There will be either a hole or a vertical asymptote at the x-value that makes the denominator zero.
Holes: x=4 Vertical Asymptotes: x=-3
Practice makes perfect
Consider the given function.
y=(x-4)(x+5)/(x+3)(x-4)Recall that division by zero is not defined. Therefore, the rational function is undefined where (x+3)(x-4)=0.
ccc
x+3=0 & & x-4=0
⇕ & & ⇕
x=-3 & & x=4
The function is not defined when x=-3 and x=4. This means we have either a hole or a vertical asymptote for both values. Let's now simplify the function.
Note that we have canceled out the factor x-4, but x+3 is still in the denominator. This indicates that the graph has a hole when x=4, and that x=-3 is a vertical asymptote.
