What is the vertex form of #y=(6x-2)(2x+11) #?

1 Answer
Mar 27, 2018

#y=6(x+31/12)^2-1225/24#

Explanation:

#y=(3x-1)(2x+11)#

Multiply the brackets

#y=6x^2+33x-2x-11#

#y=6x^2+31x-11 larr" Starting point"#
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#color(blue)("Discussing what is happening")#

Note that for standardised form #y=ax^2+bx+c# we intend to make this #y=a(x+b/(2a))^2+k+c color(white)(.) larr" completed square format"#

If you multiply out the whole thing we get:

#y=ax^2+b x color(red)(+ a( b/(2a))^2)+k+c#

The #color(red)(+ a( b/(2a))^2)+k# is not in the original equation.

To 'force' this back to the original equation we

set #color(red)(+ a( b/(2a))^2)+k=0#
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#color(blue)("Returning to the solution")#

#y=6x^2+31x-11 color(white)("d")->color(white)("d")y=6(x+31/(6xx2))^2 +k-11#

However:
#color(red)(+ a( b/(2a))^2)+k=0 color(white)("d")->color(white)("dddd") color(red)(6(31/(2xx6))^2)+k=0#

#color(white)("dddddddddddddddd")->color(white)("dddd")31^2/(4xx6)+k=0#

#color(white)("dddddddddddddddd")->color(white)("dddd") k=-961/24#

So we now have:

#y=6x^2+31x-11 color(white)("d")->color(white)("ddd")y=6(x+31/(6xx2))^2 -1225/24#

#color(white)("dddddddddddddddd")->color(white)("dddd") y=6(x+31/12)^2-1225/24#

Tony B