Question

given the function f(x) x^3+2x^2-3x-5 what is the resulting function when f(x) is shifted to the right 1 unit?

Answer 1

I'll give you a couple of example of rules about shifting.
\(\large\color{ blue }{\large {\bbox[5pt, lightyellow ,border:2px solid white ]{ \large\text{ }\\
\begin{array}{|c|c|c|c|}
\hline \texttt{Shifts} ~~~\tt from~~~ {f(x)~~~\tt to~~~g(x)}&~\tt{c~~~units~~~~} \\
\hline
\\f(x)= \sqrt[4]{x} ~~~~\rm{\Rightarrow}~~~~ g(x)= \sqrt[4]{x \normalsize\color{red }{ -~\rm{c}} } &~\rm{to~~the~~right~} \\
\text{ } \\
f(x)= \sqrt[4]{x} ~~~~\rm{\Rightarrow}~~~~ g(x)= \sqrt[4]{x \normalsize\color{red}{ +~\rm{c}} } &~\rm{to~~the~~left ~} \\
\text{ } \\
f(x)= \sqrt[4]{x} ~~~~\rm{\Rightarrow}~~~~ g(x)= \sqrt[4]{x} \normalsize\color{red}{ +~\rm{c} } &~\rm{up~} \\
\text{ } \\
f(x)= \sqrt[4]{x} ~~~~\rm{\Rightarrow}~~~~ g(x)= \sqrt[4]{x} \normalsize\color{red}{ -~\rm{c} } &~\rm{down~} \\ \\
\hline
\end{array} }}}\)



\(\large\color{ teal }{\large {\bbox[5pt, lightcyan ,border:2px solid white ]{ \large\text{ }\\
\begin{array}{|c|c|c|c|}
\hline \texttt{Shifts} ~~~\tt from~~~ {f(x)~~~\tt to~~~g(x)}&~\tt{c~~~units~~~~} \\
\hline
\\f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= (x \normalsize\color{red}{ -~\rm{c} })^2 &~\rm{to~~the~~right~} \\
\text{ } \\
f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= (x \normalsize\color{red}{ +~\rm{c} })^2&~\rm{to~~the~~left ~} \\
\text{ } \\
f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= x^2 \normalsize\color{red}{ +~\rm{c} } &~\rm{up~} \\
\text{ } \\
f(x)= x^2 ~~~~~\rm{\Rightarrow}~~~~ g(x)= x^2 \normalsize\color{red}{ -~\rm{c} } &~\rm{down~} \\ \\
\hline
\end{array} }}}\)




`ABSOLUTE VALUE`
Rules of \(\large\color{black}{ \rm shifts }\) from \(\large\color{black}{ \rm f(x) }\) to \(\large\color{black}{ \rm g(x) }\).


\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\left| x \color{blue}{ -~\rm{c} }\right| }\)
\(\large\color{blue}{ ~\rm {c} }\) units to the \(\normalsize\color{blue}{ \rm right }\).


\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\left| x \color{blue}{ +~\rm{c} }\right| }\)
\(\large\color{blue}{ ~\rm {c} }\) units to the \(\normalsize\color{blue}{ \rm left }\).


\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\left| x \right| \color{blue}{ +~\rm{c} }}\)
\(\large\color{blue}{ ~\rm {c} }\) units \(\normalsize\color{blue}{ \rm up }\).


\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\left| x \right| \color{blue}{ -~\rm{c} }}\)
\(\large\color{blue}{ ~\rm{c} }\) units \(\normalsize\color{blue}{ \rm down }\)

Answer 2

thanks a lot

Answer 3

You that you add 1 to the x, "inside the parenthesis"

in other words, to shift it 1 unit left, you are to find f(x+1)