# Find the period y= 9 \tan \left( x - \frac {\pi}{4} \right)

## Question:

Find the period {eq}y= 9 \tan \left( x - \frac {\pi}{4} \right) {/eq}

## Period in a Tangent-Based Function:

When the function incorporates tangent in the transformation such as {eq}f(x) = a\tan(k(x+c))+d {/eq}, the period can be found by dividing {eq}\pi {/eq} by {eq}k {/eq} or {eq}\frac{\pi}{k} {/eq} because each tangent wave repeats itself after a cycle of {eq}\pi {/eq}.

## Answer and Explanation:

Given: {eq}y= 9 \tan(x-\frac{\pi}{4}) {/eq}

From the form {eq}f(x) = a\tan(k(x+c))+d {/eq}, the period of the function can be found from the formula {eq}\frac{\pi}{k} {/eq}. By distribution, it can be seen that the {eq}k {/eq} value in the equation {eq}y= 9 \tan(x-\frac{\pi}{4}) {/eq} is {eq}1 {/eq} since {eq}k {/eq} serves as a coefficient of {eq}x {/eq}.

Therefore, the period:

{eq}\begin{align*} \text{ Period } &= \frac{\pi}{1} \\ &= \pi \\ \end{align*} {/eq}

Therefore, the period of the equation is {eq}\pi {/eq}.

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from High School Precalculus: Help and Review

Chapter 21 / Lesson 8