Question

What is the vertex form of `y=17x^2+88x+1`?

Answer 1

`y=17(x+44/17)-1919/17`

Explanation:

Given -

`y=17x^2+88x+1`

Vertex
x-coordinate of the vertex

`x=(-b)/(2a)=(-88)/(2xx 17)=(-88)/34=(-44)/17`

y-coordinate of the vertex

`y=17((-44)/17)^2+88((-44)/17)+1`
`y=17((1936)/289)-3872/17+1`
`y=32912/289-3872/17+1`
`y=(32912-65824+289)/289=(-32623)/289=(-1919)/17`

The vertex form of the equation is

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

`a=17` coefficient of `x^2`
`h=(-44)/17` x coordinate of the vertex
`k=(-1919)/17` y-coordinate of the vertex

`y=17(x+44/17)-1919/17`