Calculus Examples

$y = 2 \sin (x) + 3$

Step 1

Use the form $a \sin (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

$a = 2$

$b = 1$

$c = 0$

$d = 3$

Step 2

Find the amplitude $| a |$ .

Amplitude: $2$

Step 3

Find the period using the formula $\frac{2 π}{| b |}$ .

Tap for more steps...

Step 3.1

Find the period of $2 \sin (x)$ .

Tap for more steps...

Step 3.1.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.1.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2

Find the period of $3$ .

Tap for more steps...

Step 3.2.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.2.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$2 π$

Step 4

Find the phase shift using the formula $\frac{c}{b}$ .

Tap for more steps...

Step 4.1

The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Step 4.2

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{1}$

Step 4.3

Divide $0$ by $1$ .

Phase Shift: $0$

Step 5

List the properties of the trigonometric function.

Amplitude: $2$

Period: $2 π$

Phase Shift: None

Vertical Shift: $3$

Step 6

Select a few points to graph.

Tap for more steps...

Step 6.1

Find the point at $x = 0$ .

Tap for more steps...

Step 6.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 2 \sin (0) + 3$

Step 6.1.2

Simplify the result.

Tap for more steps...

Step 6.1.2.1

Simplify each term.

Tap for more steps...

Step 6.1.2.1.1

The exact value of $\sin (0)$ is $0$ .

$f (0) = 2 \cdot 0 + 3$

Step 6.1.2.1.2

Multiply $2$ by $0$ .

$f (0) = 0 + 3$

Step 6.1.2.2

Add $0$ and $3$ .

$f (0) = 3$

Step 6.1.2.3

The final answer is $3$ .

$3$

Step 6.2

Find the point at $x = \frac{π}{2}$ .

Tap for more steps...

Step 6.2.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 2 \sin (\frac{π}{2}) + 3$

Step 6.2.2

Simplify the result.

Tap for more steps...

Step 6.2.2.1

Simplify each term.

Tap for more steps...

Step 6.2.2.1.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = 2 \cdot 1 + 3$

Step 6.2.2.1.2

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2 + 3$

Step 6.2.2.2

Add $2$ and $3$ .

$f (\frac{π}{2}) = 5$

Step 6.2.2.3

The final answer is $5$ .

$5$

Step 6.3

Find the point at $x = π$ .

Tap for more steps...

Step 6.3.1

Replace the variable $x$ with $π$ in the expression.

$f (π) = 2 \sin (π) + 3$

Step 6.3.2

Simplify the result.

Tap for more steps...

Step 6.3.2.1

Simplify each term.

Tap for more steps...

Step 6.3.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (π) = 2 \sin (0) + 3$

Step 6.3.2.1.2

The exact value of $\sin (0)$ is $0$ .

$f (π) = 2 \cdot 0 + 3$

Step 6.3.2.1.3

Multiply $2$ by $0$ .

$f (π) = 0 + 3$

Step 6.3.2.2

Add $0$ and $3$ .

$f (π) = 3$

Step 6.3.2.3

The final answer is $3$ .

$3$

Step 6.4

Find the point at $x = \frac{3 π}{2}$ .

Tap for more steps...

Step 6.4.1

Replace the variable $x$ with $\frac{3 π}{2}$ in the expression.

$f (\frac{3 π}{2}) = 2 \sin (\frac{3 π}{2}) + 3$

Step 6.4.2

Simplify the result.

Tap for more steps...

Step 6.4.2.1

Simplify each term.

Tap for more steps...

Step 6.4.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (\frac{3 π}{2}) = 2 (- \sin (\frac{π}{2})) + 3$

Step 6.4.2.1.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{3 π}{2}) = 2 (- 1 \cdot 1) + 3$

Step 6.4.2.1.3

Multiply $2 (- 1 \cdot 1)$ .

Tap for more steps...

Step 6.4.2.1.3.1

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = 2 \cdot - 1 + 3$

Step 6.4.2.1.3.2

Multiply $2$ by $- 1$ .

$f (\frac{3 π}{2}) = - 2 + 3$

Step 6.4.2.2

Add $- 2$ and $3$ .

$f (\frac{3 π}{2}) = 1$

Step 6.4.2.3

The final answer is $1$ .

$1$

Step 6.5

Find the point at $x = 2 π$ .

Tap for more steps...

Step 6.5.1

Replace the variable $x$ with $2 π$ in the expression.

$f (2 π) = 2 \sin (2 π) + 3$

Step 6.5.2

Simplify the result.

Tap for more steps...

Step 6.5.2.1

Simplify each term.

Tap for more steps...

Step 6.5.2.1.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (2 π) = 2 \sin (0) + 3$

Step 6.5.2.1.2

The exact value of $\sin (0)$ is $0$ .

$f (2 π) = 2 \cdot 0 + 3$

Step 6.5.2.1.3

Multiply $2$ by $0$ .

$f (2 π) = 0 + 3$

Step 6.5.2.2

Add $0$ and $3$ .

$f (2 π) = 3$

Step 6.5.2.3

The final answer is $3$ .

$3$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 3 \\ \frac{π}{2} & 5 \\ π & 3 \\ \frac{3 π}{2} & 1 \\ 2 π & 3 \end{array}$

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Amplitude: $2$

Period: $2 π$

Phase Shift: None

Vertical Shift: $3$

$\begin{array}{c} x & f (x) \\ 0 & 3 \\ \frac{π}{2} & 5 \\ π & 3 \\ \frac{3 π}{2} & 1 \\ 2 π & 3 \end{array}$

Step 8