Question

What is the the vertex of `x =-1/2(y-2)^2-4y+12`?

Answer 1

Vertex `->(x,y)=(12,-2)`

Explanation:

`color(blue)("General introduction")`

Instead of a quadratic in `x` this is a quadratic in `y`

If the `y^2` term is positive then the general shape is `sub`
If the `y^2` term is negative then the general shape is `sup`

If you expand the brackets we end up with `-1/2y^2` which is negative. So the general shape is `sup`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

I choose to opt for the 'completed square' form of equation

Expanding the brackets we have:

`x=-1/2(y^2-4y+4)-4y+12`

`x=-1/2y^2-2y+10`

`x=-1/2(y +2)^2+12" "......................Equation(1)`
.........................................................
Check
`x=-1/2y^2-2y-2+12" "->" "color(green)(x=-1/2y^2-2y+10)`

Original eqn: `x=-1/2(y-2)^2-4y+12`

`x=-1/2y^2+2y-2-4y+12`
`color(green)(x=-1/2y^2 -2y+10) color(red)(larr" Thay match")`
................................................................

From `Equation(1)`

`y_("vertex")=(-1)xx2=-2`

`x_("vertex")=+12`

Vertex `->(x,y)=(12,-2)`

Tony B