Question

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

Answer 1

`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