To graph this tangent, we've realized there's a stretch of two by the leading coefficient. The phase shift is π tenths from setting the parenthesis equal to 0, The period for tangent is π, and the coefficient in front of the X is 1. So Π / 1 is π for the period vertical translation. The number that's added or subtracted at the end, that's not being multiplied by the tangent. So down one in this case, we're going to then do the asymptotes by sticking negative π halves less than the parenthesis portion, less than π halves. So I'm going to add π tenths to each side. We get -4 Pi tenths less than X less than six Pi tenths, or -2π fifths less than X less than three Pi fifths. So those are my two asymptotes. I'm going to draw on the asymptotes. Then I'm going to take that whole period and divide it by 4, starting with one of the asymptotes. I'm then going to add π force and get -3 Pi twentieths. Then I'm going to take that -3 Pi twentieths and add π force and get 2π twentieths, which is Pi 10s. I'm going to take the Pi 10s and add π force and get 7 Pi twentieths. I'm going to take the seven Pi twentieths, add π force and get 12π twentieths, which is 3 pieths. So those are going to be my five key locations X equal -2π fifths -3 Pi 20th. Because of the vertical translation down one, we're going to end up at -3 the Pi tenths would have been zero, but because of the vertical translation, it's going to be π tenths -1. The seven Pi twentieths would have been two, but be the because of the vertical translation down one it's going to be one and X equal 3 Pi fifths is going to be our other vertical asymptote.