Question

What are the vertical asymptotes and holes for the graph of `y=((x-5)(x-2))/((x-2)(x+4)`?

Answer 1

`x=-4 =>` vertical asymptote
`x=2 =>` removable discontinuity

Explanation:

`y= ((x-5)(x-2))/((x-2)(x+4))`

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

`(x-2)(x+4) = 0=> x=2, -4`

There is a common factor of (x-2), so we can cancel that common factor:

`y= ((x-5)cancel(x-2))/(cancel(x-2)(x+4))`

But note we can only do that provided `x-2!=0`, ie `x!=2`

We have already established that when `x=2` the denominator is zero, so there is a removable discontinuity at that point (or a "hole")

When `x=-4` the denominator is also zero and this is the only vertical asymptote.

graph{((x-5)(x-2))/((x-2)(x+4)) [-21.08, 18.92, -9.52, 10.48]}

Answer 2

The vertical asymptote : `x = -4`
The horizontal asymptote: y = 1.

Explanation:

By actual division and rearrangement,

`(y-1)(x+4)=-9`

This represents the rectangular hyperbola (RH), with the

perpendicular asymptotes `y-1=0 and x+4=0 and`

center at `(-4, 1)`.

The graph of the RH is already included, in another answer. .