Respuesta :
[tex]\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}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }{{ D}}\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
now, with that template above in mind, let's take a peek of yours
[tex]\bf \begin{array}{lcllcclll} f(x)=&-3&sin(&4x&-\pi)&+2\\ &\uparrow &&\uparrow &\uparrow &\uparrow\\ &A&&B&C&D \end{array}[/tex]
now, as far as the midline, well, the midline for sin(x) is just the x-axis, notice this one, has a D vertical shift, so the midline moved up that much
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }{{ D}}\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
now, with that template above in mind, let's take a peek of yours
[tex]\bf \begin{array}{lcllcclll} f(x)=&-3&sin(&4x&-\pi)&+2\\ &\uparrow &&\uparrow &\uparrow &\uparrow\\ &A&&B&C&D \end{array}[/tex]
now, as far as the midline, well, the midline for sin(x) is just the x-axis, notice this one, has a D vertical shift, so the midline moved up that much
Answer:
[tex]Amplitude=3[/tex]
[tex]Period=\frac{\pi}{2}[/tex]
[tex]\text{Phase shift}=\frac{\pi}{4}[/tex]
[tex]Midline=2[/tex]
Step-by-step explanation:
The general sine function is defined as
[tex]g(x)=A\sin (Bx+C)+D[/tex] .... (1)
where, |A| is amplitude, [tex]\frac{2\pi}{B}[/tex] is period, -C/B is phase shift and D is midline.
The given function is
[tex]f(x)=-3\sin (4x-\pi)+2[/tex] .... (2)
On comparing (1) and (2), we get
[tex]A=-3[/tex]
[tex]B=4[/tex]
[tex]C=-\pi[/tex]
[tex]D=2[/tex]
Using these values we can say that
[tex]Amplitude=|A|=|-3|=3[/tex]
[tex]Period=\frac{2\pi}{B}\Rightarrow \frac{2\pi}{4}=\frac{\pi}{2}[/tex]
[tex]\text{Phase shift}=-\frac{C}{B}\Rightarrow -\frac{-\pi}{4}=\frac{\pi}{4}[/tex]
[tex]Midline=D=2[/tex]
Therefore, [tex]Amplitude=3[/tex], [tex]Period=\frac{\pi}{2}[/tex], [tex]\text{Phase shift}=\frac{\pi}{4}[/tex] and [tex]Midline=2[/tex].
