Find the vertical asymptotes, if any, for the following function. Please show all of your work.
f(x) = x+11/x^2-16x

Question

anonymous · Answer

x^2 - 16x = 0

jim_thompson5910 · Answer

the denominator cannot be zero


if it were, then 


x^2-16x = 0


x(x-16) = 0


x=0 or x-16=0


x=0 or x=16


So the values x=0 or x=16 cause a division by zero error


So the vertical asymptotes are x=0 and x=16

anonymous · Answer

james I have a question..after your done with this one..can you come back to my question?
thank you

anonymous · Answer

never mind..thank you for helping me with those questions

anonymous · Answer

i can help you. which question do i go bck to?

anonymous · Answer

Thank you Jamesm!!