y=x^2-11x+30
For what values of x is y greater than or equal to 0?
You are looking at a parabola which opens upwards
Find the x-intercepts ( the solutions to x^2 - 11x + 11x + 30 = 0 , it factors very nicely)
for the region between the two intercepts, the y will be < 0,
for the values to the right and to the left of these intercepts, y >0
let me know what your answer is
To find the values of x for which y is greater than or equal to 0, we need to solve the quadratic equation y = x^2 - 11x + 30.
First, let's factorize the equation: y = (x - 5)(x - 6).
Now, set each factor equal to zero and solve for x:
x - 5 = 0
x = 5
x - 6 = 0
x = 6
The quadratic equation intersects the x-axis at x = 5 and x = 6. These are the critical points where y changes sign.
So the solution to y ≥ 0 is when x ≤ 5 and x ≥ 6.
Therefore, the values of x for which y is greater than or equal to 0 are x ≤ 5 and x ≥ 6.