##### #11 2.09 Precalculus questions

label Mathematics
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May 14th, 2015

$f(x)=\frac{x^2-9}{x-3}=\frac{(x+3)(x-3)}{(x-3)}=x+3$

So, it has a removable discontinuity at x = 3 as (x - 3) gets canceled. The graph of f(x) will have a hole at x = 3 .

$g(x)=\frac{x-9}{x-3}$

g(x) has a non removable discontinuity at x = 3 as (x - 3) does not cancel out.

x = 3 is the vertical asymptote of g(x).

May 14th, 2015

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May 14th, 2015
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May 14th, 2015
Jun 25th, 2017
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