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

Aug 10, 2015

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:
${x}^{2} - 9 x + 18 = 0$ solve using the Quadratic Formula:
${x}_{1 , 2} = \frac{9 \pm \sqrt{81 - 72}}{2} = \left(9 \pm 3\right) 2$ 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: