Question

How do you graph $y = \sin 2x + 1$ ?

Answer 1

The question is the form of $y = a\sin \left( {bx - c} \right) + d$. Here, we will find the amplitude, periods, phase shift, and vertical shift. The formula of amplitude is $\left| a \right|$, the formula of the period is $\dfrac{{2\pi }}{{\left| b \right|}}$ , the formula of phase shift is $\dfrac{c}{b}$ , and the formula of vertical shift is d. Find all the above values. Then take different values of x and find the values of y. After that, graph the trigonometric function using amplitude, period, phase shift, vertical shift, and the points.

Complete step by step answer:
In this question, the trigonometric function is given.
$\Rightarrow y = \sin 2x + 1$
Compare the above equation with $y = a\sin \left( {bx - c} \right) + d$. And find all the values.
The value of ‘a’ is 1.
The value of ‘b’ is 2.
The value of ‘c’ is 0.
The value of ‘d’ is 1.
Now, let us find the value of amplitude.
As we already know, the value of the amplitude is $\left| a \right|$.
Therefore,
$\Rightarrow amplitude = \left| a \right|$

Put the value of ‘a’ is 1.
$\Rightarrow amplitude = \left| 1 \right|$
$\Rightarrow amplitude = 1$
Hence, the amplitude is 1.
Now, let us find the value of the period.
As we already know, the value of the period is$\dfrac{{2\pi }}{{\left| b \right|}}$.
Therefore,
$\Rightarrow period = \dfrac{{2\pi }}{{\left| b \right|}}$

Put the value of ‘b’ is 2.
$\Rightarrow period = \dfrac{{2\pi }}{{\left| 2 \right|}}$
$\Rightarrow period = \dfrac{{2\pi }}{2}$
Cut the common factors from the numerator and the denominator.
$\Rightarrow period = \pi$
Hence, the period is$\pi$.
Now, let us find the value of phase shift.
As we already know, the value of phase shift is$\dfrac{c}{b}$.
Therefore,
$\Rightarrow phase - shift = \dfrac{c}{b}$
Put the value of ‘b’ is 2, and the value of ‘c’ is 0.
$\Rightarrow phase - shift = \dfrac{0}{2}$
Therefore,
$\Rightarrow phase - shift = 0$
Hence, the phase shift is 0.
Now, let us find the value of vertical shift.
As we already know, the value of vertical shift is d.
Therefore,
$\Rightarrow vertical - shift = d$

Put the value of ‘d’ is 1.
$\Rightarrow vertical - shift = 1$
Hence, the vertical shift is 1.
Now, let us find the points.
For $x = 0$
$\Rightarrow y = \sin 2x + 1$
$\Rightarrow y = \sin 2\left( 0 \right) + 1$
That is
$\Rightarrow y = \sin 0 + 1$
As we already know, $\sin 0 = 0$.
So,
$\Rightarrow y = 1$
For $x = \dfrac{\pi }{4}$
$\Rightarrow y = \sin 2\left( {\dfrac{\pi }{4}} \right) + 1$
That is
$\Rightarrow y = \sin \left( {\dfrac{\pi }{2}} \right) + 1$
As we already know, $\sin \dfrac{\pi }{2} = 1$.
So,
$\Rightarrow y = 2$
For $x = \dfrac{\pi }{2}$
$\Rightarrow y = \sin 2\left( {\dfrac{\pi }{2}} \right) + 1$
That is
$\Rightarrow y = \sin \pi + 1$
As we already know, $\sin \pi = 0$.
So,
$\Rightarrow y = 1$
For $x = \dfrac{{3\pi }}{4}$
$\Rightarrow y = \sin 2\left( {\dfrac{{3\pi }}{4}} \right) + 1$
That is
$\Rightarrow y = \sin \dfrac{{3\pi }}{2} + 1$
As we already know, $\sin \dfrac{{3\pi }}{2} = - 1$.
So,
$\Rightarrow y = 0$
For $x = \pi$
$\Rightarrow y = \sin 2\left( \pi \right) + 1$
That is
$\Rightarrow y = \sin 2\pi + 1$
As we already know, $\sin 2\pi = 0$.
So,
$\Rightarrow y = 1$

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

Note: We have to remember the trigonometry formula and trigonometry ratio values at the angle $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ$. And we have to remember the trigonometric function values in all quadrants.
In the first quadrant, sine and cosine functions are positive.
In the second quadrant, the sine function is positive and the cosine function is negative.
In the third quadrant, sine and cosine functions are negative.
In the fourth quadrant, the cosine function is positive and the sine function is negative.