Question

How do you graph $y = 2\sin \left( {2x + \dfrac{\pi }{2}} \right) + 3$ ?

Answer 1

Hint : A graph of a function f is the set of ordered pairs; the equation of the graph is generally represented as $y = f\left( x \right)$ , where x and $f\left( x \right)$ are real numbers. We substitute the value of x and we determine the value of y and then we mark the points in the graph and we join the points.

Complete step by step solution:
Here, in the given question, we have to plot the graph for the given function. The values of x and y that satisfy the given function are known as the solutions. These pairs are in the form of cartesian coordinates.
First, we have to find the value of y by using the graph equation $y = 2\sin \left( {2x + \dfrac{\pi }{2}} \right) + 3$ .
Let us substitute the value of x as $0$ .
$\Rightarrow y = 2\sin \left( {2\left( 0 \right) + \dfrac{\pi }{2}} \right) + 3$
$\Rightarrow y = 2\sin \left( {\dfrac{\pi }{2}} \right) + 3$
Now, we know that the value of $\sin \left( {\dfrac{\pi }{2}} \right)$ is $1$ .
$\Rightarrow y = 2 + 3$
$\Rightarrow y = 5$
Now we consider the value of x as $\dfrac{\pi }{2}$ , the value of y is
$\Rightarrow y = 2\sin \left( {2\left( {\dfrac{\pi }{2}} \right) + \dfrac{\pi }{2}} \right) + 3$
$\Rightarrow y = 2\sin \left( {\pi + \dfrac{\pi }{2}} \right) + 3$
We know that the value of $\sin \left( {\dfrac{{3\pi }}{2}} \right)$ is $- 1$ .
$\Rightarrow y = 2\left( { - 1} \right) + 3$
$\Rightarrow y = 1$
Now we consider the value of x as $\pi$ , the value of y is
$\Rightarrow y = 2\sin \left( {2\pi + \dfrac{\pi }{2}} \right) + 3$
$\Rightarrow y = 2\sin \left( {\dfrac{{5\pi }}{2}} \right) + 3$
We know that $\sin \left( {\dfrac{{5\pi }}{2}} \right)$ is equal to $1$ . So, we get,
$\Rightarrow y = 2\left( 1 \right) + 3$
$\Rightarrow y = 5$
Now we consider the value of x as $\dfrac{\pi }{4}$ , the value of y is
$\Rightarrow y = 2\sin \left( {2\left( {\dfrac{\pi }{4}} \right) + \dfrac{\pi }{2}} \right) + 3$
$\Rightarrow y = 2\sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{2}} \right) + 3$
$\Rightarrow y = 2\sin \left( \pi \right) + 3$
Value of $\sin \left( \pi \right)$ is zero. So, we get,
$\Rightarrow y = 2\left( 0 \right) + 3$
$\Rightarrow y = 3$
Now, we draw a table for these values we have

X	$0$	$\dfrac{\pi }{2}$	$\pi$	$\dfrac{\pi }{4}$
y	$5$	$1$	$5$	$3$

We also know the nature of the graph of sine function. Hence, we can now plot the graph of the given function $y = 2\sin \left( {2x + \dfrac{\pi }{2}} \right) + 3$ . The graph plotted for these points is represented below:

Note : The sine function can be represented by the general equation $y = a\sin \left( {kx + \phi } \right)$ . There are various parameters in this equation such as the amplitude, period and phase shift of the sine function. The value ‘a’ is the amplitude of the sine function $y = a\sin \left( {kx + \phi } \right)$ . The period of the sine function can be calculated as $\left( {\dfrac{{2\pi }}{k}} \right)$ as the value of the function repeats after regular interval of $\left( {\dfrac{{2\pi }}{k}} \right)$ radians. Also, if there is a constant added in the function like $y = a\sin \left( {kx + \phi } \right) + p$ , then the graph of the function moves p units vertically upwards due to the upward shift.