Question

What is the vertex form of `y= 4x^2 + 10x + 6`?

Answer 1

`y = 4(x-(-5/4))^2+(-1/4)`

Explanation:

`y = 4x^2+10x+6`

`= 4(x^2+5/2x+3/2)`

`=4(x^2+2(x)(5/4)+(5/4)^2-(5/4)^2+6/4)`

`=4((x+5/4)^2-(5/4)^2+6/4)`

`=4(x+5/4)^2-25/4+24/4`

`=4(x+5/4)^2-1/4`

So:

`y = 4(x+5/4)^2-1/4`

Or we can write:

`y = 4(x-(-5/4))^2+(-1/4)`

This is in strict vertex form:

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

with multiplier `a=4` and vertex `(h, k) = (-5/4, -1/4)`