Use the completing the square to find the x intercept equation is y=-3x^2 +2x -1 ?

To find the x-intercepts of the equation y = -3x^2 + 2x - 1, we first set y to 0 and solve for x:

0 = -3x^2 + 2x - 1

Next, we complete the square by moving the constant term to the other side of the equation:

3x^2 - 2x + 1 = 0

Now, we need to rewrite the equation in the form (x - h)^2 = k:

3(x^2 - (2/3)x) = -1
3(x^2 - (2/3)x + (1/3)^2) = -1 + 3(1/3)^2
3(x - 1/3)^2 = -1 + 1/3
3(x - 1/3)^2 = -2/3

Now, we can solve for x by taking the square root of both sides:

x - 1/3 = ±√(-2/9)
x - 1/3 = ±(i√2)/3
x = 1/3 ± (i√2)/3

Therefore, the x-intercepts of the equation y = -3x^2 + 2x - 1 are x = 1/3 + (i√2)/3 and x = 1/3 - (i√2)/3.

y = -3x^2 + 2x - 1
y = -3(x^2 - 2/3x + 1/9) + 1 - 3(1/9)
y = -3(x - 1/3)^2 + 1 - 1/3
y = -3(x - 1/3)^2 + 2/3
To find the x-intercepts, we set y = 0 and solve for x:
0 = -3(x - 1/3)^2 + 2/3
3(x - 1/3)^2 = 2/3
(x - 1/3)^2 = 2/9
x - 1/3 = ±√(2/9)
x = 1/3 ± √(2/9)
x = 1/3 ± 2√(2)/9
So the x-intercepts are (1/3 + 2√(2)/9, 0) and (1/3 - 2√(2)/9, 0).