Question

Describe the vertical asymptote and holes for the graph of Y=(x-3)(x-1)/(x-1)(x-5)

Answer 1

Answer:

Hole=1 and vertical asymptote=5    

Step-by-step explanation:

It is given that:

$Y=\frac{(x-3)(x-1)}{(x-1)(x-5)}$

Look at the denominator and determine the values for which it is zero, therefore

(x-1)(x-5)=0⇒x=1,5

There is a common factor of (x-1), so we can cancel that common factor:

$Y=\frac{x-3}{x-5}$

But note we can only do that provided x-1≠0, x≠1.

We have already got that when  x=1 the denominator is zero, so there is a removable discontinuity at that point (or a "hole").

when x=5, he denominator is also zero and this is the only vertical asymptote.