Question

Which equation is y = –6x2 + 3x + 2 rewritten in vertex form?

Answer 1

we have

$y = -6x^{2} + 3x + 2$

we know that

the vertex form of the vertical parabola equation is equal to

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

where

(h,k) is the vertex of the parabola

To find the equation rewritten in vertex form let's factor the equation

Factor the leading coefficient

$y = -6(x^{2} - 0.5x) + 2$

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

$y = -6(x^{2} - 0.5x+0.0625-0.0625) + 2$

$y = -6(x^{2} - 0.5x+0.0625) + 2 +0.375$

$y = -6(x^{2} - 0.5x+0.0625) + 2.375$

Rewrite as perfect squares

$y = -6(x-0.25)^{2} + 2.375$

therefore

the answer is

$y = -6(x-0.25)^{2} + 2.375$

Answer 2

The answer is $y = -6( x- \frac{1}{4}) ^{2}+3$

Regular form: y = ax² + bx + c
Vertex form: y = a(x - h)² + k
(h, k) - vertex

$y = -6 x^{2} +3x+2 \\ y -2=-6 x^{2} +3x \\ y-2+6*=-6* x^{2} +6* \frac{1}{2} x \\ \\ y-2-6* \frac{1}{16} =-6* x^{2} +6* \frac{1}{2} x -6* \frac{1}{16} \\ \\ y -2 -\frac{6}{16} =-6( x^{2} -\frac{1}{2} x+\frac{1}{16}) \\ \\ y- \frac{2*16}{16} -\frac{6}{16} =-6( x^{2} -\frac{1}{2} x+(\frac{1}{4})^{2} ) \\ \\ y - \frac{32}{16} -\frac{6}{16} = -6( x- \frac{1}{4}) ^{2} \\ \\ y - \frac{32+6}{16} = -6( x- \frac{1}{4}) ^{2} \\ \\ y - \frac{48}{16} = -6( x- \frac{1}{4}) ^{2}$

$y - 3= -6( x- \frac{1}{4}) ^{2} \\ \\ y = -6( x- \frac{1}{4}) ^{2}+3$