Vertical Asymptotes

Guest:

How do I get the denominator to equal zero for the problem 2x^2-3x over 4x^2+1?

you mean ((2(x^2))-3x) / ((4(x^2))+1) = 0

then the answer is x = 2/3 and x = 0 to find asymptote

Guest:

How do I get the denominator to equal zero for the problem 2x^2-3x over 4x^2+1?

You want to know when 4x ² + 1 = 0 so that you can determine the asymtotes? Is that correct?

Think about this
x ² >= 0 (because any number squared can never be negative)
therefore,
4x ² >= 0
therefore
4x ² +1 >= 1

Hence the denominator cannot be zero.

This is what the graph looks like.
140115 Vertical asymtotes.JPG