Question

the graph of a cosine function is drawn. one full cycle goes from x = 0 to x = 1 and the low point on that cycle is (\frac{1}{2}, -5). which of the following functions could have this graph?\
\bigcirc y = -5 \cdot \cos 2\pi x\
\bigcirc y = -5 \cdot \cos \pi x\
\bigcirc y = 5 \cdot \cos \pi x\
\bigcirc y = 5 \cdot \cos 2\pi x

Explanation:

Step1: Recall the general form of cosine function

The general form of a cosine function is \(y = A\cos(Bx)\), where \(A\) is the amplitude (\(\vert A\vert\) gives the maximum distance from the mid - line) and the period \(T=\frac{2\pi}{B}\).

Step2: Analyze the amplitude

Since the low - point is \((\frac{1}{2},- 5)\), the amplitude \(A=-5\) (because the mid - line of \(y = \cos x\) is \(y = 0\) and the vertical stretch/compression factor \(A\) affects the maximum and minimum values. The formula for the maximum/minimum of \(y=A\cos(Bx)\) is \(y = A\) (when \(\cos(Bx)=1\)) and \(y=-A\) (when \(\cos(Bx)= - 1\)). Here, the minimum value is \(-5\), so \(A=-5\)).

Step3: Analyze the period

The period of a cosine function \(y = A\cos(Bx)\) is \(T=\frac{2\pi}{B}\). We know that one full - cycle goes from \(x = 0\) to \(x = 1\), so \(T = 1\). Using the formula \(T=\frac{2\pi}{B}\), we have \(1=\frac{2\pi}{B}\), then \(B = 2\pi\).

Answer:

\(y=-5\cdot\cos(2\pi x)\)