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
asked 2020-11-09
Explain why the basic transformations of the parent function \(\displaystyle{y}={x}^{{{5}}}\) will only generate functions that can be written in the form \(\displaystyle{y}={a}{\left({x}\ -\ {h}\right)}^{{{5}}}\ +\ {k}\)

Answers (1)

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, \(\displaystyle{\color{red}{{a}}}={1}\)
0

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