Question

Describe the vertical asymptote(s) and hole(s) for the graph of postin pic

Answer 1

This is where the bottom half can't equal 0 right?

Answer 2

ello

Answer 3

ello again?

Answer 4

is that equation the postin pic here ?

Answer 5

I think he spontaneously fell asleep.

Answer 6

I've got this too.

Answer 7

There would be a hole at x = 1 because the (x-1)s cancel but x-1 would cause the function to be undefined.

There would be a vertical asymptote at any value that would cause the denominator to be zero.

Answer 8

So the vertical asymptotes will come from the bottom equaling zero. However, like Hunus said, the (x-1) cancels leaving a hole at x=1.

Answer 9

Awww...I wanted to do this one...

Answer 10

The horizontal asymptote comes from the ratio of the top to the bottom at infinity. The highest-level term in X will dominate the equation (because infinity squared >>>> any finite number of infinity). Multiply out the terms to find how many x squared you have and then compare only the x squared to find the value of the horizontal asymptote. If there were a higher power of x in the numerator, this equation would blow up and not have a horizontal asymptote. If the denominator had a higher power of x, the asymptote would be at zero. Only when equal powers of x are above and below the dividing line does the ratio of the highest powers matter.

Answer 11

sorry yall had the answer already :)

Answer 12

lol