Question

What is the the vertex of `y = (x+6)(x+4) -x+12`?

Answer 1

`y_{min} = 63/4` at `x = - 9/2`

Explanation:

`y = (x+6)(x+4) -x+12`
`y = x^2 + 10x + 24 -x+12`
`y = x^2 + 9x + 36`
`y = (x + 9/2)^2 - 81/4 + 36`
`y = (x + 9/2)^2 + 63/4`

`y_{min} = 63/4` at `x = - 9/2`

Answer 2

The vertex is `(-9/2;63/4)`

Explanation:

let's rewrite the equation in the equivalent form:

`y=x^2+4x+6x+24-x+12`

`y=x^2+9x+36`

Then let's find the vertex coordinates by the following:

`x_V=-b/(2a)`

where a=1; b=9

so

`x_V=-9/2`

and

`y_V=f(-9/2)`

that's

`y=(-9/2)^2+9(-9/2)+36`

`y=81/4-81/2+36`

`y=(81-162+144)/4`

`y=63/4`