Question

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

Answer 1

Answer:

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

Step-by-step explanation:

The vertex form of an equation of a parabola:

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

(h, k) - vertex

We have

$y=2x^2+4x-1=2\left(x^2+2x-\dfrac{1}{2}\right)$

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

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

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

$2(x+1)^2+2\left(-\dfrac{3}{2}\right)=2(x+1)^2-3$