Problem 18 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=-3 \tan \left(\frac{1}{3} x-\frac{\pi}{3}\right) $$

Step-by-Step Solution

Answer

Period: \( 3\pi \); Asymptotes every \( 3\pi \), beginning at \( 2\pi \).

Step 1: Identify the Basic Properties of the Tangent Function

The basic tangent function, \( \tan(x) \), has a period of \( \pi \). It has vertical asymptotes at odd multiples of \( \frac{\pi}{2} \), and it intercepts the origin (0,0) when plotted.

Step 2: Determine the Period of the Transformed Function

For the function \( y = -3 \tan\left( \frac{1}{3}x - \frac{\pi}{3} \right) \), the coefficient of \( x \) inside the tangent function affects the period. The formula for the period of \( \tan(bx) \) is \( \frac{\pi}{|b|} \). Here, \( b = \frac{1}{3} \), so the period is \( 3\pi \).

Step 3: Find the Phase Shift

The phase shift is calculated by setting the inside of the tangent function equal to zero: \( \frac{1}{3}x - \frac{\pi}{3} = 0 \). Solving for \( x \), we get \( x = \pi \). This means the graph is shifted \( \pi \) units to the right.

Step 4: Determine the Vertical Asymptotes

The standard vertical asymptotes of \( \tan(x) \) are at \( x = \frac{\pi}{2} + n\pi \). For \( y = -3 \tan\left( \frac{1}{3}x - \frac{\pi}{3} \right) \), solve \( \frac{1}{3}x - \frac{\pi}{3} = \frac{\pi}{2} + n\pi \). Solving this equation gives asymptotes at \( x = 2\pi + 3n\pi \), or each \( 3\pi \) apart.

Step 5: Sketch the Graph

Start by drawing the vertical asymptotes found in Step 4. The function has a period of \( 3\pi \), so sketch a tangent curve touching zero at \( x = \pi \) and being vertically stretched by a factor of -3. Remember it moves downward due to the negative coefficient. Reflect the standard tangent shape about the x-axis to account for the negative sign, and repeat this pattern along the x-axis, considering the phase shift and vertical stretch.

Key Concepts

Period of a Function

In trigonometry, the period of a function is the interval after which the function starts repeating its values. For the basic tangent function, \( \tan(x) \), the period is \( \pi \) because it repeats every \( \pi \) units along the x-axis.

When we have a transformed tangent function like \( \tan(bx) \), the period becomes \( \frac{\pi}{|b|} \). This is due to the fact that the coefficient \( b \) inside the function changes how frequently the pattern repeats.

• If \( b \) is greater than 1, the period decreases, meaning the function repeats more often.
• If \( b \) is less than 1, as in our function \( -3 \tan \left( \frac{1}{3} x - \frac{\pi}{3} \right) \), the period increases. For this function, \( b = \frac{1}{3} \), so the period is \( 3\pi \). This means the waves of the function extend over a larger interval before repeating.

Understanding the period is crucial for sketching the graph accurately and predicting the function's behavior over different intervals.

Phase Shift

The phase shift in trigonometric functions shifts the entire graph horizontally along the x-axis. It is determined by the horizontal displacement needed to set the inside of the function to zero.

For our function, \( y = -3 \tan\left( \frac{1}{3}x - \frac{\pi}{3} \right) \), we find the phase shift by solving \( \frac{1}{3}x - \frac{\pi}{3} = 0 \). Solving for \( x \) gives us \( x = \pi \), which means the graph is shifted \( \pi \) units to the right. Here’s how it works:

• Positive phase shifts indicate moving the graph to the right.
• Negative phase shifts would move it to the left.

This rightward shift alters where the tangent curve begins its repeating pattern, affecting how we sketch the graph over its period.

The phase shift does not change the period or the amplitude, but it’s essential for positioning the graph correctly along the x-axis.

Vertical Asymptotes

Vertical asymptotes in the graph of a tangent function occur where the function is undefined, or where it shoots up to infinity. In a basic \( \tan(x) \) function, these vertical asymptotes are at every odd multiple of \( \frac{\pi}{2} \) (e.g., \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots \)).

For transformed functions like \( y = -3 \tan\left( \frac{1}{3}x - \frac{\pi}{3} \right) \), finding the vertical asymptotes involves solving the equation inside the tangent function for these multiples. Here’s how you do it:

• Set \( \frac{1}{3}x - \frac{\pi}{3} = \frac{\pi}{2} + n\pi \), where \( n \) is an integer.
• Solving gives us asymptotes at \( x = 2\pi + 3n\pi \).

This tells us that the distance between each asymptote is consistent; they appear every \( 3\pi \) units.

Vertical asymptotes are key to sketching the graph as they define the boundaries between individual periods. Visualizing these boundaries helps in plotting the correct shape and behavior of the tangent function over its period.