Graph g(x)=4sin(pix)-3

$g (x) = 4 \sin (π x) - 3$

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

$a = 4$

$b = π$

$c = 0$

$d = - 3$

Find the amplitude $| a |$ .

Amplitude: $4$

Find the period of $4 \sin (π x)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| π |}$

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Cancel the common factor of $π$ .

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Cancel the common factor.

$\frac{2 π}{π}$

Divide $2$ by $1$ .

$2$

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: $\frac{0}{π}$

Divide $0$ by $π$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $- 3$

List the properties of the trigonometric function.

Amplitude: $4$

Period: $2$

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

Vertical Shift: $- 3$

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.

$f (0) = 4 \sin (π (0)) - 3$

Simplify the result.

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Simplify each term.

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Multiply $π$ by $0$ .

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

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

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

Multiply $4$ by $0$ .

$f (0) = 0 - 3$

Subtract $3$ from $0$ .

$f (0) = - 3$

The final answer is $- 3$ .

$- 3$

Find the point at $x = \frac{1}{2}$ .

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Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Simplify the result.

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Simplify each term.

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Combine $π$ and $\frac{1}{2}$ .

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

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

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

Multiply $4$ by $1$ .

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

Subtract $3$ from $4$ .

$f (\frac{1}{2}) = 1$

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.

$f (1) = 4 \sin (π (1)) - 3$

Simplify the result.

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Simplify each term.

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Multiply $π$ by $1$ .

$f (1) = 4 \sin (π) - 3$

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

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

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

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

Multiply $4$ by $0$ .

$f (1) = 0 - 3$

Subtract $3$ from $0$ .

$f (1) = - 3$

The final answer is $- 3$ .

$- 3$

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

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

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

Simplify the result.

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Simplify each term.

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Combine $π$ and $\frac{3}{2}$ .

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

Move $3$ to the left of $π$ .

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

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.

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

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

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

Multiply $- 1$ by $1$ .

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

Multiply $4$ by $- 1$ .

$f (\frac{3}{2}) = - 4 - 3$

Subtract $3$ from $- 4$ .

$f (\frac{3}{2}) = - 7$

The final answer is $- 7$ .

$- 7$

Find the point at $x = 2$ .

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

$f (2) = 4 \sin (π (2)) - 3$

Simplify the result.

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Simplify each term.

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Move $2$ to the left of $π$ .

$f (2) = 4 \sin (2 \cdot π) - 3$

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

$f (2) = 4 \sin (0) - 3$

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

$f (2) = 4 \cdot 0 - 3$

Multiply $4$ by $0$ .

$f (2) = 0 - 3$

Subtract $3$ from $0$ .

$f (2) = - 3$

The final answer is $- 3$ .

$- 3$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & - 3 \\ \frac{1}{2} & 1 \\ 1 & - 3 \\ \frac{3}{2} & - 7 \\ 2 & - 3 \end{array}$

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

Amplitude: $4$

Period: $2$

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

Vertical Shift: $- 3$

$\begin{array}{c} x & f (x) \\ 0 & - 3 \\ \frac{1}{2} & 1 \\ 1 & - 3 \\ \frac{3}{2} & - 7 \\ 2 & - 3 \end{array}$

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