Question

Analyzing the graphs of a periodic functions (need help)

Answer 1

`\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\ % function transformations for trigonometric functions \begin{array}{rllll} % left side templates f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}} \\\\ f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\ f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}} \end{array} \\\\ -------------------\\\\`


`\bf \bullet \textit{ stretches or shrinks}\\ \left. \qquad \right. \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis}`


`\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }{{ D}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\`


`\bf \bullet \textit{function period or frequency}\\ \left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)`

now, with that template above in mind, let's see.

reflected over the x-axis, that means A is negative.

vertically shrunk by 0.25 or 1/4, that means A is negative 4, or -4.

shifted to the left, that means C/B is positive

shifted by 65°, that means, we could use the default B = 1, and C = 65°, that way we end with C/B = 65/1 or just +65

and shifted downwards by 1 unit, that means D = -1.


`\bf f(x)=-4sin(1x+65^o)-1\implies f(x)=-4sin(x+65^o)-1`

and looks more or less like the picture below.