How do you find vertical, horizontal and oblique asymptotes for #y = (x^2+6x-7)/(x-1)#?
1 Answer
Apr 6, 2017
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]}