Question

What is the vertex form of `y=1/2x^2-1/6x+6/13`?

Answer 1

`y=1/2(x-1/6)^2+409/936` (assuming I managed the arithmetic correctly)

Explanation:

The general vertex form is
`color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)`
for a parabola with vertex at `(color(red)(a),color(blue)(b))`

Given:
`color(white)("XXX")y=1/2x^2-1/6x+6/13`

`rArr`
`color(white)("XXX")y=1/2(x^2-1/3x)+6/13`

`color(white)("XXX")y=1/2(x^2-1/3x+(1/6)^2)+6/13-1/2*(1/6)^2`

`color(white)("XXX")y=1/2(x-1/6)^2+6/13-1/72`

`color(white)("XXX")y=1/2(x-1/6)^2+(6*72-1*13)/(13*72)`

`color(white)("XXX")y=color(green)(1/2)(x-color(red)(1/6))^2+color(blue)(409/936)`
which is the vertex form with vertex at `(color(red)(1/6),color(blue)(409/936))`

The graph below of the original equation indicates that our answer is at least approximately correct.

graph{1/2x^2-1/6x+6/13 [-0.6244, 1.0606, -0.097, 0.7454]}