Answer: The correct option is
(C) [tex]y=-6\left(x-\dfrac{1}{4}\right)^2+\dfrac{19}{8}.[/tex]
Step-by-step explanation: We are given to select the equation that is the vertex form of he following equation :
[tex]y=-6x^2+3x+2~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
the vertex form of a function y = f(x) is written as
[tex]y=a(x-h)^2+k,[/tex] where (h, k) is the vertex.
From equation (i), we have
[tex]y=-6x^2+3x+2\\\\\\\Rightarrow y=-6\left(x^2-\dfrac{1}{2}x\right)+2\\\\\\\Rightarrow y=-6\left(x^2-2\times x\times\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right)+2+6\times \left(\dfrac{1}{4}\right)^2\\\\\\\Rightarrow y=-6\left(x-\dfrac{1}{4}\right)^2+2+\dfrac{6}{16}\\\\\\\Rightarrow y=-6\left(x-\dfrac{1}{4}\right)^2+2+\dfrac{3}{8}\\\\\\\Rightarrow y=-6\left(x-\dfrac{1}{4}\right)^2+\dfrac{16+3}{8}\\\\\\\Rightarrow y=-6\left(x-\dfrac{1}{4}\right)^2+\dfrac{19}{8}.[/tex]
Thus, the required vertex form of the given equation is
[tex]y=-6\left(x-\dfrac{1}{4}\right)^2+\dfrac{19}{8}.[/tex]
Option (C) is CORRECT.
