Precalculus Examples

$y = \arcsin (- \frac{1}{2} x)$

Step 1

Select a few points to graph.

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Step 1.1

Find the point at $x = - 2$ .

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Step 1.1.1

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

$f (- 2) = \arcsin (- \frac{- 2}{2})$

Step 1.1.2

Simplify the result.

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Step 1.1.2.1

Divide $- 2$ by $2$ .

$f (- 2) = \arcsin (1)$

Step 1.1.2.2

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

$f (- 2) = \frac{π}{2}$

Step 1.1.2.3

The final answer is $\frac{π}{2}$ .

$\frac{π}{2}$

Step 1.2

Find the point at $x = - 1$ .

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Step 1.2.1

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

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

Step 1.2.2

Simplify the result.

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Step 1.2.2.1

Move the negative in front of the fraction.

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

Step 1.2.2.2

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$f (- 1) = \frac{π}{6}$

Step 1.2.2.3

The final answer is $\frac{π}{6}$ .

$\frac{π}{6}$

Step 1.3

Find the point at $x = 0$ .

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Step 1.3.1

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

$f (0) = \arcsin (- \frac{0}{2})$

Step 1.3.2

Simplify the result.

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Step 1.3.2.1

Divide $0$ by $2$ .

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

Step 1.3.2.2

Multiply $- 1$ by $0$ .

$f (0) = \arcsin (0)$

Step 1.3.2.3

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

$f (0) = 0$

Step 1.3.2.4

The final answer is $0$ .

$0$

Step 1.4

Find the point at $x = 1$ .

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Step 1.4.1

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

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

Step 1.4.2

Simplify the result.

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Step 1.4.2.1

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$f (1) = - \frac{π}{6}$

Step 1.4.2.2

The final answer is $- \frac{π}{6}$ .

$- \frac{π}{6}$

Step 1.5

Find the point at $x = 2$ .

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Step 1.5.1

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

$f (2) = \arcsin (- \frac{2}{2})$

Step 1.5.2

Simplify the result.

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Step 1.5.2.1

Cancel the common factor of $2$ .

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Step 1.5.2.1.1

Cancel the common factor.

$f (2) = \arcsin (- \frac{2}{2})$

Step 1.5.2.1.2

Rewrite the expression.

$f (2) = \arcsin (- 1 \cdot 1)$

Step 1.5.2.2

Multiply $- 1$ by $1$ .

$f (2) = \arcsin (- 1)$

Step 1.5.2.3

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

$f (2) = - \frac{π}{2}$

Step 1.5.2.4

The final answer is $- \frac{π}{2}$ .

$- \frac{π}{2}$

Step 1.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 2 & \frac{π}{2} \\ - 1 & \frac{π}{6} \\ 0 & 0 \\ 1 & - \frac{π}{6} \\ 2 & - \frac{π}{2} \end{array}$

Step 2

The trig function can be graphed using the points.

$\begin{array}{c} x & f (x) \\ - 2 & \frac{π}{2} \\ - 1 & \frac{π}{6} \\ 0 & 0 \\ 1 & - \frac{π}{6} \\ 2 & - \frac{π}{2} \end{array}$

Step 3