x+17#

from the numerical coefficients
#a=3# and #b=-4# and #c=17#

The vertex can be computed

#h=(-b)/(2a)#
#h=(-(-4))/(2*3)#

#h=2/3#

for the #k#
#k=c-b^2/(4a)#

#k=17-(-4)^2/(4*3)#

#k=47/3#

#p=1/(4a)=1/(4*3)=1/12#

The vertex form

#(x-h)^2=+4p(y-k)#

#(x-3/2)^2=4(1/12)(y-47/3)#

#(x-3/2)^2=1/3(y-47/3)#

#y=3x^2-4x+17#
#y=3(x^2-4/3x)+17#

#y=3(x^2-4/3x+4/9-4/9)+17#

#y=3((x^2-4/3x+4/9)-4/9)+17#

#y=3((x-2/3)^2-4/9)+17#

#y=3(x-2/3)^2-4/3+17#

#y=3(x-2/3)^2+47/3#

#y-47/3=3(x-2/3)^2#

#1/3(y-47/3)=(x-2/3)^2#

The Vertex Form

#color(red)((x-2/3)^2=1/3(y-47/3))#

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