If y=x²-6x+9, Find the range of value of x for which y ≥ 0.

Pls I need working answer

We need all values of x for which y is positive or zero.

So, x²-6x+9 is either positive or zero.

x²-6x+9 ≥ 0
x²-6x ≥ -9
(x - 3)^2 - 9 ≥ -9
(x - 3)^2 ≥ 0
x - 3 ≥ 0
x ≥ 3

(x - 3)^2 ≥ 0 for all x, since it is a squared value.

The graph is a parabola just touching the x-axis at x=3.

To find the range of values of x for which y ≥ 0, we need to determine the values of x that make the equation y = x² - 6x + 9 greater than or equal to zero.

To solve this inequality, we can start by finding the x-intercepts of the equation, which occur when y equals zero. In other words, we solve the equation x² - 6x + 9 = 0.

Factoring this quadratic equation, we can rewrite it as (x - 3)² = 0. By taking the square root of both sides, we get x - 3 = 0, which simplifies to x = 3.

Thus, the equation x² - 6x + 9 = 0 has one unique solution, x = 3. This means the parabola associated with y = x² - 6x + 9 intersects the x-axis at a single point, namely x = 3.

Since this quadratic opens upwards (the coefficient of x² is positive), it means that the function is positive on either side of the x-intercept at x = 3. This implies that the range of values for x where y ≥ 0 is x ≤ 3 or x ≥ 3.

In conclusion, the range of values for x that satisfy y ≥ 0 is (-∞, 3] or [3, +∞).