calculator\nwhich function is shown on the graph?\n$f(x)=\frac{1}{2}cos x$\n$f(x)=-\frac{1}{2}sin…

Explanation:

Step1: Recall properties of sine and cosine

The general form of a sine - function is $y = A\sin(Bx - C)+D$ and of a cosine - function is $y = A\cos(Bx - C)+D$, where $A$ is the amplitude, $B$ affects the period ($T=\frac{2\pi}{|B|}$), $C$ is the phase - shift, and $D$ is the vertical shift. For the given options, $B = 1$, $C = 0$, and $D = 0$.

Step2: Check the amplitude

The amplitude of a function $y = A\sin x$ or $y = A\cos x$ is $|A|$. The maximum value of the given graph is $\frac{1}{2}$ and the minimum value is $-\frac{1}{2}$, so $|A|=\frac{1}{2}$.

Step3: Evaluate at $x = 0$

Evaluate each function at $x = 0$.

• For $y=\frac{1}{2}\cos x$, when $x = 0$, $y=\frac{1}{2}\cos(0)=\frac{1}{2}\times1=\frac{1}{2}$.
• For $y =-\frac{1}{2}\sin x$, when $x = 0$, $y=-\frac{1}{2}\sin(0)=0$.
• For $y =-\frac{1}{2}\cos x$, when $x = 0$, $y=-\frac{1}{2}\cos(0)=-\frac{1}{2}$.
• For $y=\frac{1}{2}\sin x$, when $x = 0$, $y=\frac{1}{2}\sin(0)=0$. The graph passes through the point $(0,-\frac{1}{2})$.

Answer:

C. $f(x)=-\frac{1}{2}\cos x$