Question

trig function given features (sin/cos) l1
0/3 penalty: none
estion
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ite a cosine function that has an amplitude of 4, a midline of ( y = 5 ) and a period of ( \frac{7pi}{6} ).
answer attempt 4 out of 5

Explanation:

Step1: Recall the general form of a cosine function

The general form of a cosine function is \(y = A\cos(Bx - C)+D\), where \(A\) is the amplitude, \(D\) is the mid - line, the period \(T=\frac{2\pi}{|B|}\), and \(C\) is the phase shift (we assume \(C = 0\) since there is no information about phase shift).

Step2: Determine the value of \(A\)

Given that the amplitude \(|A|=4\). We can take \(A = 4\) (the sign of \(A\) affects the reflection, but without more information about the direction of the graph, we can choose \(A = 4\)).

Step3: Determine the value of \(D\)

Given that the mid - line \(y = D\), and \(y=5\) is the mid - line, so \(D = 5\).

Step4: Determine the value of \(B\)

We know that the period \(T=\frac{2\pi}{|B|}\). Given \(T=\frac{7\pi}{6}\). Then \(\frac{2\pi}{|B|}=\frac{7\pi}{6}\). Cross - multiply gives \(7\pi|B|=12\pi\). Divide both sides by \(\pi\): \(7|B| = 12\), so \(|B|=\frac{12}{7}\). We can take \(B=\frac{12}{7}\) (since the period formula only cares about the magnitude of \(B\) for non - phase - shifted functions in the form \(y = A\cos(Bx)+D\)).

Answer:

\(y = 4\cos(\frac{12}{7}x)+5\)