Can you tell me what each part does in this function f(x)=2tan(x+(pi/2))
also how would i go about graphing this f(x)=(1/2)sec(x-(pi/6)). DO i change the sec into 1/cos and graph it from there?
Given $\displaystyle y = a \tan k (x - b) + c$ or $\displaystyle y = a \cot k(x - b) + c$ the following are true:
$\displaystyle \bold {a}$ - a constant, doesn't really do much when graphing these functions, for some other trig ratios, $\displaystyle |a|$ gives the amplitude, that is not applicable with the tangent or cotangent
$\displaystyle \bold {Period} = \frac {\pi}{|k|}$
$\displaystyle \bold {Phase Shift} = b$
$\displaystyle \bold {Vertical Shift} = c$
provided you know the definitions of all these terms, these rules aid you in graphing the trig function
Given $\displaystyle y = a \sec k(x - b) + c$ or $\displaystyle y = a \csc k(x - b) + c$, the following are true:
Pretty much everything above i sthe same, except the period. for these functions, the period is $\displaystyle \frac {2 \pi}{|k|}$
The graph of these functions will look like their original functions, except, you have to apply these conditions (that is, you may need to change the period by stretching the graph, or whatever)
Hello, johntuan!
Here's my baby-talk version of these translations . . .
Can you tell me what each part does in this function?
. . $\displaystyle f(x)\;=\;2\tan\left(x + \frac{\pi}{2}\right)$
The coefficient (2) "stretches" the graph vertically.
. . Points above the x-axis are twice as high.
. . Points below the x-axis are twice as low.
A number add to or subtracted from the $\displaystyle x$ is a phase-shift.
It shifts the entire graph horizonally in the opposite direction.
. . We have: $\displaystyle +\,\frac{\pi}{2}$; the graph is shifted $\displaystyle \frac{\pi}{2}$ units to the left.
We're expected to know the graph for $\displaystyle f(x) \:=\:\tan x$
Sketch it, then make the required modifications.
Also how would i go about graphing this?
. . $\displaystyle f(x) \;=\;\frac{1}{2}\sec\left(x - \frac{\pi}{6}\right)$
We are expected to know the graph of $\displaystyle f(x) \,=\,\sec(x)$
The coefficient $\displaystyle \left(\frac{1}{2}\right)$ "shrinks" the graph vertically.
. . All points are "half as high" or "half as low".
The phase shift is $\displaystyle \frac{\pi}{6}$ unit to the right.