Question:
Which statement is true about the discontinuities of the function [tex] f(x) = \frac{x-5}{3x^2-17x-28} [/tex]
A)There are holes at x = 7 and .
B)There are asymptotes at x = 7 and .
C)There are asymptotes at x = –7 and .
D)There are holes at (–7, 0) and .
Answer:
B)There are asymptotes at x = 7 and [tex] (x = \frac{-4}{3}) [/tex]
Step-by-step explanation:
Given:
[tex] f(x) = \frac{x-5}{3x^2-17x-28}[/tex]
Required:
Find the true statement
[tex] f(x) = \frac{x-5}{3x^2-17x-28}[/tex]
We'll first factorize the denominator.
[tex] f(x) = \frac{x-5}{(3x+4)(x-7)}[/tex]
Make x subject of the formula in (3x+4) and (x-7):
3x + 4 =
3x = -4
Divide both sides by 3:
[tex] x = \frac{-4}{3}[/tex]
x - 7
x = 7
Now check for the limit when [tex] (x = \frac{-4}{3}) [/tex] and (x = 7)
lim f(x) when [tex] (x = \frac{-4}{3}) [/tex] = ±∞
lim f(x) when (x=7) = ±∞
Sinve they both make the denominator tend to zero, they are asymptotes
Therefore, there are asymptotes at [tex] (x = \frac{-4}{3}) [/tex] and x=7
Option B is correct
