Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2?
x = pi over 4, x = 3 pi over 4
x = 0, x = pi over 4
x = pi over 2, x = 3 pi over 2
x = 3 pi over 8, x = 5 pi over 8

Thus the asymptotes are π/8 and 3π/8  

Step-by-step explanation:

An asymptote is a line that a graph approaches without touching. The asymptotes are where the graph is undefined.

tan(x)= sin(x)/cos(x), where cos(4x-π) = 0

cos(4x-π) = 0 when inside is - π/2 , π/2 , 3π/2

4x-π = π/2

Add π at both sides:

4x-π+π = π/2 +π

4x= π+2π/2

4x= 3π/2

x= 3π/8

4x-π =  3π/2

4x = 3π/2 + π

4x = 5π/2  

x = 5π/8

This one is outside the interval.

4x-π = -π/2

4x = -π/2 +π

4x = -π+2π/2

4x=π/2

x = π/8

Thus the asymptotes are π/8 and 3π/8

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

-81 -56 -27 0 81

step-by-step explanation: