which function is graphed below?\no $f(x)=-\\cos(x)$\no $f(x)=\\cos(x)$\no $f(x)=\\sin(x)$\no $f(x)=-\\sin(x)$

Explanation:

Step1: Recall key - points of trigonometric functions

The general form of a sine function is $y = A\sin(Bx - C)+D$ and for cosine is $y = A\cos(Bx - C)+D$. The standard sine function $y=\sin(x)$ has a value of $0$ at $x = 0$, and the standard cosine function $y = \cos(x)$ has a value of $1$ at $x = 0$.

Step2: Evaluate the function at $x = 0$

For the given graph, when $x = 0$, $y=0$.

• For $y =-\cos(x)$, when $x = 0$, $y=-\cos(0)=- 1$.
• For $y=\cos(x)$, when $x = 0$, $y=\cos(0)=1$.
• For $y=\sin(x)$, when $x = 0$, $y=\sin(0)=0$.
• For $y =-\sin(x)$, when $x = 0$, $y=-\sin(0)=0$.

Step3: Check the slope at $x = 0$

The derivative of $y=\sin(x)$ is $y'=\cos(x)$, and at $x = 0$, $y'(0)=\cos(0)=1$ (positive slope). The derivative of $y =-\sin(x)$ is $y'=-\cos(x)$, and at $x = 0$, $y'(0)=-\cos(0)=-1$ (negative slope). The graph has a positive - slope at $x = 0$.

Answer:

$f(x)=\sin(x)$