Question

Rewrite the quadratic function in vertex form.
Y=2x^2+4x-1

Answer 1

`\large\boxed{y=2(x+1)^2-3}`

Explanation:

The vertex form of an equation of a parabola:

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

(h, k) - vertex

We have

`y=2x^2+4x-1=2\left(x^2+2x-(1)/(2)\right)`

We must use the formula:
`(a+b)^2=a^2+2ab+b^2\qquad(*)`

`2\left(x^2+2(x)(1)-(1)/(2)\right)=2\bigg(\underbrace{x^2+2(x)(1)+1^2}_((*))-1^2-(1)/(2)\bigg)\\\\=2\left((x+1)^2-1-(1)/(2)\right)=2\left((x+1)^2-(3)/(2)\right)`

Use the distributive formula a(b + c) = ab + ac

`2(x+1)^2+2\left(-(3)/(2)\right)=2(x+1)^2-3`