Question

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

Answer 1

vertex`=(6,-5)`

Explanation:

Start by expanding the brackets, then simplifying the terms:

`y=2(x-4)^2-x^2+4x-1`
`y=2(x-4)(x-4)-x^2+4x-1`
`y=2(x^2-8x+16)-x^2+4x-1`
`y=2x^2-16+32-x^2+4x-1`
`y=x^2-12x+31`

Take the simplified equation and rewrite it in vertex form:

`y=x^2-12x+31`
`y=(x^2-12x)+31`
`y=(x^2-12x+(12/2)^2-(12/2)^2)+31`
`y=(x^2-12x+(6)^2-(6)^2)+31`
`y=(x^2-12x+36-36)+31`
`y=(x^2-12x+36)+31-(36*1)`
`y=(x-6)^2+31-36`
`y=(x-6)^2-5`

Recall that the general equation of a quadratic equation written in vertex form is:

`y=a(x-h)^2+k`

where:
`h=`x-coordinate of the vertex
`k=`y-coordinate of the vertex

So in this case, the vertex is `(6,-5)`.