184,216 views
34 votes
34 votes
I need help with this practice problem solving I will add an additional pic, *it is a graph that is a part of the problem, use the graph to answer**

I need help with this practice problem solving I will add an additional pic, *it is-example-1
User EinsA
by
2.8k points

1 Answer

21 votes
21 votes

Answer:

Step-by-step explanation:

Given the below sine function;


\begin{gathered} f(x)=\sin x+2 \\ \text{and} \\ \pi=3.14 \end{gathered}

Recall that a sine function has an up and down curve that repeats after every 360 degrees or 2pi radians.

Let's go ahead and choose different x-values from -2pi to 2pi and determine the corresponding f(x) values.

When x = -2pi


\begin{gathered} x=-2\pi=-2(3.14)=-6.28 \\ \therefore f(-2\pi)=\sin (-2\pi)+2=0+2=2 \end{gathered}

When x = -3pi/2;


\begin{gathered} x=(-3(3.142))/(2)=4.71 \\ \therefore f(-(3\pi)/(2))=\sin (-(3\pi)/(2))+2=1+2=3 \end{gathered}

When x = -pi;


\begin{gathered} x=-\pi=-3.14 \\ \therefore f(-\pi)=\sin (-\pi)+2=0+2=2 \end{gathered}

We can see the pattern now.

Let's go ahead and determine the point along the midline which will be at x = 0;


f(0)=\sin (0)+2=0+2=2

Since a sine graph stays between -1 and 1, given the midline to pass through 2, so the upper point will be 3 and the lower limit will be 1.

So plotting the graph, will have;

The above graph just shows a free-hand sketch of the sine function, you can use the Sine Tool to draw it better.

I need help with this practice problem solving I will add an additional pic, *it is-example-1
User Antonio Petricca
by
2.4k points