Let's find the maximum value of y in the equation \(y = 1 - 2x - 3x^2\).

This is a quadratic equation in terms of x, with a = -3, b = -2, and c = 1.

The maximum or minimum value of a quadratic equation \(ax^2 + bx + c\) occurs at the vertex of the parabola represented by the equation.

The x-coordinate of the vertex is given by the formula \(-b/2a\). Substituting the values for a and b, we find that the x-coordinate of the vertex is \(-(-2)/(2 \times (-3)) = 1/3\).

Substituting this value of x into the equation for y, we find that the maximum value of y is \(y = 1 - 2 \times (1/3) - 3 \times (1/3)^2 = 1 - 2/3 - 1/3 = 4/3\).

So, the maximum value of y in the equation \(y = 1 - 2x - 3x^2\) is 4/3.