With precalculus questions. no unrelated

1)The vertical asymptotes of the function f(Ф)=4 tan Ф, in the interval 0≤Ф≤2π are:  Ф=π/2 and Ф=3π/2.

Ф=π/2 and Ф=3π/2.

2)
r(Ф)=tan (Ф+2)      if  0≤Ф≤2π;
Ф+2=π/2 ⇒    Ф=(π/2)-2=(π-4)/2
Ф+2=3π/2 ⇒  Ф=3π/2-2=(3π-4)/2

Ф=(π-4)/2  and Ф=(3π-4)/2

3)
f(Ф)=tan Ф  or  f(Ф)=2tan Ф  or  f(Ф)=3 tan Ф

4)
slope-intercept form:
y=mx+b
b=y-intercept
m=slope

In this case:
m=tan Ф=tan 5π/6=-(√3)/3
b=7

y=-(√3)/3 x+7

Answer₁:  y=-(√3)/3 x+7

x-intercept;
when y=0; then:
0=-(√3)/3 x+7
(√3)/3 x=7
x=7*3 /√3
x=21/√3
x=21√3/3
x=7√3

Answer₂: x=7√3

You didn’t put the necessary image to solve the question : p