sketch y = (x-3)^2 /((x-2)(x+2) )
your solution will be all the x's when the graph is above the x-axis
I can see it there is a zero at x = 3
and there are 2 vertical asymptotes at
x = 2 , and x = -2
How about thinking about this intuitively
the numerator is always ≥ 0, since we squared something
Also we can see that x^2 - 4 is negative for values of x between -2 and 2, and x cannot be +2 or -2
so it looks like the expression is above the x-axis for x > 2 OR x < -2
solve the following inequality.
(x-3)^2/(x^2-4)is greater than or equal to 0.
2 answers
how do you write that in interval notation?