Question

Identify the middle line for the function:

Answer 1

ahemmm I'd say, graph it :)

Answer 2

i did and nothing came up

Answer 3

well... is the secant graph really
with a few shifts here and there

for the midline, the relevant shift is the vertical shift though

\(\textit{function transformations}
\\ \quad \\

\begin{array}{llll}

\begin{array}{llll} shrink\ or\\ expand\\ by\ {\color{purple}{ A}}\cdot {\color{blue}{ B}}\end{array}
\qquad
\begin{array}{llll} vertical\\ shift\\ by \ {\color{green}{ D}} \end{array}

\begin{array}{llll}{\color{green}{ D}} > 0& Upwards
\\ \quad \\
{\color{green}{ D}} < 0 & Downwards\end{array}
\\

\qquad \downarrow\qquad\qquad\quad\ \downarrow\\
% template start
f(t) = {\color{purple}{ 5}}sec ( {\color{blue}{ 3}}t + {\color{red}{ \pi }} ) + {\color{green}{ 1}}\\
% template ends
\qquad\qquad\quad\ \uparrow \\

\qquad\begin{array}{llll} horizontal\\ shift\\ by \ \frac{{\color{red}{ C}}}{{\color{blue}{ B}}}\end{array}

\begin{array}{llll}\frac{{\color{red}{ C}}}{{\color{blue}{ B}}} > 0 & to\ the\ left
\\ \quad \\
\frac{{\color{red}{ C}}}{{\color{blue}{ B}}} < 0& to\ the\ right\end{array}
\end{array}\)

Answer 4

but basically, the midline for that cyclical function with all its shifts, would be the same as the parent function of sec(t)

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzZWMoeCkiLCJjb2xvciI6IiMzODIzRDkifSx7InR5cGUiOjAsImVxIjoiNXNlYygzeC10aGV0YSkrMSIsImNvbG9yIjoiI0Q0MUUxRSJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi0xOS44MzY0MjU3ODEyNSIsIjE5LjgzNjQyNTc4MTI1IiwiLTEyLjIwNzAzMTI1IiwiMTIuMjA3MDMxMjUiXX1d

Answer 5

hmmmm shold have been \(\pi\) rather than theta there anyhow, that's just a horizontal shift anyway, won't afffect the midline

Answer 6

well how do u get a middle point

Answer 7

so its A