Question

In the following items, you will analyze how several transformations affect the graph of the function \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}}}}\). Investigate the graphs of \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}}}},{g{{\left({x}\right)}}}={f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}+{2}}}},{h}{\left({x}\right)}={\frac{{{1}}}{{{x}-{2}}}},{p}{\left({x}\right)}={\frac{{{1}}}{{{x}-{4}}}}\ \text{and}\ {z}{\left({x}\right)}={\frac{{{1}}}{{{x}^{{{2}}}+{1}}}}\). If you use a graphing calculator, select a viewing window \(\displaystyle\pm{23.5}\) for x and \(\displaystyle\pm{15.5}\) for y. At what values in the domain did vertical asymptotes occur for each of the functions? Explain why the vertical asymptotes occur at these values.

Answer 1

Step 1 We graph f(x), as well as g(x) (dotted line), h(x) (dashed line), and p(x) (solid line). Step 2 We graph z(x): Step 3 We find that the vertical asymptotes are \(\displaystyle{f{{\left({x}\right)}}}:{x}={0}\)
\(\displaystyle{g{{\left({x}\right)}}}:{x}=-{2}\)
\(\displaystyle{h}{\left({x}\right)}:{x}={2}\)
\(\displaystyle{p}{\left({x}\right)}:{x}={4}\)
\(\displaystyle{z}{\left({x}\right)}:\) None Each vertical asymptote is found by finding the value(s) of x for which the denominator equals zero, and therefore the function is undefined. Note that there is no such real value of x in the case of z(x).