Question

What is the vertex form of `6y = 18x^2+18x+42`?

Answer 1

Answered the wrong question: Typo must have double tap of the 2 key . One with shift and one without inserting a spurious 2: Error not spotted and carried through!!!
`color(blue)("vertex equation"->y=9/13(x+(color(red)(1))/2)^(color(green)(2))+ 337/156`

`color(brown)(y_("vertex")=337/156~=2.1603" to 4 decimal places")`

`color(brown)(x_("vertex") = (-1)xx1/2 = -1/2 = -0.5)`

Explanation:

Given:`" "26y=18x^2+18x+42`

Divide both sides by 26

`y=18/26 x^2+18/26x+42/18`

`y=9/13x^2+9/13 x+7/3`..................(1)

Write as:`" "y=9/13(x^(color(green)(2))+x)+7/3`.....(2)

`x ->color(red)( 1)xx x`

Change equation (2) to be

`y=9/13(x+(color(red)(1))/2)^(color(green)(2))+ 7/3 + k` ......(3)

The correction constant `k` is needed because we have changed the value of the whole RHS by changing the bracketed part as we did.

To find the value of k equate equation (1) to equation (3) through y

`9/13x^2+9/13 x+7/3 = y = 9/13(x+(color(red)(1))/2)^(color(green)(2))+ 7/3 + k`

`9/13x^2+9/13 x+7/3 = 9/13(x^2+x+1/4)+7/3+k`

`cancel(9/13x^2)+cancel(9/13 x)+cancel(7/3) = cancel(9/13x^2)+cancel(9/13x)+9/52+cancel(7/3)+k`

`k=-9/52`

So equation (3) becomes

`color(blue)("vertex equation"->y=9/13(x+(color(red)(1))/2)^(color(green)(2))+ 337/156`

`color(red)("As in the graph")`

`y_("vertex")=337/156~=2.1603` to 4 decimal places

`x_("vertex") = (-1)xx1/2 = -1/2 = -0.5`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 2

Correct answer this time. Other solution left in place as an extended example of method.

`color(blue)(" "y=3(x+1) +4)`

Explanation:

I have built this in the way I would do it for myself. The previous solution (incorrect question) shows the method in detail.

Given:`" " 6y=18x^2+18x+42`

Divide both sides by 6

`" "y=3x^2+3x+42/6`

`" "y=3(x+1)^2+k+42/6`

`" "k= -3" and " 42/6=7`

`color(blue)(" "y=3(x+1) +4)`