Question

Which of the following functions has a hole at (1,4)? ANSWER CHOICES IN THE IMAGE BELOW! PLEASE PROVIDE WORK WITH YOUR ANSWER:)!!! THANK YOU!

Answer 1

Hello!

The answer is:

d) $\frac{(x-1)(11x+1)}{(x-1)(x+2)}$

Why?

A hole is a point where rational functions lose its continuity, meaning that in that point, there is a discontinuity condition.

We can find the hole of a rational function if there are similar terms on the numerator and the denominator by finding:

First (x-component): The values of x that makes the function equal to 0 in both numerator and denominator.

Second (y-component): Re-evaluating the same term in the other factors of the function to know the y-component.

Finding the x component we have:

$f(1)=\frac{(1-1)(11*1+1)}{(1-1)(1+2)}=\frac{(0)(12)}{(0)(3)}=\frac{0}{0}$

So, the x-component is 1,

Then, re-evaluating the function:

$f(1)=\frac{(x-1)(11*1+1)}{(x-1)(1+2)}=\frac{(12)}{(3)}=\frac{12}{3}=4$

Therefore, the y-component is 4,

Hence,

The function has a hole at (1,4)

Have a nice day!