How do you identify the vertical asymptotes of #f(x) = (x+6)/(x^2-9x+18)#?

1 Answer
Aug 10, 2015

Answer:

I found:
#x=6#
#x=3#

Explanation:

You identify the vertical asymptotes by setting the denominator equal to zero: this allows you to see which #x# values the function cannot accept (they would make the denominator equal to zero).
So, set the denominator equal to zero:
#x^2-9x+18=0# solve using the Quadratic Formula:
#x_(1,2)=(9+-sqrt(81-72))/2=(9+-3)2# so you get:
#x_1=6#
#x_2=3#
So your function cannot accept these values for #x#;
The two vertical lines of equations:
#x=6#
#x=3#
will be your "forbidden" lines or vertical asymptotes.

Graphically:
enter image source here