Question

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

Answer 1

The vertex is at `(1/8,63/16)`

Explanation:

Your quadratic equation is of the form

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

The vertex is at the point `(h,k)`

Rearrange your equation to obtain a form similar to that of the quadratic equation.

`y=4x^2-x+4`

`y=4x^2-x+ color(red)(4/64) - color(red)( 4/64)+4`

`y=(4x^2-x+ color(red)(4/64) )- color(red)( 4/64)+4`

Take `color(red)4` as a common factor.

`y=4(x^2-1/4x+ color(red)(1/64) )- color(red)( 4/64)+4`

`y=4(x - 1/8 )^2 + (4xx64-4)/64`

`y=4(x - 1/8 )^2 + 252/64`

`y=4(x - 1/8 )^2 + 63/16`

The vertex is at `(1/8,63/16)`

graph{4*x^2-x+4 [-7.8, 8.074, -1.044, 6.896]}