Question

VERTICAL ASYMPTOTES? AND CHECK MY WORK
y=tan(2x)-1
a. list two vertical asymptotes
b. period- pi/2
c. phase shift- none
d. vertical translation- down 1

Answer 1

y=wat ? wat is tan

Answer 2

its a trig question

Answer 3

b. period- pi/2 \(\checkmark\)
c. phase shift- none \(\checkmark\)
d. vertical translation- down 1 \(\checkmark\)

Answer 4

@jdoe0001 do you know how to find the vertical asymptotes, could you explain it

Answer 5

\(\bf tan(2x)=\cfrac{sin(2x)}{cos(2x)}\qquad \textit{let's say we use }x={\color{red}{ \frac{\pi}{4}}}
\\ \quad \\
tan(2x)=\cfrac{sin(2\cdot {\color{red}{ \frac{\pi}{4}}})}{cos(2\cdot {\color{red}{ \frac{\pi}{4}}})}

\implies tan(2x)=\cfrac{sin(2\cdot {\color{red}{ \frac{\pi}{4}}})}{cos(\frac{\pi}{2})}\quad \textit{so, what's }cos(\frac{\pi}{2})\quad ?\)

Answer 6

recall that for a rational, vertical asymptotes occur when the denominator turns to 0, thus making the rational "undefined"

so if you just set cos(2x) = 0 and find "x" angle, those are the spots where the vertical asymptotes will occur

Answer 7

im sorry that's a really good explanation im am just really really lost @jdoe0001

Answer 8

\(\bf tan(2x)=\cfrac{sin(2x)}{cos(2x)}\qquad \textit{let's say we use }x={\color{red}{ \frac{\pi}{4}}}
\\ \quad \\
tan(2x)=\cfrac{sin(2\cdot {\color{red}{ \frac{\pi}{4}}})}{cos(2\cdot {\color{red}{ \frac{\pi}{4}}})}

\implies tan(2x)=\cfrac{sin(2\cdot {\color{red}{ \frac{\pi}{4}}})}{cos(\frac{\pi}{2})}
\\ \quad \\
tan(2x)\cfrac{sin(2\cdot {\color{red}{ \frac{\pi}{4}}})}{0}\)

so let's use set the denominator to 0 and see what we get for the angle "x"

\(\bf cos(2x)=0\implies cos^{-1}[cos(2x)]=cos^{-1}(0)\implies 2x=cos^{-1}(0)
\\ \quad \\
x=\cfrac{cos^{-1}(0)}{2}\Leftarrow\textit{vertical asymptotes occur at those angles}\)

Answer 9

so would I plug the bottom part into my calculator?

Answer 10

you can just look at the Unit Circle, there are 2 conspicuous spots where the cosine is 0

Answer 11

90 degrees and 270 degrees

Answer 12

so that means, at those points, at those angles
the cos(2x) = 0
and therefore \(\bf tan(2x)=\cfrac{sin(2x)}{cos(2x)}\) will end up with a 0 in its denominator
and thus a vertical asymptote

vertical asymptotes occur in a fraction when the denominator is 0

Answer 13

okay how to I find those points on the graph, like not in degrees on the circle but as in the graph