Question

What is the vertex form of `y=1/3x^2+5/6x+7/8`?

Answer 1

`y=1/3(x+5/4)^2-11/16`
Have a look at the explanation to see how it is done!

Explanation:

Given:`color(white)(....)y=1/3x^2+5/6x+7/8`

Consider the part inside the brackets:`color(white)(....)y=(1/3x^2+5/6x)+7/8`

Write as: `1/3(x^2 + {5/6 -: 1/3}x)`

`1/3(color(red)(x^2)+color(blue)(5/2color(green)(x)))`

If we halve `5/2` we get `5/4`

Change the bracketed bit so that have

`1/3(color(red)(x)+color(blue)(5/4))^2`

We have changed `color(red)(x^2)` to just `color(red)(x)`; halved the coefficient of `color(green)(x)-> color(blue)(1/2 xx 5/2=5/4)` and totally removed the single `color(green)(x)`

So we know write the equation as:

`y-> 1/3(x+5/4)^2+7/8`

The thing is; we have introduced an error that results from squaring the bracket. The error is when we square the `(+5/4)` bit. This error means that the right no longer ='s the left. That is why I have used `y->`

`color(blue)("To correct for this we write:")`

`y-> 1/3(x+5/4)^2color(blue)(-(5/4)^2)+7/8`

The correction now means that the `color(red)("left does = right.")`

`ycolor(red)(=) 1/3(x+5/4)^2color(blue)(-(5/4)^2)+7/8`

So the arithmetic now gives:

`y=1/3(x+5/4)^2-11/16`