How do you graph $ y = 5 + \sin (\dfrac{1}{2}x) $ ?

Hint: We know that the general form of the sine function is $ y = A\sin (Bx + C) + D $ where
A is the amplitude, that is, A tells us the value of the peaks of the function.
B is the frequency, that is, B tells us the number of oscillations a function does in a fixed interval.
C and D denote the horizontal and vertical shift of a function respectively. The value of C and D is zero for the given function, so there is no horizontal or vertical shift.
On comparing the given equation with the standard form, we can find the graph of this equation.

Complete step-by-step answer:
We are given that $ y = 5 + \sin (\dfrac{1}{2}x) $ and we know that the standard form is $ y = A\sin (Bx + C) + D $ ,
Comparing these two equations, we get –
 $ A = 1,\,B = \dfrac{1}{2},\,C = 0\,and\,D = 5 $
Thus, the given function has a vertical shift of 5, and the amplitude of this function is 1, so the given sine function has peak values at 1+5=6 and -1+5=4, that is, it oscillates between 6 and 4 and the given sine function completes half oscillations between 0 and $ 2\pi $ , but for plotting the graph, we have to find out the period of the function. Period of a sine function is –
 $
  p = \dfrac{{2\pi }}{B} \\
   \Rightarrow p = \dfrac{{2\pi }}{{\dfrac{1}{2}}} = 4\pi \;
  $
That is the given function repeats the oscillation after every $ 4\pi $ radians, it completes 1 oscillation in the interval 0 to $ 4\pi $ .
Now, we know all the quantities for plotting the graph of the given function, so the graph of $ y = 5 + \sin (\dfrac{1}{2}x) $ is plotted as follows –


Note: In this question, we are given a sine function and we have to plot its graph. To plot the graph of a function, we have to first convert the function to the standard form, and then we compare it with the standard equation. This way we can find out how the graph of any should be plotted and thus can be plotted easily.