Question

Part of the graph of a sine or cosine curve is given below. Write four different equations that represent the curve. The first two in the form:
y=a sin k(x-b), one with a positive a, and the other with a negative a.
The other two in the form:
y=a cos k(x-b), one with a positive a, and the other with a negative a.

Answer 1

bear in mind that
a negative "a" coefficient, flips it upside-down
notice the sine -> http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzaW4oeCkiLCJjb2xvciI6IiNEMTEwMTAifSx7InR5cGUiOjAsImVxIjoiLXNpbih4KSIsImNvbG9yIjoiIzI2MTdEMSJ9LHsidHlwZSI6MTAwMH1d

Answer 2

also bear in mind that
all you're doing is transforming that original equationn
\(\large {
\textit{function transformations}
\\ \quad \\

\begin{array}{llll}

\begin{array}{llll} shrink\ or\\ expand\\ by\ {\color{purple}{ A}}\cdot {\color{blue}{ k}}\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
y = {\color{purple}{ A}} sin( {\color{blue}{ k}}x + {\color{red}{ b}} ) + {\color{green}{ D}}\\
% template ends
\qquad\qquad\quad\ \uparrow \\

\qquad\begin{array}{llll} horizontal\\ shift\\ by \ \frac{{\color{red}{ b}}}{{\color{blue}{ k}}}\end{array}

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

Answer 3

also bear in mind that for cosine as well as sine, a "transformed" function will have a period of \(\bf \cfrac{2\pi }{{\color{blue}{ k}}}\)

Answer 4

ok

Answer 5

so the idea being
whatever "a" and "k" and "c" you use
the graph should look like the one given

Answer 6

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzaW4oeCkiLCJjb2xvciI6IiNEMTEwMTAifSx7InR5cGUiOjAsImVxIjoiLXNpbih4LXBpKSIsImNvbG9yIjoiIzI2MTdEMSJ9LHsidHlwZSI6MTAwMH1d

notice that same functions as before
I simply gave a \(-\pi\) that is, the -sin(x) was shifted to the right, by that much
thus matching the sin(x) due to the shift