Answer:
○ [tex]\displaystyle f(x) = 10cos\:\frac{\pi}{2}x + 10[/tex]
Step-by-step explanation:
[tex]\displaystyle \boxed{f(x) = 10sin\:(\frac{\pi}{2}x + \frac{\pi}{2}) + 10} \\ \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 10 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-1} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{2}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{4} \hookrightarrow \frac{2}{\frac{\pi}{2}}\pi \\ Amplitude \hookrightarrow 10[/tex]
OR
[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 10 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{4} \hookrightarrow \frac{2}{\frac{\pi}{2}}\pi \\ Amplitude \hookrightarrow 10[/tex]
The cosine graph in the photograph on the right is the OPPOCITE of the cosine graph in the photograph on the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.
**Moreover, both graphs have the same maximum and minimum value, but we are looking for the equation that will not reflect over the midline of [tex]\displaystyle y = 10,[/tex]which means we do NOT want a negative in front of the amplitude value, therefore we have this equation:
[tex]\displaystyle f(x) = 10cos\:\frac{\pi}{2}x + 10[/tex]
If you are having trouble figuring out which graph is which, just remember that the parent function of [tex]\displaystyle y = cos\:x[/tex]never touches the origin unless it shifts one unit south or gets reflected from somewhere. In this case, the photograph on the left touches the origin because of a reflection [negative symbol], therefore the photograph on the right matches the equation selected.
I am delighted to assist you at any time.
