Find any vertical asymptote(s) of the graph of f(x)=x(x−3)2.
Solution The domain of f is {x|x≠3}. Since 3 is the only number for which the denominator of f equals zero, we construct Table 13 and investigate the one-sided limits of f as x approaches 3. Table 13 suggests that lim
So, x=3 is a vertical asymptote of the graph of f.
\underrightarrow{x~\hbox{approaches 3 from the left}} | \underleftarrow{x~\hbox{approaches 3 from the right}} | ||||||||
---|---|---|---|---|---|---|---|---|---|
x | 2.9 | 2.99 | 2.999 | \ \rightarrow | 3 | \leftarrow | 3.001 | 3.01 | 3.1 |
f( x) =\dfrac{x}{(x-3)^{2}} | 290 | 29{,}900 | 2{,}999{,}000 | f(x) becomes unbounded | 3{,}001{,}000 | 30{,}100 | 310 |
120
Figure 52 shows the graph of f( x) =\dfrac{x}{( x-3)^{2}} and its vertical asymptote.