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# For what values of is

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For what values of  is

Note: Be thorough and explain why all points in your answer are solutions and why all points outside your answer are not solutions.

Guest Oct 26, 2014

#2
+92495
+5

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

CPhill  Oct 26, 2014
#1
0

Guest Oct 26, 2014
#2
+92495
+5

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

CPhill  Oct 26, 2014