Question 215672
This is a tricky one. To find vertical asymptotes one normally just figures out what values for x make a denominator 0. And for g(x), if x = 3 then the denominator is zero. But, as it turns out, x = 3 is <b>not</b> a vertical asymptote for g(x).<br>
The reason x = 3 is not an asymptote, is that one should reduce the fraction before searching for vertical asymptotes. Let's factor g(x) and reduce the fraction. As usual, we start factoring with the Greatest Common Factor (GCF):
{{{g(x) = (7(3-x))/(3(x-3))}}}
Since (3-x) is the negative of (x-3), I will factor out a -1 from the numerator:
{{{g(x) = (-7(x-3))/(3(x-3))}}}
Now the (x-3)'s cancel leaving
{{{g(x) = (-7)/3}}}
g(x), even though its simplified form has no x's, still has the domain it has had since the start. The domain of g(x) is all Real numbers except 3 (because 3 makes the denominator of the original g(x) zero). Graphing the simplified equation with the domain we get a horizontal line with a "hole" (aka discontinuity) at x = 3. There is no vertical asymptote for g(x).