Answer:
[tex]\large\boxed{y=2(x+1)^2-3}[/tex]
Step-by-step explanation:
The vertex form of an equation of a parabola:
[tex]y=a(x-h)^2+k[/tex]
(h, k) - vertex
We have
[tex]y=2x^2+4x-1=2\left(x^2+2x-\dfrac{1}{2}\right)[/tex]
We must use the formula: [tex](a+b)^2=a^2+2ab+b^2\qquad(*)[/tex]
[tex]2\left(x^2+2(x)(1)-\dfrac{1}{2}\right)=2\bigg(\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-\dfrac{1}{2}\bigg)\\\\=2\left((x+1)^2-1-\dfrac{1}{2}\right)=2\left((x+1)^2-\dfrac{3}{2}\right)[/tex]
Use the distributive formula a(b + c) = ab + ac
[tex]2(x+1)^2+2\left(-\dfrac{3}{2}\right)=2(x+1)^2-3[/tex]
