Question

Which equation is y = 6x2 + 12x – 10 rewritten in vertex form?

Answer 1

Answer:

$y=6(x+1)^{2}-16$

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

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

where

(h,k) is the vertex of the parabola

In this problem we have

$y=6x^{2}+12x-10$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$y+10=6x^{2}+12x$

Factor the leading coefficient

$y+10=6(x^{2}+2x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$y+10+6=6(x^{2}+2x+1)$

$y+16=6(x^{2}+2x+1)$

Rewrite as perfect squares

$y+16=6(x+1)^{2}$

$y=6(x+1)^{2}-16$ -------> equation in vertex form

The vertex is the point $(-1,-16)$

Answer 2

Y = 6 x² + 12 x - 10= 6 ( x² + 2 x ) - 10 = 6 ( ( x² + 2 x + 1 ) - 1 ) - 10=
6 ( x + 1 )² - 6 - 10 = 6 ( x + 1 )² - 16 ( vertex form)