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anonymous
 one year ago
Which of the following would best represent a cosine function with an amplitude of 3, a period of pi over 2, and a midline at y = −4?
f(x) = −4 cos 4x + 3
f(x) = 3 cos(x − pi over 2) − 4
f(x) = 4 cos(x − pi over 2) + 3
f(x) = 3 cos 4x − 4
anonymous
 one year ago
Which of the following would best represent a cosine function with an amplitude of 3, a period of pi over 2, and a midline at y = −4? f(x) = −4 cos 4x + 3 f(x) = 3 cos(x − pi over 2) − 4 f(x) = 4 cos(x − pi over 2) + 3 f(x) = 3 cos 4x − 4

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The amplitude is the number before the trig function (sine, cosine, etc.). So we want a 3 before the cos in this case.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It will usually be written out like this \[\large acos(bxc)+d\\\large amplitude~=~a\\\\\large period~=~\frac{2pi}{b}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The midline is the point halfway between the top and bottom of the wave. The midline of a cos function is usually 0. So to change it to 4 smaller, we have to make the vertical change (the d in my example) equal 4. So set d=4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wrong. Did you just guess?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nevermind i figured it out thanks anyways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\large amplitude=a\\\large amplitude=3\\\large a=3\] So in \(\large acos(bx)+d\), \(\large a=3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So did you get the answer then?
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