Question

What is the vertex form of `y=4x^2+x-6`?

Answer 1

`y= 4(x-(-1/8))^2 + (-97/16)`

Explanation:

To find the vertex form of a quadratic equation we use a process called completing the square.

Our goal is the form `y = a(x-h)^2 + k` where `(h, k)` is the vertex. Proceeding, we have

`4x^2 + x - 6 = 4(x^2+1/4x)-6`

`= 4(x^2+1/4x+1/64-1/64)-6`

`= 4(x^2+1/4x + 1/64)-4/64-6`

`= 4(x+1/8)^2 - 97/16`

`= 4(x-(-1/8))^2 + (-97/16)`

Thus the vertex form is
`y= 4(x-(-1/8))^2 + (-97/16)`

and the vertex is at `(-1/8, -97/16)`