What is the vertex form of #y = -3x^2+9x+1#?

1 Answer
Dec 6, 2015

#y=-3(x-3/2)^2+31/4#

Explanation:

Given:#color(white)(..)y=-3x^2+9x+1...........(1)#

Write as:#color(white)(..)y=-3(x^2color(green)(-3x))+1#

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Consider the RHS only

Write as: #-3(x-3/2)^2+1.............................(2)#

The #(-3/2)# comes from halving the coefficient of #x " in "color(green)( -3x)#

Expression (2) has an inherent error which we need to correct

#-3(x-3/2)^2#
#=-3( x^2 -3x+9/4)#
#= -3x^2+9x-27/4...................(3)#

Add the constant of +1 as shown in equation (1) giving

#= -3x^2+9x-27/4 + 1...................(3_a)#

When you compare #(3_a)# to (1) you see that the error introduced is #-27/4#

We correct for this by removing it from the vertex form equations using #color(blue)(+27/4)#

Thus the #underline(color(red)("incorrect"))# form of #y=-3(x-3/2)^2+1 color(blue)(" is adjusted by:")#

#y=-3(x-3/2)^2+1color(blue)(+27/4)#

Giving:

#y=-3(x-3/2)^2color(brown)(+31/4)#