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.
Answer: 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
Answer: Ф=(π-4)/2 and Ф=(3π-4)/2
3)
Answer: 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
