Hello S2Krazy Originally Posted by
S2Krazy I was wondering how do i do the phase shift for
all i know is:
??
≤ 3x
≤ ??
The graph of $\displaystyle y = \tan(\theta+ \tfrac{\pi}{6})$ is the same as the graph of $\displaystyle y =\tan(\theta)$, shifted a distance $\displaystyle \tfrac{\pi}{6}$ to the left.
If you want to see why, just look at the zero values. When $\displaystyle \theta = 0, y=\tan(\theta)=\tan(0) = 0$. So the graph 'starts' at $\displaystyle (0,0)$.
But when $\displaystyle \theta=0, y=\tan(\theta+\tfrac{\pi}{6})=\tan(\tfrac{\pi}{6})$. So this graph 'starts' at $\displaystyle (0,\tfrac{\pi}{6})$, and it's when $\displaystyle \theta = -\tfrac{\pi}{6}$ that we get the value $\displaystyle y = 0$.
Now if we multiply the independent variable by $\displaystyle 3$, and replace $\displaystyle \theta$ with $\displaystyle 3x$, we get the graphs of $\displaystyle y = \tan(3x)$ and $\displaystyle y = \tan(3x+\tfrac{\pi}{6})$. Now, in the second graph, we shall need to take $\displaystyle 3x = -\tfrac{\pi}{6}$ to get our zero value of $\displaystyle y$. So that's $\displaystyle x = -\tfrac{\pi}{18}$. So the graph has now been shifted $\displaystyle \tfrac{\pi}{18}$ to the left.
Finally, note that multiplying by $\displaystyle 5$, to give $\displaystyle y = 5\tan(3x+\tfrac{\pi}{6})$, makes no difference to the phase-shift: it merely 'stretches' the graph in a vertical direction either side of the x-axis by a factor of $\displaystyle 5$.
Grandad