Explain why the basic transformations of the parent function y=x^{5} will only generate functions that can be written in the form y=a(x - h)^{5} + k

Question
Transformations of functions
Explain why the basic transformations of the parent function $$\displaystyle{y}={x}^{{{5}}}$$ will only generate functions that can be written in the form $$\displaystyle{y}={a}{\left({x}\ -\ {h}\right)}^{{{5}}}\ +\ {k}$$

2020-11-10
Sterp 1
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is stretched vertically by a factor of $$\displaystyle{\color{red}{{a}}},$$ we get the graph for $$\displaystyle{y}={\color{red}{{a}}}{f{{\left({x}\right)}}}$$
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is compressed vertically by a factor of$$\displaystyle{\color{red}{{a}}}$$, we get the graph for $$\displaystyle{y}={\color{red}{{a}}}{f{{\left({x}\right)}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is stretched/compressed vertically by a factor of}\ {\color{red}{{a}}},\ \text{we get the graph for}\ {y}={\color{red}{{a}}}{x}^{{{5}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ Note that we can also stretch the graph horizontall,
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is stretched horizontally by a factor of m, we get the graph for $$\displaystyle{y}={f{{\left({m}{x}\right)}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is stretched horizontally by factor of m, we get the graph for}\ {y}={m}^{{{5}}}{x}^{{{5}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Which is the same as a vertical shift by a factor $$\displaystyle{\color{red}{{a}}}={m}^{{{5}}}$$
Therefore, by vertical and horizontal stretch of $$\displaystyle{y}={x}^{{{5}}}$$ we can obtain the polynomial $$\displaystyle{y}={\color{red}{{a}}}{x}^{{{5}}}$$
Step 2 When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is reflected across the x-axis, we get the graph for $$\displaystyle{y}=\ -{f{{\left({x}\right)}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is refleced across the x-axis, we get the graph for}\ {y}=\ -{x}^{{{5}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is reflected across the y-axis, we get the graph for $$\displaystyle{y}={f{{\left(-{x}\right)}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is refleced across the x-axis, we get the graph for}\ {y}=\ -{x}^{{{5}}}=\ -{x}^{{{5}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Note that sign of $$\displaystyle{\color{red}{{a}}}$$ accounts for the reflection about x and y axes
Step 3
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is vertically translated by $$\displaystyle{\color{blue}{{k}}}$$ units in the upward direction, we get the graph for $$\displaystyle{y}={f{{\left({x}\right)}}}\ +\ {\color{blue}{{k}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is vertically translated by}\ {\color{blue}{{k}}}\ \text{units in the upward direction, we get the graph for}\ {y}={x}^{{{5}}}\ +\ {\color{blue}{{k}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Note that if $$\displaystyle{\color{blue}{{k}}}$$ is negative, the graph is translated vertically downwards
Step 4
When $$\displaystyle{y}={f{{\left({x}\right)}}}$$ is horizontally translated by h nits in the right direction, we get the graph for $$\displaystyle{y}={f{{\left({x}\ -\ {h}\right)}}}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}\right\rbrace}{h}{l}\in{e}\text{When}\ {y}={x}^{{{5}}}\ \text{is horizontally translated by h units in the right direction, we get the graph for}\ {y}={\left({x}\ -\ {h}\right)}^{{{5}}}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Note that If h is negative the translation is in the left direction
Step 5
Therefore, When all the transformations are applied we get $$\displaystyle{y}={\color{red}{{a}}}{\left({x}\ -\ {h}\right)}^{{{5}}}\ +\ {\color{blue}{{k}}}$$
If the graph is not translated, then h and $$\displaystyle{\color{blue}{{k}}}$$ are zero
If the graph is not stretched/compressed or reflected, $$\displaystyle{\color{red}{{a}}}={1}$$

Relevant Questions

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.
h is related to one of the six parent functions.
a) Identify the parent function f.
b) Describe the sequence of transformations from f to h.
c) Sketch the graph of h by hand.
d) Use function notation to write h in terms of the parent function f.
$$\displaystyle{h}{\left({x}\right)}={\left(-{x}\right)}^{{{2}}}-{8}$$
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.
$$\displaystyle{g{{\left({x}\right)}}}=\ -{2}{\left|{x}\ -\ {1}\right|}\ -\ {4}$$
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.$$\displaystyle{g{{\left({x}\right)}}}={\frac{{{1}}}{{{3}}}}{\left({x}-{2}\right)}^{{{3}}}$$
Standard transformations can be used to help graph rational functions of the form $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{A}}}{{{B}{x}-{C}}}}+{D}$$ Explain how the parameters A, B, C, and D relate to the graph of the rational functions.
Describe the transformations of
$$\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{3}}}{\left\lbrace{x}\right\rbrace}$$ and $$\displaystyle{h}{\left({x}\right)}={x}^{{{3}}}$$
that result in the functions listed below.
$$\displaystyle{g{{\left({x}\right)}}}=\ -\sqrt{{{3}}}{\left\lbrace{x}\ -\ {3}\right\rbrace}$$ and $$\displaystyle{j}{\left({x}\right)}=\ -{\left({x}\ -\ {3}\right)}^{{{3}}}$$
$$\displaystyle{p}{\left({x}\right)}={4}\sqrt{{{3}}}{\left\lbrace-{x}\right\rbrace}\ -\ {1}$$ and $$\displaystyle{q}{\left({x}\right)}={0.25}{\left(-{4}{x}\right)}^{{{3}}}\ -\ {1}$$
Begin by graphing
$$\displaystyle{f{{\left({x}\right)}}}={2}^{{{2}}}$$
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.
$$\displaystyle{g{{\left({x}\right)}}}={2}^{{-{x}}}$$
Sketch a graph of the function. Use transformations of functions when ever possible. $$\displaystyle{f{{\left({x}\right)}}}=\ {\frac{{{1}}}{{{3}}}}{\left({x}\ -\ {5}\right)},\ {2}\ \leq\ {x}\ \leq\ {8}$$
$$\displaystyle\text{quotient}+{\frac{{{r}{e}{m}{a}\in{d}{e}{r}}}{{\div{i}{s}{\quad\text{or}\quad}}}}$$
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}}}}$$
$$\displaystyle{g{{\left({x}\right)}}}={\frac{{{2}{x}+{7}}}{{{x}+{3}}}}$$
For $$\displaystyle{y}=\ -{{\log}_{{{2}}}{x}}$$.
a) Use transformations of the graphs of $$\displaystyle{y}={{\log}_{{{2}}}{x}}$$ and $$\displaystyle{y}={{\log}_{{{3}}}{x}}$$ o graph the given functions.