Graph f(x)=-1+3sin(pi/2x)

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

Rewrite the expression as $3 \sin (\frac{π x}{2}) - 1$ .

$3 \sin (\frac{π x}{2}) - 1$

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

$a = 3$

$b = \frac{π}{2}$

$c = 0$

$d = - 1$

Find the amplitude $| a |$ .

Amplitude: $3$

Find the period of $3 \sin (\frac{π x}{2})$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $\frac{π}{2}$ in the formula for period.

$\frac{2 π}{| \frac{π}{2} |}$

$\frac{π}{2}$ is approximately $1.57079632$ which is positive so remove the absolute value

$\frac{2 π}{\frac{π}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$2 π \frac{2}{π}$

Cancel the common factor of $π$ .

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Factor $π$ out of $2 π$ .

$π \cdot 2 \frac{2}{π}$

Cancel the common factor.

$π \cdot 2 \frac{2}{π}$

Rewrite the expression.

$2 \cdot 2$

Multiply $2$ by $2$ .

$4$

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

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

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

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

Phase Shift: $\frac{0}{\frac{π}{2}}$

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $0 (\frac{2}{π})$

Multiply $0$ by $\frac{2}{π}$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $- 1$

List the properties of the trigonometric function.

Amplitude: $3$

Period: $4$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $- 1$

Select a few points to graph.

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Find the point at $x = 0$ .

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

$f (0) = - 1 + 3 \sin (\frac{π (0)}{2})$

Simplify the result.

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Simplify each term.

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Cancel the common factor of $0$ and $2$ .

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Factor $2$ out of $π (0)$ .

$f (0) = - 1 + 3 \sin (\frac{2 (π \cdot (0))}{2})$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (0) = - 1 + 3 \sin (\frac{2 (π \cdot (0))}{2 (1)})$

Cancel the common factor.

$f (0) = - 1 + 3 \sin (\frac{2 (π \cdot (0))}{2 \cdot 1})$

Rewrite the expression.

$f (0) = - 1 + 3 \sin (\frac{π \cdot (0)}{1})$

Divide $π \cdot (0)$ by $1$ .

$f (0) = - 1 + 3 \sin (π \cdot (0))$

Multiply $π$ by $0$ .

$f (0) = - 1 + 3 \sin (0)$

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

$f (0) = - 1 + 3 \cdot 0$

Multiply $3$ by $0$ .

$f (0) = - 1 + 0$

Add $- 1$ and $0$ .

$f (0) = - 1$

The final answer is $- 1$ .

$- 1$

Find the point at $x = 1$ .

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Replace the variable $x$ with $1$ in the expression.

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

Simplify the result.

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Simplify each term.

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Multiply $π$ by $1$ .

$f (1) = - 1 + 3 \sin (\frac{π}{2})$

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

$f (1) = - 1 + 3 \cdot 1$

Multiply $3$ by $1$ .

$f (1) = - 1 + 3$

Add $- 1$ and $3$ .

$f (1) = 2$

The final answer is $2$ .

$2$

Find the point at $x = 2$ .

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

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

Simplify the result.

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Simplify each term.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $π$ by $1$ .

$f (2) = - 1 + 3 \sin (π)$

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

$f (2) = - 1 + 3 \sin (0)$

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

$f (2) = - 1 + 3 \cdot 0$

Multiply $3$ by $0$ .

$f (2) = - 1 + 0$

Add $- 1$ and $0$ .

$f (2) = - 1$

The final answer is $- 1$ .

$- 1$

Find the point at $x = 3$ .

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Replace the variable $x$ with $3$ in the expression.

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

Simplify the result.

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Simplify each term.

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Move $3$ to the left of $π$ .

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

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.

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

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

$f (3) = - 1 + 3 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (3) = - 1 + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$f (3) = - 1 - 3$

Subtract $3$ from $- 1$ .

$f (3) = - 4$

The final answer is $- 4$ .

$- 4$

Find the point at $x = 4$ .

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Replace the variable $x$ with $4$ in the expression.

$f (4) = - 1 + 3 \sin (\frac{π (4)}{2})$

Simplify the result.

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Simplify each term.

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Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $π (4)$ .

$f (4) = - 1 + 3 \sin (\frac{2 (π \cdot (2))}{2})$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (4) = - 1 + 3 \sin (\frac{2 (π \cdot (2))}{2 (1)})$

Cancel the common factor.

$f (4) = - 1 + 3 \sin (\frac{2 (π \cdot (2))}{2 \cdot 1})$

Rewrite the expression.

$f (4) = - 1 + 3 \sin (\frac{π \cdot (2)}{1})$

Divide $π \cdot (2)$ by $1$ .

$f (4) = - 1 + 3 \sin (π \cdot (2))$

Move $2$ to the left of $π$ .

$f (4) = - 1 + 3 \sin (2 \cdot π)$

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

$f (4) = - 1 + 3 \sin (0)$

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

$f (4) = - 1 + 3 \cdot 0$

Multiply $3$ by $0$ .

$f (4) = - 1 + 0$

Add $- 1$ and $0$ .

$f (4) = - 1$

The final answer is $- 1$ .

$- 1$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & - 1 \\ 1 & 2 \\ 2 & - 1 \\ 3 & - 4 \\ 4 & - 1 \end{array}$

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

Amplitude: $3$

Period: $4$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $- 1$

$\begin{array}{c} x & f (x) \\ 0 & - 1 \\ 1 & 2 \\ 2 & - 1 \\ 3 & - 4 \\ 4 & - 1 \end{array}$

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