Dividing Rational Expressions

Question

What are the coordinates of the holes that would arise from the following multiplication problem? Select all that apply.

1. (3, -2)
2. (3, 2)
3. (2, 1/3)
4. (3, 1/2)
5. (-2, 3)

Answer 1

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.

You need to determine which values of x would cause at least one of the terms in the numerator AND at least one of the terms in the denominator to be equal to 0, which in turn would cause both the numerator and denominator to be 0.

In the numerator, x = -2, 3 or 1 would cause the numerator to be zero as one of the terms would be 0.

In the denominator x = 3, -1 or -2 would cause the denominator to be zero as one of the terms would be 0.

The holes would be the common numbers between those two which are -2 and 3, so (-2,3) is the answer.

Answer 2

From the original equation, x cannot equal 3, -1 or -2 because of the (x-3),(x+1) and (x+2) binomials in the denominator. Each of those values for x would result in zero in the denominator, an undefined answer. Those x values will either determine a hole or an asymptote.

When the expression is simplified, the (x-3) binomial cancels out because it is in the numerator and the denominator. Since (x-3) is removed from the equation, it is removed from the graph and is a "hole".

Similarly, the (x+2) binomial is also canceled out because it, too, is in the numerator and the denominator, So the value x = -2 is also removed from the graph and is another"hole".

The coefficient 2 is also in the numerator and denominator and cancels out, so the expression simplifies to (x-1) / (x + 1).

To find the y values of the hole coordinates, plug in the corresponding x values, or 3 and -2.

For x = 3, that becomes (3-1)/(3+1) = 2/4 = 1/2. That hole would be at (3,1/2).

For x = -2, the y value is (-2-1)/(-2+1) = -3/-1 = 3. That hole would be at (-2,3)

Note that x also cannot equal -1, but since (x+1) was not removed from the equation there is not a hole, but rather an asymptote at x = -1.

Hope this helps.