SOLUTION: Solve {{{7/(x^2-6x-16)<=0}}}

Algebra ->  Graphs -> SOLUTION: Solve {{{7/(x^2-6x-16)<=0}}}      Log On


   

Question 135451: Solve 7%2F%28x%5E2-6x-16%29%3C=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, let's find the vertical asymptotes

x%5E2-6x-16=0 Stet the denominator equal to zero


%28x-8%29%28x%2B2%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x-8=0 or x%2B2=0

x=8 or x=-2 Now solve for x in each case


So the vertical asymptotes are

x=-2 or x=8


This means that the critical values are x=-2 and x=8



So let's test a value that is to the left of x=-2



7%2F%28x%5E2-6x-16%29%3C=0 Start with the given inequality



7%2F%28%28-3%29%5E2-6%28-3%29-16%29%3C=0 Plug in x=-3


7%2F%2811%29%3C=0 Simplify


Since 7%2F%2811%29%3C=0 is false, this means that the interval does not work

------------------------------------

So let's test a value that is in between x=-2 and x=8



7%2F%28x%5E2-6x-16%29%3C=0 Start with the given inequality



7%2F%28%280%29%5E2-6%280%29-16%29%3C=0 Plug in x=0


-7%2F16%3C=0 Simplify


Since -7%2F16%3C=0 is true, this means that the part of the solution set is []


-------------------------------

Finally, let's test a point that is to the right of x=8



7%2F%28x%5E2-6x-16%29%3C=0 Start with the given inequality



7%2F%28%2810%29%5E2-6%2810%29-16%29%3C=0 Plug in x=10


7%2F24%3C=0 Simplify


Since 7%2F%2824%29%3C=0 is false, this means that the interval does not work



----------------------------------------

Answer:

So the solution in interval notation is []