Question

How do you write `y=-x^2+20x-80` in vertex form?

Answer 1

`y=-(x-10)^2+20" "` is the vetex form.

Explanation:

The vertex form of a quadratic equation is :
`" "`
`color(blue)(y=a(x-h)^2+k)" "` where `(h,k) " "`is the vertex
`" "`
The quadratic form is performed by factorization.
`" "`
`y=-x^2+20x-80`
`" "`
`rArry=-(x^2-20x+80)`
`" "`
`rArry=-((x)^2-2(x)(10)+80)`
`" "`
To complete the square of the above equation we recognize that the second term should be `color(red)(10^2=100` so we should add `20`
`" "`
`rArry=-((x)^2-2(x)(10)+80color(red)(+20-20))`
`" "`
`rArry=-((x)^2-2(x)(10)+100-20)`
`" "`
`rArry=-((x)^2-2(x)(10)+100)+20`
`" "`
`rArry=-((x)^2-2(x)(10)+10^2)+20`
`" "`
`rArry=-(x-10)^2+20`
`" "`
Therefore,`y=-(x-10)^2+20" "` is the vertex form of the parabola where `" "(10,20)` is its vertex.

Answer 2

`y = -(x-10)^2 +20`

Explanation:

To write a quadratic trinomial in vertex from you need to change:

`y = ax^2 +bx +c " "` into the form `" "y = p(x+q)^2 + t`

In `y = p(x+q)^2 + t` the vertex is at `(-q, t)`

`y = -x^2 +20x-80" "larr` make `x^2` positive

`y = -[x^2 -20x+80]" "larr` complete the square

`y = -[x^2 -20x color(blue)(+ 100-100)+80]" "larrcolor(blue)(+(b/2)^2 -(b/2)^2)`

`y =-[(x-10)^2 -20]`

`y = -(x-10)^2 +20`

This is now vertex form, giving the vertex as `(10,20)`