How do you find vertical, horizontal and oblique asymptotes for #y = (x^2+6x-7)/(x-1)#?

1 Answer
Apr 6, 2017

Answer:

There are no asymptotes.

Explanation:

The first step is to factorise the numerator and simplify.

#y=((x+7)cancel((x-1)))/cancel((x-1))=x+7#

Since the factor (x - 1 ) has been removed this indicates there is a hole at x = 1

#y=x+7" is the equation of a straight line "# and has no asymptotes.

graph{(x^2+6x-7)/(x-1) [-32.47, 32.48, -16.23, 16.23]}