Graph f(x)=-sin(2x)

$f (x) = - \sin (2 x)$

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

$a = - 1$

$b = 2$

$c = 0$

$d = 0$

Find the amplitude $| a |$ .

Amplitude: $1$

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

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

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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: $\frac{0}{2}$

Divide $0$ by $2$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: $1$

Period: $π$

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

Vertical Shift: $0$

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.

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

Simplify the result.

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Multiply $2$ by $0$ .

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

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

$f (0) = - 0$

Multiply $- 1$ by $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{π}{4}$ .

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Replace the variable $x$ with $\frac{π}{4}$ in the expression.

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

Simplify the result.

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Cancel the common factor of $2$ .

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Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

$f (\frac{π}{4}) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = - 1$

The final answer is $- 1$ .

$- 1$

Find the point at $x = \frac{π}{2}$ .

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

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

Simplify the result.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

$f (\frac{π}{2}) = - \sin (π)$

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

$f (\frac{π}{2}) = - \sin (0)$

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

$f (\frac{π}{2}) = - 0$

Multiply $- 1$ by $0$ .

$f (\frac{π}{2}) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{3 π}{4}$ .

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Replace the variable $x$ with $\frac{3 π}{4}$ in the expression.

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

Simplify the result.

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Cancel the common factor of $2$ .

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Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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.

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

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

$f (\frac{3 π}{4}) = - (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{4}) = 1$

The final answer is $1$ .

$1$

Find the point at $x = π$ .

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

$f (π) = - \sin (2 (π))$

Simplify the result.

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

$f (π) = - \sin (0)$

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

$f (π) = - 0$

Multiply $- 1$ by $0$ .

$f (π) = 0$

The final answer is $0$ .

$0$

List the points in a table.

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

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

Amplitude: $1$

Period: $π$

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

Vertical Shift: $0$

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

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