# Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions, and then applying the appropriate transformations. y=1-frac{1}{x}

Question
Transformations of functions
Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions, and then applying the appropriate transformations. $$\displaystyle{y}={1}-{\frac{{{1}}}{{{x}}}}$$

2021-02-19
Step 1 We will start with the graph of the standard/parent function $$\displaystyle{y}={\frac{{{1}}}{{{x}}}}$$ When we reflect the graph of $$\displaystyle{y}={f{{\left({x}\right)}}}$$ in the x-axis, we get the graph of $$\displaystyle{y}=-{f{{\left({x}\right)}}}$$ Therefore, the graph of $$\displaystyle{y}=-{\frac{{{1}}}{{{x}}}}$$ is obtained by reflecting the graph of PSKy=\frac{1}{x} in the x-axis. In the graph below: The blue dashed curve represents $$\displaystyle{y}={\frac{{{1}}}{{{x}}}}$$ The red solid curve represents $$\displaystyle{y}=-{\frac{{{1}}}{{{x}}}}$$ Step 2 When we shift the graph of $$\displaystyle{y}={f{{\left({x}\right)}}}$$ by k units in the upward direction, we get the graph of $$\displaystyle{y}={f{{\left({x}\right)}}}+{k}$$ Therefore, the graph of $$\displaystyle{y}={1}-{\frac{{{1}}}{{{x}}}}$$ is obtained by shifting the graph of $$\displaystyle{y}=-{\frac{{{1}}}{{{x}}}}$$ by 1 unit in the upward direction. In the graph below: The red dashed curve represents $$\displaystyle{y}=-{\frac{{{1}}}{{{x}}}}$$ The black solid curve represents $$\displaystyle{y}={1}-{\frac{{{1}}}{{{x}}}}$$

### Relevant Questions

Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions, and then applying the appropriate transformations.
$$\displaystyle{y}={1}-{2}\sqrt{{{x}}}+{3}$$
Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions, and then applying the appropriate transformations. $$\displaystyle{y}={x}^{{{2}}}-{4}{x}+{5}$$
Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations.
$$y=sqrt(x - 2) - 1$$
Begin by graphing
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{x}}}.$$
Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn graphs.
$$\displaystyle{g{{\left({x}\right)}}}={\frac{{{1}}}{{{2}}}}\ \cdot\ {2}^{{{x}}}$$
g is related to one of the six parent functions. (a) Identify the parent function f. (b) Describe the sequence of transformations from f to g. (c) Sketch the graph of g by hand. (d) Use function notation to write g in terms of the parent function f.$$\displaystyle{g{{\left({x}\right)}}}={\frac{{{1}}}{{{3}}}}{\left({x}-{2}\right)}^{{{3}}}$$
g is related to one of the six parent functions.
a) Identify the parent function f.
b) Describe the sequence of transformations from f to g.
c) Sketch the graph of g by hand.
d) Use function notation to write g in terms of the parent function f.
$$\displaystyle{g{{\left({x}\right)}}}=\ -{2}{\left|{x}\ -\ {1}\right|}\ -\ {4}$$
Begin by graphing
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{x}}}.$$
Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn graphs.
$$\displaystyle{g{{\left({x}\right)}}}={2}^{{{x}\ +\ {2}}}$$
Begin by graphing
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{x}}}.$$
Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range. If applicable, use a graphing utility to confirm your hand-drawn graphs.
$$\displaystyle{g{{\left({x}\right)}}}={2}^{{{x}}}={2}^{{{x}}}\ +\ {2}$$
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{x}}}$$
$$\displaystyle{g{{\left({x}\right)}}}=-{2}^{{{x}}}$$
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{2}}}$$
$$\displaystyle{g{{\left({x}\right)}}}={2}^{{-{x}}}$$