What is the vertex form of #y=(x + 21)(x + 1) #?

1 Answer
Mar 23, 2016

#color(blue)("Vertex "-> (x,y)->(-11,-100)#

Explanation:

For a more detailed explanation of method see the example of
http://socratic.org/s/asZq2L8h .Different values but the method is sound.

Given:#" "y=(x+21)(x+1)#

Let #k# be the error correcting constant

Multiply out giving

#" "y=x^2+22x+21#

#y=(x^(color(magenta)(2))+22x)+21+k" "color(brown)("No error yet so k=0 at this stage")#

Move the power to outside the bracket

#y=(x+22color(green)( x))^(color(magenta)(2)) +21+k" "color(brown)("Now we have the error "->k!=0 )#

Remove the #x# from #22color(green)(x)#

#" "y=(x+color(red)(22))^2+21+k#

Multiply #color(red)(22)" by "(1/2) =color(blue)(11 )#

#" "color(green)(y=(x+color(red)(22))^2+21+k)#

#"changes to "color(green)(y=(x+color(blue)(11))^2+21+k)#

The error introduced is #(axxb/2)^2 ->(1xx22/2)^2 =+ 121#

So #k = -121# to 'get rid' of the error
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So
#" "color (magenta)( y=(x+color(blue)(11))^2+21)-121#

#" "y=(x+11)^2-100#

#" "color(blue)("Vertex "-> (x,y)->(-11,-100)#

Tony B