Answer:
[tex]f(x)=2cos(x+\frac{\pi}{2})+3[/tex]
Step-by-step explanation:
Because the function is symmetric about the y-axis, using the cosine function is most appropriate.
Refer to the equation for a cosine function:
[tex]f(x)=acos(bx+c)+d[/tex]
Amplitude: [tex]|a|[/tex]
Period: [tex]\frac{2\pi}{|b|}[/tex]
Phase shift: [tex]-\frac{c}{b}[/tex]
Midline: [tex]y=d[/tex]
The amplitude would be the average of the maximum and minimum y-values of the function, which would be [tex]|a|=|\frac{5-1}{2}| =|\frac{4}{2}| =|2|=2[/tex].
The value of [tex]b[/tex] in [tex]\frac{2\pi}{|b|}[/tex] represents the length of the period, so since the length of the period is [tex]2\pi[/tex], this means that [tex]b=\frac{2\pi}{|2\pi|} =1[/tex].
The phase shift, [tex]-\frac{c}{b}[/tex], describes the horizontal shift of a function. Because the phase shift is [tex]-\frac{\pi}{2}[/tex], then we can set up the equation [tex]-\frac{\pi}{2}=-\frac{c}{1}[/tex] where we determine [tex]c=\frac{\pi}{2}[/tex].
The midline (or vertical shift), [tex]d[/tex], is the horizontal line that passes through between the maximum and minimum points, which the function oscillates. In this case, the midline would be located at the line [tex]y=3[/tex], therefore, [tex]d=3[/tex].
Putting all our information together, your final equation is:
[tex]f(x)=2cos(x+\frac{\pi}{2})+3[/tex]
