How do I find the vertical asymototes of....?

Question

y = (2x^2-7x-4)/(3x^2-13x+4)

Answer 1

Find values of x that cause division by zero (what causes the denominator to be zero?)

3x^2-13x+4=0 and solve for x, those are your vertical asymptotes.

(3x-1)(x-4)=0

3x-1=0 or x-4=0

3x=1 or x=4

x=1/3 or x=4

There are your vertical asymptotes!

Answer 2

(2x^2-7x-4)/(3x^2-13x+4)

[(2x + 1)(x - 4)]/[(3x - 1)(x - 4)]

(2x + 1)/(3x - 1)

A hole occurs when x - 4 = 0 which is when x = 4

A vertical asymptote occurs when 3x - 1 = 0 which is when x = 1/3