Precalculus Examples

$y = \tan (x - \frac{π}{6})$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = \tan (x - \frac{π}{6})$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $\tan (x - \frac{π}{6}) = 0$ .

$\tan (x - \frac{π}{6}) = 0$

Step 1.2.2

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x - \frac{π}{6} = \arctan (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

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

$x - \frac{π}{6} = 0$

Step 1.2.4

Add $\frac{π}{6}$ to both sides of the equation.

$x = \frac{π}{6}$

Step 1.2.5

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x - \frac{π}{6} = π + 0$

Step 1.2.6

Solve for $x$ .

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Step 1.2.6.1

Add $π$ and $0$ .

$x - \frac{π}{6} = π$

Step 1.2.6.2

Move all terms not containing $x$ to the right side of the equation.

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Step 1.2.6.2.1

Add $\frac{π}{6}$ to both sides of the equation.

$x = π + \frac{π}{6}$

Step 1.2.6.2.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} + \frac{π}{6}$

Step 1.2.6.2.3

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} + \frac{π}{6}$

Step 1.2.6.2.4

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 + π}{6}$

Step 1.2.6.2.5

Simplify the numerator.

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Step 1.2.6.2.5.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π + π}{6}$

Step 1.2.6.2.5.2

Add $6 π$ and $π$ .

$x = \frac{7 π}{6}$

Step 1.2.7

Find the period of $\tan (x - \frac{π}{6})$ .

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Step 1.2.7.1

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

$\frac{π}{| b |}$

Step 1.2.7.2

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

$\frac{π}{| 1 |}$

Step 1.2.7.3

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

$\frac{π}{1}$

Step 1.2.7.4

Divide $π$ by $1$ .

$π$

Step 1.2.8

The period of the $\tan (x - \frac{π}{6})$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \frac{π}{6} + π n, \frac{7 π}{6} + π n$ , for any integer $n$

Step 1.2.9

Consolidate the answers.

$x = \frac{π}{6} + π n$ , for any integer $n$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(\frac{π}{6} + π n, 0)$ , for any integer $n$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = \tan ((0) - \frac{π}{6})$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = \tan (0 - \frac{π}{6})$

Step 2.2.2

Remove parentheses.

$y = \tan ((0) - \frac{π}{6})$

Step 2.2.3

Simplify $\tan ((0) - \frac{π}{6})$ .

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Step 2.2.3.1

Subtract $\frac{π}{6}$ from $0$ .

$y = \tan (- \frac{π}{6})$

Step 2.2.3.2

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

$y = \tan (\frac{11 π}{6})$

Step 2.2.3.3

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

$y = - \tan (\frac{π}{6})$

Step 2.2.3.4

The exact value of $\tan (\frac{π}{6})$ is $\frac{\sqrt{3}}{3}$ .

$y = - \frac{\sqrt{3}}{3}$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - \frac{\sqrt{3}}{3})$

Step 3

List the intersections.

x-intercept(s): $(\frac{π}{6} + π n, 0)$ , for any integer $n$

y-intercept(s): $(0, - \frac{\sqrt{3}}{3})$

Step 4