Because the function in the numerator is > 0 for all real numbers (thus, the graph never intersects the x axis), we only need to concern ourselves with the denominator.....!!!
This function will be undefined when the denominator = 0
So.....setting 2x^2 + x - 6 to 0 and factoring, we have
(2x -3)(x + 2) = 0 and setting each factor to 0, we have x = -2 and x = 3/2...so the graph is undefined at these two values (it has vertical asymptotes at these two values)....we just need to look at 3 intervals
When x < -2, the rational function > 0 (test -3 to see this is true)
And when x > -2 but < 3/2, the rational function < 0 (test 0)
And when x > 3/2 , the rational function > 0 (test 2)
So, the solution is
(-∞, -2) U (3/2, ∞)
Here's the graph...(the asymptotes will be hard to determine from this!!)....https://www.desmos.com/calculator/piat8yrld0
Because the function in the numerator is > 0 for all real numbers (thus, the graph never intersects the x axis), we only need to concern ourselves with the denominator.....!!!
This function will be undefined when the denominator = 0
So.....setting 2x^2 + x - 6 to 0 and factoring, we have
(2x -3)(x + 2) = 0 and setting each factor to 0, we have x = -2 and x = 3/2...so the graph is undefined at these two values (it has vertical asymptotes at these two values)....we just need to look at 3 intervals
When x < -2, the rational function > 0 (test -3 to see this is true)
And when x > -2 but < 3/2, the rational function < 0 (test 0)
And when x > 3/2 , the rational function > 0 (test 2)
So, the solution is
(-∞, -2) U (3/2, ∞)
Here's the graph...(the asymptotes will be hard to determine from this!!)....https://www.desmos.com/calculator/piat8yrld0